Solutions to I E Irodov - Physical Fundamentals of Mechanics

Irodov Problem 1.388

2010-01-14T08:01:00.015+05:30

Suppose that a certain instant of time t the velocity of the rocket is v and its rest mass m. Suppose that the rocket ejects gas of rest mass k units/sec. Consider another instant of time t + dt, where dt is an infinitesimally small interval of time. The rocket has ejected kdt amount of rest mass. Suppose that the rocket's velocity has increased by an amount dv during this interval. Since there is no external force on the rocket, the momentum of the (ejected gas + rocket) system at t must be equal to that at t+dt.

The momentum of the rocket at time t is given by . The momentum of the rocket at t+dt is given as . The gas ejected by the rocket has a of mass and moves at a non-relativistic velocity u relative to the rocket in the opposite direction to the rocket. In other words, its velocity as seen by an observer on Earth will be v-u in the direction of the rocket (this is shown in the diagram). Hence, the momentum of the ejected gas will be .

Applying conservation of momentum at t and t+dt we have,

We can now use Taylor series expansion and retain only the first order terms in Eqn (1) to obtain,

Now we know that the mass of the rocket reduces at a rate k, in other words -kdt = dm. Hence, (2) can be rewritten as,

Irodov Problem 1.387

2010-01-13T13:48:00.009+05:30

Since the mass spontaneously breaks up into three pieces without any external forces both linear mommentum and energy must be conserved. In general, the three
pieces could move along three different directions at velocities v1,v2 and v3
as shown in the figure 1. In that case we have that,

From Eqn (1) since the sum of a linear combination of the three velocity vectors is zero (they are linearly dependent) they must be co-planar. This means that without a loss of generality we can restrict the velocity vectors to lie on a plane and choose the direction of velocity v1 to lie along the negative x-axis. This is depicted in figure 2.

Now we can simplify (1) as,

What choice of velocities and directions will maximize the energy if particle 1? To determine this answer we can use Lagrangian Multipliers to minimize the objective function,

The conditions for local extrema are then obtain by setting the partial derivatives of J with respect to each of the unknowns to zero. Hence at extrema we have,

Eqn 5 essentially says that in order for the energy of piece 1 to be maximum, both the the pieces 2 and 3 must fly off in the same direction and propel piece 3 in the opposite direction.

Now differentiating with respect to v2 and v3 we have,

Eqn 5 and 7 together essentially say that the energy of piece 1 will be maximized when pieces 2 and 3 move at the same velocity and the same direction - in other words they move together. This situation is depcited in Figure 3. Let v2=v3=v.

Now equations (1) and (2) can be rewritten as,

From (8) we obtain,

From (9) we obtain,

Eqn (12) given the maximum energy that piece 1 can possible have after the disintegration.

Irodov Problem 1.386

2010-01-12T07:30:00.001+05:30

Let the kinetic energy of the single particle be T'. In Problem 1.384 we have already determined the combined kinetic energy of the particles as,

Hence, we need to find T' such that,

Irodov Problem 1.385

2010-01-11T17:11:00.009+05:30

This is a very interesting basic problem that demonstrates how different relativistic collisions are compared to Newtonian collisions - the essential difference being that mass can change in relativity.

After the two masses collide and fuse into a new mass, its rest mass is not simply the sum of rest masses of the two constituent particles but something entirely different. In fact it depends on the energy of the collision!

Throughout the collision two important properties will always hold, i) the momentum will be conserved and ii) the total energy will be conserved. Let the final rest mass of the fused particle be . Let the initial velocity of the left particle be u and let v be the velocity of the combined particle after collision.

Since we know that the kinetic energy of the particle before collision is T we have,

Conservation of Momentum
Since there are no external forces on the system, its total momentum must be conserved and so we have,

Conservation of Energy
The total energy of the system before collision must be equal to that after collision. The total energy is the sum of the rest mass energies of each of the particles and the kinetic energy of the colliding particle.

From Eqn (3) and (4) we obtain,

Now we can substitute (5) in (4) to obtain,

Irodov Problem 1.384

2010-01-11T07:11:00.012+05:30

As seen from the stationary frame of reference (in which N2 is stationary), neutron N1 is on a collision course towards neutron N2 and has a kinetic energy of .The center of inertia of the (N1+N2) system (CG) lies in between N1 and N2 and moves with velocity vCG. As seen from the reference frame of the CG, both neutrons N1 and N2 move towards the CG at equal velocities. This is shown in the Figure above.

The total energy of N1 in the stationary reference frame is given by,

We can now compute the velocity v of neutron N1 as,

Hence, the momentum of neutron N1 and hence the entire system (N1 + N2) and hence, the momentum of the CG as seen from the staionary reference frame is given by,

The total energy of the entire system (N1 + N2) and hence that of the CG, as seen from the stationary reference frame is given by the sum of kinetic energy of the neutron N1 and the sum of the rest mass energies of each of the neutrons N1 and N2. Thus, the total energy in the system in the stationary reference frame is given by,

The invariant quantity can then be computed as,

This quantity is invariant with respect to the choice of any inertial frame and hence must be the same even in the frame of the center of inertia (CG). The momentum of the CG is 0 in the reference frame of the CG. Hence the energy of the CG, ECG can be computed as,

The kinetic energy of the system as viewed from a reference frame CG can be obtained by subtracting the rest mass energies of the two protons from total energy as,

As seen from the reference frame of the CG, both neutrons will have the same energy and hence then energy of each neutron in the CG reference frame will be,

We also know from problem 1.383 that for each neutron,

Here, is the momentum of the neutron N1 in the reference frame of the CG.

b) The total rest mass of the CG is 2mo. Hence, at velocity VCG, its energy is given by . We can thus, compute the velocity of the CG as,

Irodov Problem 1.382

2010-01-08T20:55:00.004+05:30

The momentum of the photon is given by,

From the solution found in Problem 1.381, we can compute the energy of the photon as seen in reference frame K' as,

Irodov Problem 1.383

2010-01-08T20:36:00.004+05:30

We can use the results from Problem 1.381 to prove this invariance result.

What Eqn (1) is indicates is an invariance since the quantity does not depend which inertial frame you are observing from!

The value of the constant can be determined simply by considering a frame that is moving with the particle itself. In such a frame, the particle is absolutely stationary, in other words P=0. The energy is equal to the rest mass energy and so we have,

Irodov Problem 1.381

2010-01-08T19:29:00.007+05:30

From the relativistic transformations of velocities, energy and momentum we have,

Irodov Problem 1.380

2010-01-07T19:31:00.022+05:30

a) For this problem let us denote v as the magnitude of the instantaneous velocity of the particle and as the unit vector along the instantaneous direction of motion of the particle. In other words the instantaneous velocity vector of the particle is given by .

From the fundamental laws of dynamics we can write,

Similarly, we have,

In order for (1) and (2) be parallel there are only two possible ways

1. , in which case both vectors get trivially aligned. However, this also means that the particle never changes it direction. In other words, the force has to be aligned in the instantaneous direction of motion of the particle i.e. F is parallel to velocity vector.

2. , in which case the magnitude of the velocity of the particle never changes. This can only happen if the force acting on the particle always acts perpendicular to the instantaneous direction of its motion i.e F is perpendicular to the velocity vector.

b) The solution to this part is quite evident from Eqns (1), and (2) considering each of the two cases separately.

Case 1: F is parallel to velocity vector

Case 2 : F is perpendicular to velocity vector

Irodov Problem 1.379

2010-01-07T12:44:00.003+05:30

This problem can be solved from a very simple inspection of solution to problem 1.378 and seeing that,

The constant offset is not important since the origin could always be displaced by a constant value without effecting the dynamics.

Irodov Problem 1.378

2010-01-07T12:32:00.002+05:30

We can solve this using the fundamental equation of dynamics as follows:

Irodov Problem 1.377

2010-01-07T11:15:00.012+05:30

In the frame of reference of the sphere

As seen from the frame of reference of the sphere, the sphere itself is stationary. The gas particles are however moving at a speed v towards it. Let us now consider the collision of the sphere with a single particle. As asked in the question consider a section of the sphere that is normal to the direction of velocity. Assuming that the sphere is extremely large compared to the gas particles, in case of an elastic collision, the gas particle will simply bounce off with the same velocity v in the opposite direction. The net change in momentum of the gas particle is given by,

As seen from the reference frame of the particles there are n particles per unit volume. However, the same unit volume with n particles as seen by an observer moving with the sphere is shrunk by a factor of since space itself would have shrunk by this factor in the direction of motion. In other words, the density of the gas a seen from the observer on the sphere would be

Since n' particles change their momentum each second, the net force exerted on the sphere as a reaction will be

Solution in reference frame of the particles
As seen from the view of the gas particles, they were initially stationary and the sphere was moving at a velocity v. Since as seen from the sphere's reference frame, the final velocity (velocity after collision) was 2v, their final velocity as seen by a stationary observer can be obtained by using the relativistic transformation as,

This means that the change in momentum from the point of view of a stationary observer is given by,

Since the number of particles colliding with the sphere per unit time is n, the net force exerted on the sphere is given by , which evaluates exactly the same as calculated from the point of view of an observer on the sphere!

Irodov Problem 1.376

2010-01-06T15:05:00.006+05:30

The number of particles impinging upon the target each second is given by

The momentum of each particle is given by the solution in Problem 1.375 as,

The total change in momentum of all the n particles each second is the force imparted in the target and is given by,

The total energy of the n particles getting absorbed into the target is

Irodov Problem 1.375

2010-01-06T14:39:00.005+05:30

From the solution to Problem 1.374 we know that given that the particle has a kinetic energy T and a rest mass of m0,

Also we know that the momentum of the particle is given by,

From (1), (2) and (3) we can write,

Irodov Problem 1.374

2010-01-06T13:37:00.008+05:30

Let T be the kinetic energy of the particle. Let its rest mass be mo. Let the ratio of its kinetic energy to rest mass energy be equal to,

If we used Newtonian mechanics to determine the velocity we would have,

If we used Relativistic mechanics on the other hand we would have,

The relative error in the estimation of velocity would then be,

For small values of we can consider only the linear terms in equation (4) and we have,

Irodov Problem 1.373

2010-01-06T12:16:00.003+05:30

When the particle moves at velocity v, its kinetic energy is given by

Thus, if its kinetic energy is equal to its rest mass energy, then we have,

Irodov Problem 1.372

2010-01-06T12:13:00.001+05:30

All the work done on the particle translates into kinetic energy of the particle.
The Newtonian kinetic energy is given by,

The relativistic kinetic energy is given by the change in total energy in the particle given by,

Irodov Problem 1.371

2010-01-06T12:12:00.002+05:30

The ratio of relativistic momentum to Newtonian momentum is given by,

Irodov Problem 1.370

2010-01-06T12:04:00.004+05:30

As the particle moves its mass increases. When the velocity of the particle is v, its mass is given by, .

Consequently its momentum is given by

Now we can solve for v from equation from (1) and obtain,

Irodov Problem 1.369

2009-12-31T18:44:00.010+05:30

Without loss of generality let us assume that the observer is moving at a speed with velocity v along the positive x-axis. As seen by the observer, the body is moving along the negative x-axis with a speed v. Let be its density be .

Consider an infinitesimally small cuboidal piece of the body of dimensions dx, dy and dz. The mass dm of this infinitesimally small piece is given by,

As seen by the observer, the mass of the inifinitesimally small cuboid would be

Further, also as seen by the observer, the cuboid would appear to have shrunk along the x dimension to .

Let the density of body as seen by the moving observer be then we have,

Irodov Problem 1.368

2009-12-31T18:18:00.003+05:30

The mass of the moving particle as seen by the observer is given by,

At v/c = 1 - 0.0001 = 0.9999 we have m = 70.712m0 .

Irodov Problem 1.367

2009-12-29T06:35:00.003+05:30

This problem is an extension of 1.366 and essentially asks, how long did the rocket travel according to an observer in the rocket. As the rocket accelerates and moves faster, time gets progressively slower. Let t' be the time in the reference frame of the rocket. We know that,

Irodov Problem 1.366

2009-12-29T05:29:00.005+05:30

This problem can be considered as an extension of the problem 1.365(a). We have to simply replace V=v, to obtain the relationship between w' and the acceleration of the rocket w as seen from Earth. This is because at each instant, the moving reference frame K' has the velocity v same as that of the rocket as seen from Earth. Thus, we have,

The distance traveled by the body can be obtained using (2) and the velocity of the body is obtained from (1). The fraction by which the velocity differs from light speed can be obtained by computing 1-v/c.

Irodov Problem 1.365

2009-12-29T04:30:00.006+05:30

a) Let the direction of motion of the frame K' be assigned as the x-axis. Let v', x' and t' be the instantaneous velocity, position and time of the moving body as seen by the observer in frame K'. Since, the direction of motion of the accelerating body coincides with the motion of the frame K' , x -axis and x' axis coincide. From the relativistic transformations for velocity and time we have,

The acceleration of the body as seen from frame K' is then given by,

(b) In this problem the body moves along the y-axis while the frame K' moves along the x-axis. Thus, using relativistic transformations we have,

The acceleration of the body as seen in frame K' is then given by,

Irodov Problem 1.364

2009-12-27T06:00:00.011+05:30

As seen by observer O' :
Without loss of generalization let us assume that observer O' is at the origin (x'=0,y'=0,t'=0). x',y' and t' being the coordinates in frame K' as seen by O'. As seen by O', the rod moves only along the y' axis at a speed v'. In other words y'=v't'. The rod however, has no motion along the x'-axis. Further assume that the rod is places such that the x' coordinate of its mid point is at x'=0 (this only makes solving the problem easier but does not fundamentally change the result).

At time t'=0, O' sees the two ends of the rod and finds them perfectly parallel to the x'-axis. The ends of the rod A and B as seen by O' are actually due to two different rays of light that started some times in the past t'A and t'B and both reached O' simultaneously at t'=0. Since, the rod is not moving along the x'-axis, and O ' saw the rod to be parallel to the x'-axis, both these times must have been exactly the same i.e. t'A = t'B.

The space-time diagram of the entire situation is depicted in the figure beside. The world-line of the rod is a plane tilted at an angle to the x'y' plane. The world-lines of the ends of the rod only change their y',t' coordinates but not their x' coordinates. The light rays are emitted along the surface of a cone but we depict only those rays that reach the observer O'. These rays of light move at 45 degrees to the x'y' plane but lie entirely in the world-line of the rod.

As seen by observer O :
Now lets us see the same phenomenon through the eyes of an observer O located in Frame K at the origin i.e. (x=0,y=0,t=0), where x,y,t are the space time coordinates of frame K. Let us assume that the origins of frames K and K' coincide. As seen by O, the rod moves along the x-axis with a speed V. It also moves long the y-axis with a speed (vy is less than v' because time t' in frame is running slower as seen by O, in other words the same distance along the y-axis is covered in a shorter time when seen in frame O. The same can also be derived the same way as it was derived in Problem 1.358) . In other words, the coordinates of the center of the rod are given by,

The world-line of the rod as seen by O in frame K is depicted in the figure. The light rays traveling from the ends of the rod reach the origin at the same time but unlike O', O arrives at different conclusion, namely the light ray that left end A started at a time tA, that is before the time tB, when the ray from end B started its journey. This also means that the end A was a little behind along the y-axis than the end B, in other words the rod was not parallel to x-axis.

Now let us compute the angle the rod is at to the x-axis as seen by O. First we will start by computing the difference tA - tB.

Let the length of the rod as seen in frame K be l. Thus, both the rays of light have to cover a distance of l/2 along the rod.

As we trace back the light ray starting from end A in time, the ray moving at speed c has to "catch up" with the moving rod at it moves at a rate V to the left. This means that,

The negative sign comes because, the time tA is in the past to t=0.

The situation is different when we trace back to find out when the ray from rod B started its journey. In this case, the rod is aiding the ray in catching up by moving towards it at a rate V. Hence, we have,

Thus, we have,

During this time interval tB - tA the end B moved a distance of vy(tB - tA) farther along the y-axis than end A.

As shown in the figure thus, we can compute the angle to the x-axis as,

Irodov Problem 1.363

2009-11-22T09:57:00.008+05:30

To solve this problem we can use the derived result from Problem 1.358 namely,

Here,

and so we have,

Irodov Problem 1.362

2009-11-22T07:51:00.011+05:30

In this problem we need to first determine what is the speed of the particle with respect to frame K.

For this problem we can use the solution we derived from Problem 1.358 namely,

Here, we know that v'x = 0 (2a), v'y = v' (2b) and we need to find vx and vy.

The magnitude of velocity of the particle as seen from frame K is given by,

The particle will decay in time in its own frame. In frame K however, this will seem like a much longer time interval given by,

The distance traveled by the particle during this time as seen by frame K is given by,

Irodov Problem 1.361

2009-11-21T18:24:00.003+05:30

In this problem we can use the solution we found in Problem 1.358, namely

Let us find the velocity of the particle moving along the y-axis relative to the particle moving on the x-axis. For this, in our above equations we have to substitute, V= v1 , vx=0, vy=v2. Thus, we have, the relative velocity is given by,

Irodov Problem 1.360

2009-11-21T16:50:00.004+05:30

For this problem we can use the solution found in Problem 1.359 namely,

We can determine the velocity of one of the rods relative to the other by setting v1=v2=v to obtain,

To one rod thus, the other rod would appear to be shrunk and of length,

Irodov Problem 1.359

2009-11-21T16:09:00.003+05:30

a) The position of the particles as seen in the laboratory frame of reference are given by x1 = v1t and x2 = -v2t. The relative velocity is thus simple the time derivative of x1-x2 given by v1+v2.

b) For this part of the problem we could use the result obtained from Problem 1.358 namely,

To obtain the velocity of particle 2 as seen from particle 1 we have to substitute V=v1 and vx = -v2 in the above equation. Hence, we obtain,

Irodov Problem 1.358

2009-11-21T14:48:00.013+05:30

The position of the particle in frame K as a function of time is given by,

Using Lorentz transform we can determine the position of the particle as a function of time t as seen in frame K' as,

The velocity of the particle as seen in frame K' is then given by,

Irodov Problem 1.357

2009-11-21T13:33:00.011+05:30

In Newton's view of space no matter which inertial frame we look from distance between two points in space remains the same. In other words if we have two frames K and K' with constant relative velocity w.r.t each other then .

In Einstein's world however, space shrinks as seen from the eyes of a moving observer, thus, clearly the distance between two points is not an invariant between two inertial frames. Further, time also dilates or expands for the moving observer.

It turns out that the corresponding invariant in Einstein's space-time is given by

This invariant property of Einstein's space-time can be applied to solved the problem. Let us call this invariant as the space-time distance between two events.

a) The space-time distance between events A and B as seen from the figure is given by . The space-time distance between events A and B as seen from any inertial will be the same. Thus, in a frame where the two events occur at the same point in space, we have,

b) Same as above the space-time distance between events A and C is given by
.

In a frame where the events occur at the same time, the distance between them will be given by

Irodov Problem 1.356

2009-11-21T12:37:00.008+05:30

This is best solved and explained through an example. Say a person located at x=0, fires a gun at time t=0 with respect to frame K. Say the bullet takes time in frame, travels a distance d along the positive x-axis and then hits a board. In other words, the space-time coordinates of the event corresponding to the man firing the gun are (0,0) and that of the event corresponding to the bullet hitting the board are (,d). Now consider a frame K' which is moving at a velocity v along the x-axis. As seen from frame K', using the Lorentz transform, the space-time coordinates of the event of bullet hitting the board are given by . In order for the causality to be disturbed, it should appear to the observer in K' that the bullet hit the board before it was shot. In other words,

Now suppose that the bullet traveled at a velocity vb. Then we have . From equation (1) we have the condition for causality to be contradicted as,

What equation (2) tells us is that in order for causality to be contradicted, at least one among the bullet or the frame K' must be traveling faster that the speed of light! Since nothing can travel faster than the speed of light, causality could never be contradicted.

Irodov Problem 1.355

2009-11-20T07:11:00.011+05:30

First let us start by understanding the Lorentz transform. Lorentz transformation answers the following question. There are two observers O and O' with O moving at a speed v along the positive x=axis. If according to O a certain event occurs at (x,t), according to O' where (x') and when (t') does it occur?

In classical mechanics, the transformation would be x'=x-vt , t'=t. Since, the origin of O' has moved vt units forward along the x-axis with respect to that of O, all points would appear to have moved backwards along the x-axis by vt units, with respect to O'. Time elapsed of course is the same as perceived by O and O'.

In relativistic mechanics, time as perceived by the observers O and O' is not the same. The time would tick slower in O' compared to that of O as seen by O. In other words t' is less than t. The Lorentz transform tells us that,

Let us understand (1). Suppose that the origins of the two frames (x,t) and (x',t') were initially aligned. After t seconds O' would have moved to x=vt as seen by O. Substituting this in (1) we have,

This shows, that time t' as perceived by O' would be slower. This means that .

Differentiating (1) with respect to t we have,

In order for the clocks to be in perpetual synchrony we must have that . From (2) we have,

This means that if the observation point were displaced at this rate the clocks would always seem to be in Synchrony.

Irodov Problem 1.354

2009-11-18T07:41:00.005+05:30

Since all clocks (A-G) in frame K have been synchronized to each other, they will show the same time as seen by any observer in Frame K. While all clocks (A'-G') in K' have also been synchronized to each other, as seen from Frame K, these clocks run slower as K' is moving with respect to K. Clocks D and D' are perfectly synchronized as seen from K. Thus, all clocks in Frame K' to the right of D' i.e. (E'-G') will progressively fall behind. Similarly, all clocks to the left of D' i.e. (A'-C') must be progressively ahead so as to catch up with D by the time they reach the center despite their slower rate.

Irodov Problem 1.353

2009-11-17T10:44:00.006+05:30

As seen from AB's reference frame: A'B' is moving from left to right with a speed v and so has a length of . AB is at rest has a length of l0.

The distance traveled by A'B' when A and A' align is and so the time taken is

The distance traveled by A'B' when B and B' align is l0 and so the time taken is l0/v.

As seen from A'B''s reference frame: AB is moving from right to left at a speed v and so has a length of . A'B' is at rest and has a length of l0.

The distance traveled by AB when B and B' align is and hence the time required is .

The distance traveled by AB when A and A' align is lo and so the time required is lo/v.

Isn't it interesting that according to AB, A and A' align first and then B and B' while according to A'B', B and B' align first and then A and A'. This demonstrates that in relativistic mechanics sequence of events as seen by different observers may in fact be different!

Irodov Problem 1.352

2009-11-17T08:08:00.007+05:30

a) Let the proper length of the rod be lo. Since the rod is moving at a speed v with respect to frame K, the length of the rod as seen from Frame K would be lesser than its proper length and equal to . At time instant tA, the front end of the rod was at xA. This implies that the rear end must have been at . At time tB, the rear end of the rod was at xB, this implies that the rear end of the rod moved a distance of in a time interval of tB - tA. Since the rod was moving at a speed v, we have,

b) In order to determine the time interval required to measure the proper length of the rod in equation (1) we can set xB - xA = lo, and we have,

Irodov Problem 1.351

2009-11-16T08:46:00.012+05:30

This is a problem that illustrates the classic breakdown of simultaneity in relativistic mechanics. Before we get into solving the problem let us first try to understand what is happening.

There are three important things to understand,
i) light speed as seen in any reference frame is the same and equal to c.
ii) an observer feels that an event occurred when the event reaches him and events travel at the speed of light.
iii) If the location and time of two events are equal in one frame then the location and time of the two events in some other frame will also be the equal. In other words if two cars collide in one frame (they occupy equal space and time) then the cars will collide in all other frames (they will occupy equal space and time - note that values of these may differ in different frames).

For simplicity's sake consider an observer O in the mid-point between the two particles, moving at the same velocity as the particles (note that the placement of the observer is not crucial to solving the problem, it only simplifies it - I shall explain this in more detail later in the solution). Consider another observer O' who is located in the laboratory reference frame (frame K), and as seen by O happened to be at the same position as him when the particles decayed (again this location of O' has been chosen for simplicity, and this does not alter the final answer).

Events as seen from the particle frame of reference: O sees that both the particles decayed at the same time. This means that the event of particle decay from both particles reached the observer at the same time. Imagine that each decay was accompanied by a brilliant flash - then both the flashes reached the observer at exactly the same time. Since O is at the mid-point between the particles and the particles are stationary as seen by O, O deduces that the particles decayed at exactly the same time.

Meanwhile as seen by O, an observer O' who is in the laboratory reference (frame K in the problem), is moving backwards (to the left) with a speed of v. Thus, the decay event (flash of light) from the left particle will have to travel a shorter distance to catch up while the decay event from the right particle will have to travel a longer distance. Thus, O' will sense the left particle decay event earlier compared to that from the right. So according to O, O' will see the decay of left particle before the decay of the right particle.

Events as seen by O' from the laboratory frame of reference (frame K): There can never be dis-agreement between actual observed phenomena among frames, their interpretations/deductions may however be different. Indeed, observer O' sees the decay of the left particle earlier than the right particle as deduced by O. However, since O' was at the mid point between the particles when they decayed, there is only one way to interpret this - the left particle decayed must have decayed earlier than the right particle.

Since O was moving to the right, the event (flash of light) from the left particle had to travel a longer distance to catch up with O, as compared to the event from the right particle. However, since the left particle decayed earlier than the right particle - this gave the right particle decay event enough head-start to reach O at the same time as the decay event from left particle. Thus, O' sensed the events at the same time!

In other words, both O and O' agree on
i) decay events from both particles reached O at the same time
ii) decay event from left particle reached O' before that from the right particle.

Their interpretations are different
i)O thinks that the left and right particles decayed at the same time
ii)O' thinks that the left particle decayed before the right particle

The Actual Solution to the Problem (from the point of view of O')
Let us solve this in the laboratory reference frame. In the laboratory reference frame, the distance between the particles is l.

Suppose that the particle from the left took time t1 to reach O. Then we can write,

Suppose that the particle from the right took time t2 to reach O. Then we have,

The time difference then is given by,

Irodov Problem 1.350

2009-11-14T22:48:00.007+05:30

Let the rods be AB and A'B' and have the same rest length of l0. Let the relative velocity of A'B' as seen from AB be v. As seen from AB, thus, the rod A'B' appears to have contracted in length and has a length equal to
. Hence, to someone on AB, it will appear as though the rod A'B' has to travel a distance of between the events when its right and left ends meet. Since, the corresponding time interval is , we have,

Irodov Problem 1.349

2009-10-11T13:09:00.009+05:30

Let the original length of the rod (in a frame fixed to the rod) be l0. Also, let the length of the scale's unit be 1 as measured in the frame fixed to the rod.
Suppose that the rod is moving with velocity v relative to the scale.

In the reference frame fixed to the scale:

As seen from the scale, since the rod is moving it would appear smaller. Its length would be . Thus, as measured by the scale, the length of rod will be

In the reference frame fixed to the rod:

As seen from the frame fixed to the rod, the rod is stationary and has a length of l0 , however, the scale is moving with a velocity v. Thus, each unit of the scale will shrink and will be . Thus, when measured using the scale the measured length will be,

From (1) and (2) we have,

Irodov Problem 1.348

2009-10-11T09:03:00.011+05:30

Since the two particles a traveling at the same speed v, relative to each other they are stationary. This means that as measured from a reference frame fixed to any of the particles, the distance between them l0 will be the original distance. The distance measured by someone in the Laboratory will however be a shrunk version of the original distance between the particles given by . Similarly, time will appear to be slower in the reference frame fixed to the particles compared to the time measured in the Laboratory. So while the time interval between collisions measured in the Laboratory reference frame is , the same however, when measured in the reference frame fixed to the particles will be a lesser (since time is slower) given by .

Solution to the Problem in Reference Frame of the Particles:

In the reference frame of the particles we have,

Solution to the Problem in the Laboratory Reference Frame :

Irodov Problem 1.347

2009-10-11T06:49:00.006+05:30

This problem is very similar to Problem 1.346. As seen from the Laboratory reference frame, the Muon's time would be dilated (lowed down) and appear longer. Let the proper decay time of the Muon be . In its own reference frame the Muon will decay in time , however as seen in the Laboratory it will decay in a longer time given by. The distance l traveled by the Muon in the Laboratory reference frame is given by,

b) The distances in the Muon's reference frame will seem to be shrunk as compared to Laboratory and so it will feel it has traveled a much shorter distance given by .

Irodov Problem 1.346

2009-10-11T06:29:00.007+05:30

The Particle moves very fast compared to someone in the Laboratory. Consequently, time will appear to be slower for the fast moving particle as seen from Laboratory reference frame. While in its own reference frame, the particle will decay exactly within its correct lifetime , to someone in the Laboratory reference frame this will appear to be a longer interval . The relation is given by,

This means that as seen in Laboratory, the particle would have traveled a distance of,

Irodov Problem 1.345

2009-10-10T12:56:00.006+05:30

Let K be the stationary frame and K' the frame that moves along with the rod.

As seen from K:
The rod will be shorter and have a length and moves at a velocity v to the right. Thus, the rod will pass the mark in a time interval,

As seen from K' :

The rod is stationary and has a length l but the mark moves with a velocity v to the left. Thus, the mark will pass the rod by in a time interval,

From (2) we have,

Irodov Problem 1.344

2009-10-10T12:32:00.002+05:30

The time measured in the moving reference frame t' when the time measured in a stationary reference frame is t will be,

Irodov Problem 1.343

2009-09-24T07:06:00.006+05:30

Suppose that the height, radius and the slant length of the code in a reference frame stationary with respect are code be h, R and L respectively. When the cone moves along the axis, the vertical height to h' shrinks to

but the radius R will remain the same.

a) Hence we have,

b) The lateral surface area of the cone is given by

Irodov Problem 1.342

2009-09-20T11:11:00.007+05:30

Let lo be the original length of the rod. The rod moves at a velocity v and at an angle with respect to the stationary frame and has a length l. If the rod slowed down to a velocity zero, it would have a length lo but it would not be aligned at an angle . The rod would have shrunk in the horizontal direction (direction of motion) but not in the vertical direction (perpendicular to the direction of motion). So the vertical component of the length of the rod remains unchanged whether the rod is moving or not at . The horizontal component of the rod however will ne longer when the rod is at rest and equal to . The rest length will thus be,

Irodov Problem 1.341

2009-09-13T11:16:00.007+05:30

a) Let ABC be the original triangle. Since all its sides are equal to a it is an equilateral triangle. Let CD be the perpendicular bisector along which the triangle moves with a velocity v w.r.t to frame K. From elementary trigonometry.

Let A'B'C' be the triangle as seen from a frame K. w.r.t to K, the triangle will shrink by a factor along the direction of motion (along CD). However, along the direction perpendicular to the motion of the triangle (along AB) all lengths will remain intact.

In other words,
AB = A'B' = a, AD = A'D' = a/2, C'D' would Thus, w.r.t to frame K, the triangle will no longer be equilateral since the sides A'C and C'B' would have shrunk. Hence, we have,

b) In this part the triangle is moving in the direction of one of the sides (along AB) and not the bisector. Thus, all measurements of the triangle in this direction shrink by a factor relative to frame K. The measurements in the perpendicular direction (along CD) however remain intact.

Thus, we have,

Irodov Problem 1.340

2009-09-13T10:01:00.000+05:30

Let the original length of the rod be l. When it moves lengthwise at a speed v relative to a frame K. The length of the rod relative to frame K will be , in other words it will be shorter by a fraction of its original length. Thus, we have,

Irodov Problem 1.339

2009-09-12T15:02:00.000+05:30

Let the density of steel be and that of olive oil be and let the viscosity of olive oil be. Suppose that at some instant the velocity of the sphere is v.

As the sphere sinks, there are three forces acting on it, i) the force of gravity responsible for sinking it given by ii) the buoyant force offered by the fluid given by and iii) the viscous force offered by the fluid .

The mass of the sphere is given by and so from Newton's first law we have,

From (2) it is clear that the steady state velocity will be . The velocity v will be off by from the steady state velocity by fraction n when,

From (1) we have,

Irodov Problem 1.338

2009-09-12T13:45:00.000+05:30

Let the density of the fluid be and the density of the sphere be and let the viscosity of the fluid be . Let the terminal velocity of the sphere in the fluid be v.

As the sphere falls through the fluid there are three forces acting on it, i) the force of gravity , ii) the buoyant force due to the fluid and iii) the viscous force due to fluid on the sphere is given by . Since the body is moving at terminal velocity and there is no acceleration, all these forces must cancel out. Hence, we have,

The Reynold's number for a sphere moving in fluid is given by,

Irodov Problem 1.337

2009-09-12T13:01:00.000+05:30

The Reynold's number for a sphere of radius r1 falling through a fluid of density at a velocity v1, is given by . Since at velocity v1 the flow starts to become turbulent, this is the Reynold's number corresponding to turbulent flow. For the other fluid the flow must become turbulent at the same Reynold's number. So we have,

Irodov Problem 1.336

2009-08-01T05:34:00.000+05:30

The Reynold's number for flow in a tube within a circular tube is given by . Here Q is the volumetric flow rate i.e. volume of fluid flowing per sec. Hence, for this problem with pipe of reducing cross section it is given by . Since Q does not change along the tube for two points separated by a distance the ratios of the Reynold's number will be . The Reynold.s number will increase as the tube narrows.

Irodov Problem 1.335

2009-07-19T06:36:00.001+05:30

First let us understand what is happening in the problem. As the liquid moves along the tube it has to work against the viscous forces and so it looses energy. From Bernoulli's equation we know that the energy per unit volume is given by . v is constant throughout the tube since the rate volume of water flowing should be the same at all points in the tube. h is constant since the tube is horizontal. This means, the pressure head P, is changing due to the work done by the viscous forces. Thus, as water flows along the tube, the pressure head reduces and this is reflected by reduction in height of the water columns h2 and h1.

Let the viscous force acting on the fluid be F. This means that the work done by the viscous forces as the liquid moves a distance of x is given by Fx and this will be also equal the loss in energy of the fluid. Hence, Bernoulli's equation will have an Fx term in addition as well to account for the loss due to work done against viscous forces.

Let us consider the energy per unit volume at points A and B. The fluid at point A is still i.e. has zero velocity and the pressure is . The pressure at point B is while the fluid moves with a velocity v. Also the fluid has lost an energy of Fl working against the viscous forces. Hence we have,

Similarly considering points A and D we have,

From (1) and (2) we can write,

Now let us consider points A and E. At E, the fluid enters the atmosphere and so we have,

Irodov Problem 1.334

2009-07-18T04:46:00.000+05:30

a) Consider an infinitesimally thin cylindrical section of thickness dr of radius r around the axis. The area of cross-section of this section is . The volume of water flowing per unit time through this infinitesimally thin section is given by . The total volume of water flowing through the tube per unit time V can be found by integrating over all such cross-sections as,

b) The mass of water flowing through the infinitesimally thin cross-section is given by . This entire mass of water flows at a speed v(r), hence its kinetic energy is given by . The total kinetic energy if the water in the cylinder is given by,

c) Let F be the viscous force acting (and hence the driving force that is responsible for pushing the water) . The shear stress at the tube due to this force is given by . Hence we have,

(note I have ignored the negative sign here since it simply gives the direction of the force).

d) The pressure difference between the ends of the pipe is what is causing the water to flow and is given by the force per unit area as,

Irodov Problem 1.333

2009-07-08T18:16:00.000+05:30

Let the viscous torque opposing the rotation of the cylinder be T. The viscous force acting at a distance r from the axis would be F(r)=T/r. Let the length of the cylinder be l, then the viscous stress at a distance r would be,

The outer cylinder rotates at and is located at radius R2. So from (3) we have,

The angular velocity of the inner cylinder is 0 at radius R1 hence from (4) we have,

Also from (5) it is clear that the torque per unit length due to viscous forces is,

Irodov Problem 1.332

2009-07-02T21:24:00.000+05:30

Let us assume that the shearing force pushing the inner cylinder is F acting in the direction of the axis. Suppose the length of the cylinder is l. The shearing stress at this inner cylinder is given by . At a distance of r from the axis the shear stress will be

From the basic law of viscosity we can determine the gradient of fluid velocity at a distance r as ,

We know that the velocity of the fluid at surface of the outer cylinder is 0, in other words v(R2)=0, hence from (2) we have,

Irodov Problem 1.331

2009-06-28T16:14:00.000+05:30

The viscosity of the fluid causes generates a resistive torque T that tries to slow down the rotating disc. The power generated by by this torque as the disc rotates at an angular velocity of w, is given by Tw.

Now let us compute this resistive torque generated by viscosity. Consider an infinitesimally thin ring of radius r and width dr on the disc. The area of this ring is . Let the resistive viscous force acting on this thin ring be dF.

The velocity of the oil close to this ring section of the disc will be the same as that of the rotating disc equal to wr. The velocity of the oil close to walls of cavity will be 0 since the walls are stationary. Thus, the velocity gradient of the oil in the vertical region between this thin ring section of disc and walls of the cavity will be wr/h.

From the definition of coefficient of viscosity the viscous force dF generated by fluid between the upper cavity wall and the thin ring section of the disc is given by,

The same amount of resistive torque is generated by the fluid between the lower wall of the cavity and the thin ring section as well and so the net resistive viscous force is 2(dF). This resistive force 2(dF) results in a resistive torque dT = 2(dF)r given by,

The work done by the viscous forces is given by Tw equal to .

Irodov Problem 1.330

2009-06-28T09:34:00.002+05:30

Essentially the problem asks us to find the function h = f(r), i.e. how the height of the fluid depends on the distance from the axis of rotation.

Basically what happens is that as the fluid rotates, each section of the rotating fluid exerts a centrifugal force (as seen from the frame rotating with the axis) on the next section of the fluid. Consequently the next section of the fluid increases in height to generate enough hydrostatic pressure to compensate for the centrifugal force.

Let us first start by computing the hydrostatic force exerted by an infinitesimally thin section of rotating fluid of height h and width dr.

At a depth x from the surface the hydrostatic pressure exerted by the fluid is . The force exerted over a thin section of height dx at a this depth x is given by . The net hydrostatic force exerted by the entire section of liquid is given by,

The centrifugal force exerted by this section of liquid as it rotates is given by . Here dm is the mass of the infinitesimally thin section of the liquid and is given by . The centrifugal force exerted by this section of the liquid is thus given by .

Now let us consider too adjacent infinitesimally thin sections of the liquid of height h and h+dh.
The right (lower height column) exerts a hydrostatic force of F(h) to the left while the left (higher column) exerts hydrostatic force F(h+dh) to the right. The right column also exerts a centrifugal force of towards the left. For equilibrium we have,

As dh->0 and dr->0 we can neglect second order terms in (1) such as dh squared and obtain,

Here h0 is the height of the fluid at the center of the cylinder. In practice knowing the total volume of the liquid in the container ho can be found.

b) The hydrostatic pressure at the bottom of the cylinder any radius r is simply

Irodov Problem 1.329

2009-06-28T09:34:00.001+05:30

We assume that just before the tube in the tank the water is still i.e its velocity is 0. Let the velocity of water as it exits out of the spout be vs. The internal pressure of water just before the tube in the tank is given by the pressure due to water column and the atmospheric pressure PA as .
As the water exists out of the spout the pressure is equal to the atmospheric pressure. Applying Bernoulli's equation for just before the tube and as it exists out of the tube we can write,

Now let us consider a cross-section within the tube that has cross-section area a. Let the velocity of the water at this point be v. Since the rate of volume of the water flowing out the tube must be equal at all cross-sections within the tube we have,

Let the internal pressure at this cross-section within the tube be P. Appllying Bernoulli's equation at this cross-section within the tube and at the exit point of the tube we have,

The water exerts pressure P normal to the surface of the tube outwards and the atmosphere exerts pressure PA pushing it inwards. The net pressure exerted on the tube outwards is thus P-PA. The horizontal component of the force due to this pressure (dF) that is trying to rip this section of the tube out of the tank is given by the pressure times the vertical component of the area of the tube given by,

Irodov Problem 1.328

2009-05-30T09:24:00.000+05:30

The area of cross-section of the tube is . Since the fluid flows out of the tube at a rate Q, the velocity of the flowing water v must be .
The reaction force generated by the water as it exits out of the tube is given by . The moment generated by this flowing water seen from the entrance of the tube is given by,

Irodov Problem 1.327

2009-05-29T18:49:00.001+05:30

Consider an infinitesimally thin section of water of width dx escaping out of and at depth x. Let the velocity of this flow be v. The reaction pressure exerted by this section of flowing water is given by . The cross section of this infinitesimally thin section of water flow is bdx. Thus, the reaction force exerted by the infinitesimally thin section of flow is given by . Similar to as solved in Problem 1.327,

Hence we have,

Irodov Problem 1.326

2009-05-29T18:01:00.000+05:30

Suppose that v1 and v2 are the speeds with which the water flows out of the lower and the higher hole respectively.
The reactions exerted by the water flowing out by these two escaping jets is given by and . The net reaction on the container is given by,

Suppose that the depth of the lower hole is h1. The velocity of water just before the entrance of the hole is assumed to be 0. The fluid pressure here is , where Po is the atmospheric pressure. Just outside the hole, the water escapes into open atmosphere i.e. into a region of atmospheric pressure. Applying Bernoulli's equation to points just before and after the hole we have,

Irodov Problem 1.325

2009-05-29T17:01:00.000+05:30

Eulerian equation is a very general equation. There are three basic assumptions that restrict Eulerian equation into Bernoulli's equation. i) the fluid must be non-rotational (there should be no vorticity), or mathematically , ii) the force f must be a conservative force field in other words sum of potential energy and kinetic energy is conserved (such as gravity), and iii) the fluid must be incompressible (water can be considered incompressible for most practical purposes).

In order for the flow to be non-rotational any pressure change can only occur along the direction of the flow (otherwise a rotation will be induced - this not very different from if you apply force in a direction not aligned with velocity of a particle its path will exhibit curvature), in other words the direction of the gradient of pressure must coincide with the direction of the flow of the fluid or where s is along the flow direction of the fluid.

If the only force acting is gravity then along the z-axis, the negative sign comes because gravity acts aong the negative z -axis.

Irodov Problem 1.324

2009-05-24T05:09:00.001+05:30

As the tube rotates, the centrifugal force (as seen in the rotating frame of the tube) creates a pressure head causing the water to be jetted out of the orifice at the end of the tube.

Let us first compute the pressure at the end of the tube just before the orifice. For this we must consider the centrifugal force exerted by each vertical section of water on the next.

Consider, an infinitesimally thin slice of water of width dx at a distance x from the one end of the tube. Let the area of cross-section of the tube be S. The mass of water contained in this infinitesimally thin slice of water is given by , here is the density of water. The centrifugal force exerted by this slice of water on the next is given by,

The net force acting at the end of the tube just before the orifice can be obtained by integrating over all such thin slices as,

In (2) P is the pressure head generated by the centrifugal force acting on the water. The net pressure at the end of the tube just before the orifice is give by P + Po, where Po is the atmospheric pressure. We assume that at this point the water is still i.e. its velocity is zero. Right outside the tube, the water jets into an area of pressure Po (atmospheric pressure), say at a velocity v. Using Bernoulli's equation at points just before and just outside the orifice we have,

Irodov Problem 1.323

2009-05-24T04:29:00.001+05:30

Suppose that at some instant the height of water in the cylinder is h. At the bottom of the cylinder, just before the water jets out, the pressure is given by , here Po is the atmospheric pressure. We assume that water here is still i.e. its velocity is 0.

As soon as the water jets out it enters the atmosphere where the pressure is Po. Let the velocity of the water shooting out be v. Then using Bernoulli's equation for water at points just before the orifice and just outside the orifice we have,

The rate (volume per unit time) at which water flows out is given by sv, where s is the area of the small hole. As water flows out of the cylinder, the height of the water also drops. The rate of change of volume in the cylinder must be equal to the rate at which water flows out of the cylinder. Since the area of the circular section of the cylinder is S, the volume of water contained in the cylinder is given by Sh. Hence, we can find the time taken for the tank to be empty as follows,

Irodov Problem 1.322

2009-05-23T07:08:00.001+05:30

The force F acting upon the fluid to push it out is constant and hence the pressure exerted on the fluid is also constant. This in turn means that the velocity with which the water jets out of the orifice will also be constant (due to Bernoulli's equation, since pressure head does not change). Let this velocity be v. We also know that the surface area of the orifice is s, so the rate at which water is flowing out of the orifice is given by vs (volume of flow per unit time). Since it took a t seconds for the water to flow out completely and the total volume of water was V, we have,

The entire work done by the force is essentially converted into the kinetic energy of the water jetting out of the cylinder. Since the mass of the water is , the total kinetic energy imparted by to the entire volume of water and hence the work done given by,

Irodov Problem 1.321

2009-05-09T13:05:00.000+05:30

The total pressure at point C, at the bottom of the cylinder is given by , where h is the height of the water level and Po is the atmospheric pressure. At point C, the water is assumed to be still and so its velocity is 0. At point A, the water freely falls and the only pressure is once again the atmospheric pressure Po. Let the velocity of the water immediately after the orifice be v. Then, applying Bernoulli's equation for points C and A, we have,

The amount of water flowing per unit time at any point in the orifice can be determined by multiplying the surface area of section through which the water times the velocity of the water.

Near point A, the water powers through a cylindrical section of height and radius R1. In others words the total volume of water flowing out per unit time is given by .

consider a point B within the orifice that is t a radial distance r from the central axis of the cylinder. The surface area of the section through which water is flowing at this point is given by . Let the velocity of the water at point B be vr. Since the total amount of water flowing through any cross-section per unit time must be the same, we have,

Let the internal pressure of the fluid at point B be Pr. Then applying Bernoullis equation for points A and B we have,

Irodov Problem 1.320

2009-05-09T06:07:00.001+05:30

At the surface level of the water within the tube the velocity of the water is v and directed vertically upwards (this will be same as the velocity of the water at the surface in the rest of the area). Let the atmospheric pressure be Po. At the maximum height the velocity of the water is 0 and so we can use the Bernoulli equation to determine the maximum height as,

Irodov Problem 1.319

2009-05-07T08:17:00.000+05:30

Suppose that the hole is made at a height h. The fluid pressure P at this height is given by . Let the atmospheric pressure be Po.The water just to the left of the hole is assumed to be still, in other words it has no velocity.

Just to the right of the hole, say the water shoots out with a horizontal velocity v. The water coming out of the hole to enter a zone with pressure equal to atmospheric pressure and so from Bernoulli's equation we have,

Gravity pulls the water down with an acceleration of g, and so the time t taken by water to reach the surface from a height h is given by

The horizontal distance traveled by the water before its hits the ground is given by,

The extrema (maxima/minima) of l lies at,

Also at this value of h, it can be verified that the second derivative is negative indicating a maxima.

The distance traveled by the water for this value of h is given by substituting h = H/2 in l and is given by H.

Irodov Problem 1.318

2009-05-07T07:01:00.000+05:30

The fluid pressure at the bottom of the container, just above the hole is given by the sum of pressures due to the columns of kerosene and water given by,

Let the atmospheric pressure be Po.

The fluid just above the hole at the bottom is assumed to be still in other words, velocity of the fluid just above the hole is 0.

Just below the hole, say water gushes out with a speed v. The fliud is now flowing into a region of of atmospheric pressure.

Applying Bernoulli's equation we have,

Irodov Problem 1.317

2009-05-07T06:27:00.000+05:30

The streamlines of the gas are depicted by the straight line arrows indicated in the figure. Consider the a line in the path of the gas ABC. At points between A through B along this line the velocity of the gas is v. However, along point between B to C, the velocity of the gas is 0, since it has nowhere to flow.

Bernoulli's equation must apply to all points along a line ABC. Let the pressure of the gas in the region AB be P1 and that in the region BC (inside the pilot tube) be P2.

Hence, we have,

The total amount of flow is given by,

Irodov Problem 1.316

2009-05-03T04:31:00.000+05:30

Since the entire tube is at the same horizontal level, from Bernoulli's equation we have,

The mass of water flowing per unit time through cross sections 1 and 2 must be exactly the same since there is no leakage. Hence, we have,

The difference is pressures P2 - P1 is given be the difference in heights to which the water columns rise in the two cross-sections since both of them are exactly balanced by the same atmospheric pressure. Hence we have,

From (1), (2) and (3) we obtain,

The volume of water flowing per unit time is given by,

Irodov Problem 1.315

2009-05-03T03:34:00.001+05:30

At any point in the bend, as the water travels through the bend of the tube it experiences a change in the direction of its velocity vector by 90 degrees as shown in the figure from v to v'. This means that the acceleration vector of the water in this bend region dv/dt is oriented along the direction from point 1 to 2 as shown in the figure.

We know from the Eulerian equation of hydrodynamics that,

Here, P is the fluid pressure, is the density of the fluid and f is the external force per unit volume acting on the fluid. Since the tube is horizontal, there is no force of gravity acting on the fluid, nor there is any other external force mentioned and hence f=0. Hence, we have,

This means that the direction of increase of pressure P (gradient of P) will be in the opposite direction to that of dv/dt. Since dv/dt is directed from 1 to 2, the pressure must increase at all points as we move from 2 towards 1. This in turn means that P1 > P2.

To compare velocities at points 1 and 2 let us consider the Bernoulli's equation. We know that,

Since the tube is horizontal, h1 = h2. Thus, (1) becomes,

Since P1 > P2 then from equation (2) it is clear that .

Irdov Problem 1.314

2009-05-02T13:48:00.000+05:30

The work done dW in compressing a fluid by an infinitesimally small amount dV is given by PdV, where P is the pressure acting on the fluid (this comes from the fact that dW = Fdx = PAdx = PdV). The compressibility is defined as . Thus, the work done dW in increasing the pressure by dP over a volume of fluid V, is given by. The energy of elastic deformation stored in water of volume V is the work done in increasing pressure to P from 0 and is given by,

Since, at a depth h, the pressure of water is pgh, where density of water is p.

Irodov Problem 1.313

2009-05-02T12:50:00.000+05:30

This problem, in my opinion, should actually be a sub-part of problem 1.312, I am really not sure why Irodov decided to make it a separate problem. As seen from eqn (1) in solution to problem 1.312, at a distance of r from the axis the strain is given by,

Thus, the strain energy per unit volume at a distance r is given by,

Irodov Problem 1.312

2009-05-02T11:15:00.001+05:30

Since one end is fixed all points on that end will be at the same place after the distortion. Consider a line OA on the fixed side and a corresponding parallel line O'B on the free end. After the distortion, B moves to a location B' - this induces a shear strain of magnitude BB'/l. Another point X on AB moves to location X' and so this area experiences a shear XX'/l. Thus, the cylinder experiences different shear strains at different distances from the axis.

Suppose that the point X is at a distance x from the axis of the cylinder. The length of XX' is approximately and so the shear strain experienced at point X is given by,

The energy per unit volume due to this shear strain is given by . Consider an infinitesimally thin cylindrical section of thickness dx at point X. The volume of this infinitesimally thin cylindrical section is . The total energy can be computed by integrating the energy contained in all such infinitesimally thin sections and hence this given by,

Irodov Problem 1.311

2009-05-01T18:22:00.001+05:30

As the steel band shown beside is bent into a hoop, the deformation causes strain in the hoop. Suppose that the steel band was made into a hoop of radius ro. Then we have,

Consider an infinitesimally thin section of radius r and thickness dr in the hoop. The length of this section of the hoop is . Hence the change in length experienced by this section is given by . The strain seen by this section of the hoop is given by,

The volume of this infinitesimally thin ring is given by . The elastic energy per unit volume is given by . Hence, the total elastic energy contained in this infinitesimally thin hoop section is given by,

The total energy contained in the hoop is thus given by,

The work done on the steel band to deform it into the hoop will be exactly equal to this elastic energy due to deformation.

Irodov Problem 1.310

2009-05-01T18:22:00.000+05:30

The part of the cylinder AX will be pulled downward at point X (which is at a height x from the bottom end B), by the weight of the cylinder in part XB given by where p is the density of the cylinder. Hence, the stress experienced at the cross-section of point X will be

Thus, at each height the cylinder experiences a different amount of stress. The strain at this point X is then given by,

The net elongation of the cylinder can be found as,

The elastic energy per unit volume at the location X is given by . Consider an infinitesimally thin disc at location X of thickness dx. The volume of this infinitesimally thin disc will be The elastic energy in an infinitesimally thin disc of thickness dx is given by given by,

The total elastic energy contained in the cylinder is given by,

Irodov Problem 1.309

2009-04-23T19:10:00.000+05:30

The volume density of energy due to elastic deformation is given by . So the volume of the steel rod is V, then the total energy in the steel rod is given by . If the mass of the steel rod is m then we can rewrite it as.

Irodov Problem 1.308

2009-04-23T18:00:00.000+05:30

Consider an infinitesimally thin ring section of radius r and thickness dr. The mass of this thin ring section is given by where p is the density of the ring. Each point in this thin ring section experiences a tangential linear acceleration of . When viewed from the reference frame fixed to the ring, this translates to a pseudo force . This pseudo force translates to a pseudo torque . The net torque is then given by,

Irodov Problem 1.307

2009-04-14T04:58:00.000+05:30

The work done by acconstant torque T in turning an angle is given by . The power generated by a constant torque T is thus given by. We already know from Irodov Problem 1.305 part b that the twisting moment generated when a solid cylindrical shaft turns by an angle is given by,

and so the power generated is simply.

Irodov Problem 1.306

2009-04-14T04:50:00.000+05:30

This problem differs very slightly from Irodov Problem 1.305 part b in that instead of being a solid cylinder the cylinder is a shell. Thus, the twisting moment can be derived as,

Irodov Problem 1.305

2009-04-14T03:50:00.000+05:30

a)

The shearing moment acting on the cylinder, causes shear stress (check the definition of shear modulus here) resulting in the cylinder to twist and thus in a shear strain. As the cylinder twists
an angle resulting in the upper surface to shift a distance of relative to the lower surface as shown in the figure. Thus, the shear strain is given by . This, shear strain will result in a shear stress acting tangential to the surface given by . The net tangential force acting on the cylinder is given by the stress times the area of cross section given by,

Since this force acts at a distance r from the axis, the twisting moment is given by,

b)

For this part consider an infinitesimally thin vertical section of the solid cylinder at a distance x from the center and of thickness dx. As solved in part a of the problem, the twisting moment acting on this section will be given by,

Irodov Problem 1.304

2009-04-13T08:19:00.001+05:30

Consider an infinitesimally thin (of width dx) strip of the square plate at a distance x from the axis of rotation. These points are experiencing a linear tangential acceleration of . If the density of the plate is p and its thickness is h, then the mass of this infinitesimally think strip will be phldx. As seen from a frame that is rotating, this will translate to a pseudo force of magnitude . This pseudo force results in a pseudo torque of magnitude in a direction opposite to the direction of rotation of the plate.

Now let us consider the torques a section of the plate XB (as indicated in the figure). The net torque acting on this section of the plate due to the pseudo torque can be computed as,

This pseudo torque will be exactly balanced by the torque due to elastic forces N(x) acting at the boundary of plate section at X. From Irodov Problem 1.301 (or this link) we know that the elastic curve of the plate is given by,

Where, for this plate, as solved in Irodov Problem 1.303 part a,

Hence, we have,

Irodov Problem 1.303

2009-04-11T08:46:00.000+05:30

a)

Let the width of the cantilever be b, its height h, length l and density p.
Consider the torques acting on a section of the cantilever OB as shown in the figure. There are two moments acting on this section of the cantilever, i) due to its own weight phb(l-x)g acting at the center of the section at a distance (l-x)/2 from O in a clockwise direction and ii) the elastic moment N(x) generated by the curvature of the cantilever. From Irodov problem 1.301 (derivation at this link) we know that,

Since the cantilever is at equilibrium, these moments must cancel each other. Thus, N(x) must act in the anti-clockwise direction. This in turn means that the upper surface of the cantilever OB must be stretched more than the lower surface O'B' as shown in the figure. In other words, the elastic curve of the cantilever must be a concave function i.e. .

Hence, we have,

Now the moment of inertia I about the neutral axis of the
cantilever can be obtained as,

Using (1) we can obtain the depression y as, . The negative sign indicating depression.

b)

If the width of the beam is b as in part a), the total weight of the beam is 2bphlg. In order for the beam to be in equilibrium, the net reaction force from the end points must be equal to 2bphlg acting upwards. Further, the reaction from each of the end points should be exactly same since otherwise a net torque will be induced in the beam about the center of the beam. Thus, each end point provides a reaction of bplgh as shown in the figure.

Now let us consider a small section of the beam AO that is x units long, as shown in the figure. The net weight of this section of the beam is bpghx and acts at the center of this section at a distance of x/2 from the end point A.

There are three torques acting on the beam about point O, i) due to reaction from the end point A given by lpbghx in the clockwise direction, ii) due the weight of the beam given by and acts in the anti-clockwise direction and finally iii) the elastic moment N(x) due to the curvature in the beam.

N(x) must completely balance the sum of the other two torques and thus must act in the anti-clockwise direction (since the torque acting at the end point is greater than the torque due to the weight). This can only happen if the bottom surface of the beam AO is stretched more than the top surface of the beam A'O' as seen in the figure. In other words, the elastic curve of the beam will be a convex function i.e. .

Hence we have,

The negative sign indicates depression.

Irodov Problem 1.302

2009-04-11T04:18:00.000+05:30

In order to keep the beam in equilibrium, the two ends must provide a total reaction force of F in the upward direction. The reaction forces offered by each of the two ends must exactly be the same since otherwise a torque will be induced in the beam causing it to rotate about its center. Hence, each of the ends offer a force of F/2 in the upward direction.

Now let us consider the torque acting on a section AO of the beam, such that O is at coordinate x from the left end. The reaction force F/2 causes a torque in the clockwise direction of magnitude Fx/2. This must be compensated by the elastic moment in the beam N(x). In other words N(x) must act in the counter-clockwise direction and must be equal to Fx/2 in magnitude. The elastic moment in the beam can be in the counter-clockwise direction only if the lower surface of the beam AO is stretched more than the lower surface A'O' as shown in the figure. This means that the elastic curve of the beam must be a convex function i.e. . We also know that the elastic cruve follows the equation,

(as given in Irodov problem 1.301 or the link).

Hence we have,

The negative sign indicates depression.

Irodov Problem 1.301

2009-04-10T20:08:00.000+05:30

a) Consider the section of the cantilever OB such that point O is located at a distance x from the rigid end point to which the rod is attached.

If the moment generated at point O is N(x), we know that the shape of cantilever obeys the equation,

The derivation of this equation is given at this link.

For this part, the bending moment is independent of x and equal to No in the clockwise direction. In order for the cantilever part OB to be steady (no turning under the influence of the couple No), the moment generated at the end O i.e. N(x), must exactly counter-balance No. In order to generate an elastic moment in the counter-clockwise direction the upper surface OB must be stretched more than the lower surface of the beam O'B' as shown in the figure. This in turn means that the shape of the beam must be a concave function i.e. . Thus we have,

The negative sign indicates depression.

b) For the second part since a force F acts at the end, the moment of this force about point O is given by in the clockwise direction. Hence, as in part a we have,

The negative sign indicates depression.

Irodov Problem 1.300

2009-04-10T10:47:00.001+05:30

Let us first understand whats happening in the problem. The force of gravity acts on the cantilever and induces a torque that aims to turn the cantilever in the clockwise direction. As a result the cantilever bends and curvature is induced into the cantilever. Due to the curvature in the cantilever, the upper surface of the cantilever is stretched while the lower surface is compressed relative to the center line. As seen in the diagram, top surface AB is stretched and is longer than OO' (the center line) and A'B' the bottom surface is smaller than OO'. This distortion results in strain and hence stress in the cantilever. This stress results in a torque in the counter-clockwise direction and balances out the torque due to gravity.

Let the width of the cantilever be b, its density be p. The force of gravity acts at a distance from l/2 from the wall and hence the torque due to gravity will be .

Suppose that due to the curvature in the cantilever, the end surface bends an angle as shown in the figure.

Consider an infinitesimally thin section of the cantilever that is at a height y from the center line, and has a thickness of dy. The change in length of this section will be as shown in the figure. As a result the strain induced in this section we will be,

The stress induced in this section will thus be,

The area of this infinitesimally thin section of the cantilever is bdy. The net force due to the stress in this section will thus be . Since the force acts at a distance y from the center, the torque induced by this section will be . The net torque due to the stress in the cantilever can be found as,

This torque T due to stress must balance the torque due to gravity for the cantilever to be stable and so we have,

Let the radius of curvature seen by the cantilever be R, then as seen from the figure,

Irodov Problem 1.298

2009-04-08T23:45:00.000+05:30

Consider a section of the rod XB that is x units long from the lowest end of the rod. Let the density of the rod be p and its cross-section area be S. There are two forces acting on this part of the rod, i) the force of gravity pxSg pulling it down and ii) the force due to stress in the rod at point X given by . Since the section of the rod is not falling down we have,

The strain experienced by the section XB of the rod is thus given by,

Suppose that the change in length for the section XB of the rod is , then we have,

The net strain in the rod is thus given by,

(b) In addition to the longitudinal strain that elongates the rod, the rod also experiences a lateral strain (dictated by Poisson's ratio) that leads to a reduction in the rod's radius given by . In other words if the radius of the rod is R, then the change in the radius is given by,

The change in volume of the rod is given by,

Irodov Problem 1.297

2009-04-08T23:15:00.000+05:30

Let the radius of the cylinder be R then the pressure seen by the cylinder is given by . The longitudinal strain seen by the cylinder is . In other words . The pressure also causes a lateral strain dictated by the Poisson's ratio making the cylinder to widen (radius increases) given by . In other words,. We can now calculate the change in volume as,