Irodov Problem 1.132

As in problem 1.131, the force is the negative gradient of the potential. In Cartesian coordinates we have,




For a force field to be central it must depend only on - the magnitude of the radius vector from the center of the force. This is clearly not true for the force above. In other words this is not a central force field.

Equi-potential surface is the surface (or curve) on which the value potential all points have the same potential. In other words,


This is the equation of an ellipse centered at the origin.

Irodov Problem 1.131

Force is the negative gradient of potential i.e. . In spherical coordinates this is given by



Since in this problem, U is only a function or r, i.e. depends only on the distance from the center, the force acting on it must be radial (since all other partial derivatives will be zero in the above expression). Thus, the force acting radially outwards (away from the center) is given by,



At the equilibrium point the net force will be zero, this means





The concept of stability has been explained in the solution to problem 1.109. The idea being that under small perturbations the body must have a tendency to return to its original equilibrium point. Consider a small perturbation about the equilibrium point. The force acting on the body will be,




The easiest way to check stability is to linearize the above expression for small perturbation about the equilibrium point and see if the force is restoring. So using Taylor's expansion and ignoring all terms with power higher than 1 we have,






The force thus always acts in the opposite direction of the perturbation (as seen by the negative sign), this means that the force acts in a manner so as to restore the particle to its equilibrium position. In fact for small perturbations the motion is simple harmonic. In other words the equilibrium point is stable.

b)At the extrema the first derivative should be zero so at the point where magnitude if attractive force is maximum, so we have,





The value of this force can be obtained by simply substituting in the expression for force and is given by,

Irodov Problem 1.130

Frankly the wording of the problem confused me a bit initially since the positive direction of F was not clarified in the problem. However application of some common sense will tell us that F is considered to be in the same direction as that of gravity. It is evident from the problem that the value of F when y=0 i.e. the mass is at the ground, is -2mg. At this point the mass is supposed to accelerate upwards. This can only happen if -2mg means 2mg force directed upwards. In other words the positive direction of F must be aligned with the direction of gravity.

At any given time, there are two forces acting on the mass, the force of gravity mg and the force F, so that the total force acting on the mass is F+mg. This means that when the body starts its motion upwards, there is a net force of mg - 2mg = -mg (acting upwards). As the body moves upwards and its height y increases the magnitude of F keeps decreasing. At a height the magnitude of the force F becomes zero. After this point, F reverses its direction and starts to act in the same direction as the force of gravity. The action of the force is completely symmetric (like a mirror image) about the height as shown in the figure. In other words if you take any two heights and , the force F at these heights will be equal and opposite. From this symmetry it is evident that the maximum height achieved h will be . This is indicated in the figure.

The work done by force F during the first half of the accent will be positive since the direction of motion (up) is same as the direction of the force (up). This can be evaluated as,











The reason we use the negative sign in (-dy) is since the direction of increase of dy is in the upwards direction whereas we have chosen the positive direction to be downwards for F.

The gain in potential energy is simply given by since at the end of half its accent its at a height of 1/2a.

Irodov Problem 1.129

Let and be the extensions in the two springs. The total extension of both the springs combined together is
. Let F be the force which is being applied to stretch the springs.

The most important thing is to notice is that both the springs will experience the same tension. Why? Suppose that the tension in spring 1 was F1 then spring 2 would experience a force F at one end while F1 at the other end. This would mean that there would be a net force of F-F1 acting on the spring. But this cannot be true since the springs are not accelerating, they are simply stretched!! This means F-F1=0 or F=F1. In other words tension in both springs must be the same. This actually holds for any number of springs connected in series.

Hence we have,


In other words,









The work done by this force will be entirely the energy stored in the springs given by,

Irodov Problem 1.128

The centrifugal force of F = will always act radially outwards in the rotating reference frame. As the mass moves the work done by the centrifugal force is . The vector dr will have two components a radial component dr (corresponding to the radial motion of the body) and (corresponding to the tangential motion of the body). Since, F acts only radially, the dot product of the tangential part of the bodies motion and F will be 0. Hence we have,

Irodov Problem 1.127

Say the mass acquires a velocity v0 at time t=0, the friction force Nk acts in the direction opposite to the direction of motion and hence decelerates it, N being the normal reaction from the surface. Along the vertical direction there are two forces acting on the mass, the normal reaction N and the force of gravity mg. There is no acceleration in the vertical direction and so we have,

N-mg=0 (1)


The mass hence experiences a deceleration Nk/m at any instant of time. The work done by friction at any instant is -Nkv, where v is the instantaneous horizontal velocity of the mass. The negative sign appears since the direction of motion of the mass is exactly in the opposite direction to the force of friction.

a) In this part the force of friction is constant given by Nk = mgk and hence experiences a constant deceleration of -gk. The time taken for the block to come to complete rest is v0/gk.
Initially the mass had a kinetic energy of and due to friction it will loose all of it when it stops. The total negative work done by friction will then be equal to the initial kinetic energy of the mass i.e. . The average power generated by friction is total work done by total time given by,





b) In this part the force of friction depends on distance from initial position as . Thus, as any given time,









The instantaneous power generated by friction is given by -mgkv and is thus given by,




The extrema (maxima or minima) occurs when











From (4) and (3) we have,