Irodov Problem 1.243

Let the tension in the string holding mass m be T. Suppose that the acceleration with which the mass m falls down is given by a. Also suppose that the angular acceleration of the cylindrical pulley about its axis of rotation is given by .

There are two forces acting on mass m in the vertical direction, i) the tension in the string T pulling it up and ii) the force of gravity mg pulling it down. So from Newton's laws we have,



Now let us consider the cylinder. The tension in the string creates a torque of TR on the cylinder and causes it to rotate about its axis of rotation and induces the angular acceleration. So if I is the moment of inertia of the cylinder we have,



The linear acceleration of the mass m and the angular acceleration of the cylinder are however related to each other as,



since the thread is not slipping over the cylinder.


Now solving for a from (1), (2) and (3) we get,





a) The angular velocity of the cylinder as a function of time is thus given by,





b) If v is the velocity of the mass m at some time t. The kinetic energy of the whole system is given by,

Irodov Problem 1.242

The moment of inertia of a solid sphere of radius R and uniform density p is given by,

(see this page).


Now let us consider as slightly larger sphere of radius . The moment of inertia of the shell if the moment of inertia of the larger sphere minus that of the smaller sphere and is given by,

Irodov Solution 1.241

The mass of the remaining disc is m. Suppose that the mass of the original disc before it was cut out was M. The mass of portion that was cut out (of radius R/2) will be M/4.
In other words,

M - M/4 = m or,

M = (4/3)m.

The moment of inertia of a solid disc (before it was cut out) about axis AB that passes through its center and perpendicular to the plane of the disc is given by (see problem 1.240) .



The moment of inertia of the portion of disc that was cut out about the axis CD (passing through its center) is given by,




The moment of inertia of the cutout portion of the disc about an axis passing through AB can now be determined using the parallel axis theorem as,







The moment of inertia of the remaining portion of the disc about AB can now be computed by simply subtracting the moment of inertia of the cut out portion disc from the that of the entire disc. In other words,







Now the final twist to the tale. Irodov asks for moment of inertia about an axis that passes through the CG of the remaining portion of the disc and not the center of the disc. Since the portion of the disc was cut out, the CG of the remaining portion of the disc would have shifted to the right as indicated by O' in the figure. The mass of the portion of the disc that was cutout was M/4 and its CG is located at -R/2 (negative because its to the left) from O (the center of the original disc is considered as the origin). Suppose that the CG of the remaining part of the disc to O' located at x units from O. Then we have,






Using the parallel axis theorem, now we can compute the moment of inertia of the remaining portion of the disc about EF (the axis passing through O') is given by,

Irodov Problem 1.240

Essentially in this problem Irodov asks you to prove the famous perpendicular axis theorem for moment of inertia. The proof is quite simple and given in the same location.










From perpendicular axis theorem we know that Iz = Ix + Iy. Also since its a disc, the moment of inertia should not depend on which diameter is chosen, in other words Ix = Iy. Hence,
Ix = Iy = Iz/2.

In problem 1.239a I already found the moment of inertia Iz of a disc given by,




Thus we have,

Irodov Problem 1.239

a) Let the density of the disc be p. Consider a infinitesimally thin slice of the disc in the shape of a ring of thickness dr and width b -same as that of the disc. The mass of this ring will be . The reason we chose this particular shape (ring) for the infinitesimally thin ring is because all points of this ring are equidistant from the axis of rotation being at a distance r.
The moment of inertia can now be calculated as,





b)

We have already determined the moment of inertia of a disc of radius r and thickness b in the part a) of the problem. A cone is simply a stack of several infinitesimally thin disks of gradually decreasing radii, stack on top of one another. Consider an infinitesimally thin disc of thickness dh, at a distance h from the apex of the cone O. Let the height of the cone be H. The radius of this infinitesimally thick disc will be . Let the density of the cone be p. Based on our solution in part a), the moment of intertia of this infinitesimally thin disc will be,



The total moment of inertia of the entire cone can be determined by integrating (summing) over all infinitesimally thin discs as,








Isn't it interesting to see that the moment of inertia of a cone does not depend on its height but only on the radius of its base!!

Irodov Problem 1.238

Let us consider an infinitesimally small piece of the rod of length dx at a distance x from the axis of rotation. Its mass is given by as shown in the figure. The moment of inertia of the rod is given by,





(b)
Consider an infinitesimally small piece of the plate whose width and length are dx and dy respectively located at coordinates (x,y) if we consider the corner through which the axis passes as the origin as shown in the figure. The distance of this point from the orgin is given by .
The mass of this infinitesimally small piece is given by . So the moment of inertia of the plate is given by,

Irodov Problem 1.237

The three forces acting at A, B and D are given by,







The net force acting on the square is given by,



What the question essentially asks is that if you were asked to replace the three forces acting on square by a single force of magnitude Fnet, then at what point on the side BC, O would it be acting to create the same effect as the three forces?

If Fnet = 2Fi were acting at this point O, then the torque on the square about point O must be exactly 0, since the moment arm of Fnet is 0 about point O. In other words, we have to find a point O on side BC about which the net torque due to all the three forces is 0.

Suppose that the length of the side of the square is a. Let the origin (0,0) be at the center of the square. Then the coordinates of the corners A, B and D will be,








The equation of the line BC is . Based on this, let the coordinate of O be,




Then, we can compute the position vectors of point A, B and D with respect to O as,







The net torque due to all the three forces about point O is given by,





If the net torque about point O is to be 0 then, setting No = 0 we have . This corresponds to the mid point of the line BC.

Irodov Problem 1.236

The net torque N acting on the system is given by r1 x F1 + r2 x F2. This is given by,






The net force acting on the system is given by F = F1 + F2 = Bi + Aj. The lever arm l is given by,