Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

Class 12 : Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

The document Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev is a part of the Class 12 Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 266. Determine the kinetic energy of a tractor crawler belt of mass m if the tractor moves with velocity v (Fig. 1.69). 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Solution. 266. Since the lower part of the belt is in contact with the rigid floor, velocity of this part becomes zero. The crawler moves with velocity v, hence the velocity of upper part of the belt becomes 2v by the rolling condition and kinetic energy of upper part  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev which is also the sought kinetic energy, assuming that the length of the belt is much larger than the radius of the wheels.


Q. 267. A uniform sphere of mass In and radius r rolls without sliding over a horizontal plane, rotating about a horizontal axle OA (Fig. 1.70). In the process, the centre of the sphere moves with velocity v along a circle of radius R. Find the kinetic energy of the sphere. 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Solution. 267. The sphere has two types of motion, one is the rotation about its own axis and the other is motion in a circle of radius R. Hence the sought kinetic energy

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev    (1)

where is the moment of inertia about its own axis, and I2 is the moment of inertia about the vertical axis, passing through O, 

But, Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev  (using parallel axis theorem,)  (2)

In addition to

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev     (3)

Using (2) and (3) in (1), we get Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 268. Demonstrate that in the reference frame rotating with a constant angular velocity ω about a stationary axis a body of mass m experiences the resultant
 (a) centrifugal force of inertia Fcf = mω2Rc, where Rc is the radius vector of the body's centre of inertia relative to the rotation axis;
 (b) Coriolis force Fcor = 2m [v'cω], where is the velocity of the body's centre of inertia in the rotating reference frame

Solution. 268. For a point mass of mass dm, looked at from C rotating frame, the equation is

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev radius vector in the rotating frame with respect to rotation axis and Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev velocity in the same frame. The total centrifugal force is clearly

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev is the radius vector of the C.M. of the body with respect to rotation axis, also

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

where we have used the definitions

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 269. A midpoint of a thin uniform rod AB of mass m and length l is rigidly fixed to a rotation axle OO' as shown in Fig. 1.71. The rod is set into rotation with a constant angular velocity ω. Find the resultant moment of the centrifugal forces of inertia relative to the point C in the reference frame fixed to the axle OO' and to the rod.

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Solution. 269. Consider a small element of length dx at a distance x from the point C, which is rotating in a circle of radius r = x sin θ

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Now, mass of the element  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

So, centrifugal force acting on this element Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev and moment of this force about C, 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

and hence, total moment

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 270. A conical pendulum, a thin uniform rod of length l and mass m, rotates uniformly about a vertical axis with angular velocity ω (the upper end of the rod is hinged). Find the angle θ between the rod and the vertical.

Solution. 270. Let us consider the system in a frame rotating with the rod. In this frame, the rod is at rest and experiences not only the gravitational force  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev the reaction force  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev but also the centrifugal force Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

In the considered frame, from the condition of equilibrium i.e.   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

or,    Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev        (1)

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

where Ncf is the moment of centrifugal force about O. To calculate Ncf, let us consider an element of length dx, situated at a distance x from the point O. This element is subjected to a horizontal pseudo force  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev The moment of this pseudo force about the axis of rotation through the point O is

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

So   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev     (2)

It follows from Eqs. (1) and (2) that, 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev         (3)


Q. 271. A uniform cube with edge a rests on a horizontal plane whose friction coefficient equals k. The cube is set in motion with an initial velocity, travels some distance over the plane and comes to a stand- still. Explain the disappearance of the angular momentum of the cube relative to the axis lying in the plane at right angles to the cube's motion direction. Find the distance between the resultants of gravitational forces and the reaction forces exerted by the supporting plane. 

Solution. 271. When the cube is given an initial velocity on the table in some direction (as shown) it acquires an angular momentum about an axis on the table perpendicular to the initial velocity and (say) just below the C.G.. This angular momentum will disappear when the cube stops and this can only by due to a torque. Frictional forces cannot do this by themselves because they act in the plain containing the axis. But if the force of normal reaction act eccentrically (as shown), their torque can bring about the vanishing of the angular momentum. We can calculate the distance Ax between the point of application of the normal reaction and the C.G. of the cube as follows. Take the moment about C.G. of all the forces. This must vanish because the cube does not turn or tumble on the table. Then if the force of friction is fr

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

But N = mg and fr = kmg, so

Δx = ka/2


Q. 272. A smooth uniform rod AB of mass M and length l rotates freely with an angular velocity ω0, in a horizontal plane about a stationary vertical axis passing through its end A. A small sleeve of mass m starts sliding along the rod from the point A. Find the velocity v' of the sleeve relative to the rod at the moment it reaches its other end B. 

Solution. 272. In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary. Hence, it follows that

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(ω is the final angular velocity of the rod)

and   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

From these equations we obtain

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 273. A uniform rod of mass m = 5.0 kg and length l = 90 cm rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse J = 3.0 N•s in a horizontal direction perpendicular to the rod. As a result, the rod obtains the momentum p = 3.0 N.s. Find the force with which one half of the rod will act on the other in the process of motion. 

Solution. 273. Due to hitting of the ball, the angular impulse received by the rod about the C.M. is equal to  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev If ω is the angular velocity acquired by the rod, we hav

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev        (1)

In the frame of C.M., the rod is rotating about an axis passing through its mid point with the angular velocity ω. Hence the force exerted by one half on the other = mass of one half x acceleration of C.M. of that part, in the frame of C.M 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 274. A thin uniform square plate with side l and mass M can rotate freely about a stationary vertical axis coinciding with one of its sides. A small ball of mass m flying with velocity v at right angles to the plate strikes elastically the centre of it. Find:
 (a) the velocity of the ball v' after the impact;
 (b) the horizontal component of the resultant force which the axis will exert on the plate after the impact. 

Solution. 274.  (a) In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary. Hence it follows that

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

and  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

From these equations we obtain

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(b) Obviously the sought force provides the centripetal acceleration to the C.M. of the rod and is

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 275. A vertically oriented uniform rod of mass M and length l can rotate about its upper end. A horizontally flying bullet of mass m strikes the lower end of the rod and gets stuck in it; as a result, the rod swings through an angle α. Assuming that m << M, find:
 (a) the velocity of the flying bullet;
 (b) the momentum increment in the system "bullet-rod" during the impact; what causes the change of that momentum;
 (c) at what distance x from the upper end of the rod the bullet must strike for the momentum of the system "bullet-rod" to remain constant during the impact.

Solution. 275. (a) About the axis of rotation of the rod, the angular momentum of the system is conserved. Thus if the velocity of the flying bullet is v.

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev
Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev     (1)

Now from the conservation of mechanical energy of-the system (rod with bullet) in the uniform field of gravity

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev    (2)

[because C.M. of rod raises by the height  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Solving (1) and (2), we get

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

where , ωl is the velqcdty of the bullet and Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev equals the velocity of C.M. of the rod after the impact. Putting the value of v and ω we get

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

This is caused by the reaction at the hinge on the upper end.

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Final momentum is

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

So,   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

This vanishes for   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 276. A horizontally oriented uniform disc of mass M and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass m. A light thread running down through the hollow axle of the disc is tied to the body. Initially the body was located at the edge of the disc and the whole system rotated with an angular velocity ω0. Then by means of a force F applied to the lower end of the thread the body was slowly pulled to the rotation axis. Find:
 (a) the angular velocity of the system in its final state;
 (b) the work performed by the force F. 

Solution. 276. (a) As force F on the body is radial so its angular momentum about the axis becomes zero and the angular momentum of the system about the given axis is conserved. Thus

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(b) From the equation of the increment of the mechanical energy of the system :

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Putting the value of ω from part (a) and solving we get

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 277. A man of mass ml  stands on the edge of a horizontal uniform disc of mass m2 and radius R which is capable of rotating freely about a stationary vertical axis passing through its centre. At a cer- tain moment the man starts moving along the edge of the disc; he shifts over an angle φ' relative to the disc and then stops. In the process of motion the velocity of the man varies with time as v' (t). Assuming the dimensions of the man to be negligible, find:
 (a) the angle through which the disc had turned by the moment the man stopped;
 (b) the force moment (relative to the rotation axis) with which the man acted on the disc in the process of motion. 

Solution. 277. (a) Let z be the rotation axis of disc and <p be its rotation angle in accordance with right-hand screw rule (Fig.). (φ and φ' are to be measured in the same sense algebraically.) As Mz of the system (disc + man) is conserved and Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev we have at any instant,

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

or,    Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

or,   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev    (1)

This gives the total angle of rotation of the disc.

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 278. Two horizontal discs rotate freely about a vertical axis passing through their centres. The moments of inertia of the discs relative to this axis are equal to l1  and l2, and the angular velocities to oh and ω1. When the upper disc fell on the lower one, both discs began rotating, after some time, as a single whole (due to friction). Find:
 (a) the steady-state angular rotation velocity of the discs;
 (b) the work performed by the friction forces in this process. 

Solution. 278. (a) Frome the law of conservation of angular momentum of the system relative to vertical axis z, it follows that:

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Hence  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Not that for ωz > 0, the corresponding vector   Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev with the poitive direction to the z axis, and vice versa. As both discs rotates about the same vertical axis z, thus in vector form.

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

However, the problem makes sense only if  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(b) From the equation of increment of mechanical energy of a system: Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Using Eq. (1)

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 279. A small disc and a thin uniform rod of length l, whose mass is η times greater than the mass of the disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity v, after which it elastically collides with the end of the rod. Find the velocity of the disc and the angular velocity of the rod after the collision. At what value of η will the velocity of the disc after the collision be equal to zero? reverse its direction? 

Solution. 279. For the closed system (disc + rod), the angular momentum is conserved about any axis. Thus from the conservation of angular momentum of the system about the rotation axis of rod passing through its C.M. gives :

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev       (1)

(v' is the final velocity of the disc and co angular velocity of the rod) For the closed system linear momentum is also conserved. Hence

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(where vc is the velocity of C.M. of the rod) From Eqs (1) and (2) we get

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Applying conservation of kinetic energy, as the collision is elastic

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Vectorially, noting that we have taken  Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev


Q. 280.  A stationary platform P which can rotate freely about a vertical axis (Fig. 1.72) supports a motor M and a balance weight N. The mo- ment of inertia of the platform with the motor and the balance weight relative to this axis is equal to I. A light frame is fixed to the motor's shaft with a uniform sphere A rotating freely with an angular velocity ω0 about a shaft BB' coinciding with the axis OO'. The moment of inertia of the sphere relative to the rotation axis is equal to I. Find:
 (a) the work performed by the motor in turning the shaft BB' through 90'; through 180°;
 (b) the moment of external forces which maintains the axis of the arrangement in the vertical position after the motor turns the shaft BB' through 90°. 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Solution. 280. See the diagram in the book (Fig. 1.72) (a) When the shaft BB' is turned through 90° the platform must start turning with angular velocity Ω so that the angular momentum remains constant. Here

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

The work performed by the motor is therefore

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

If the shaft is turned through 180°, angular velocity of the sphere changes sign. Thus from conservation of angular momentum,

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRevis the complete angular momentum of the sphere i. e. we assume that the angular velocity of the sphere is just  - ω0). Then 

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

and the work done must be,

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

(b) In the case (a), first part, the angular momentum vector of the sphere is precessing with angular velocity Ω. Thus a torque,

Irodov Solutions: Dynamics of A Solid Body- 3 Notes | EduRev

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