NCERT Physics Class 11 Exemplar Ch 7 System Of Particles And Rotational Motion Part 5

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Q.17. A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in Fig. Match the following (most appropriate choice):

Block on horizontal surface

Block on Horizontal Surface

Block on horizontal surface

Slipping of Cube
Slipping of cube

(a)

(i) Cube will move up.

(b)

(ii) Cube will not exhibit motion.

(c)

(iii) Cube will begin to rotate and slip at A.

(d)

(iv) Normal reaction effectively at from A, no motion.

Answer: (a) ii, (b) iii, (c) i, (d) iv

Q. 18. A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig. 7.9). The sphere is struck horizontally at a height h from the floor. Match the following:

Rolling of Sphere
Rolling of sphere

(a)

(i) Sphere rolls without slipping with a constant velocity and no loss of energy.

(b)

(ii) Sphere spins clockwise, loses energy by friction.

(c)

(iii) Sphere spins anti-clockwise, loses energy by friction.

(d)

(iv) Sphere has only a translational motion, looses energy by friction.

Answer: (a) iii, (b) iv (c) ii (d) i

Q.19. The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Answer: No. Given

The sum of torques about a certain point ‘0

The sum of torques about any other point O′,

Here, the second term need not vanish.

Q.20 A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Answer: The centripetal acceleration in a wheel arise due to the internal elastic forces which in pairs cancel each other; being part of a symmetrical system In a half wheel the distribution of mass about its centre of mass (axis of rotation) is not symmetrical. Therefore, the direction of angular momentum does not coincide with the direction of angular velocity and hence an external torque is required to maintain rotation.

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