EXERCISES

Author:Jharkhand Board

Jharkhand Board Physics Solutions for EXERCISES

Simple step-by-step solutions to EXERCISES questions of Systems of Particles and Rotational Motion from PHYSICS PART 1 TEXTBOOK FOR CLASS XI. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from EXERCISES with Hints & Solutions

EASY
11th Jharkhand Board
IMPORTANT

Read the statement below carefully, and state, with reasons, if it is true or false:

During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

EASY
11th Jharkhand Board
IMPORTANT

Read the statement below carefully, and state, with reasons, if it is true or false:

The instantaneous speed of the point of contact during pure rolling is zero.

EASY
11th Jharkhand Board
IMPORTANT

Read the statement below carefully, and state, with reasons, if it is true or false:

The instantaneous acceleration of the point of contact during rolling is zero.

EASY
11th Jharkhand Board
IMPORTANT

Read the statement below carefully, and state, with reasons, if it is true or false:

For perfect rolling motion, work done against friction is zero.

EASY
11th Jharkhand Board
IMPORTANT

Read the statement below carefully, and state, with reasons, if it is true or false:

A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

HARD
11th Jharkhand Board
IMPORTANT

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

Show K=K'+12MV2
where K is the total kinetic energy of the system of particles, K' is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and 12MV2 is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system).

HARD
11th Jharkhand Board
IMPORTANT

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

Show L=L'+R×MV
where L'=Σri'×pi' is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember ri'=ri-Rri is the position of ith particle with respect to origin and R and V is the position and velocity of centre of mass with respect to origin, respectively.

Note, L' and R×MV can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

HARD
11th Jharkhand Board
IMPORTANT

Separation of Motion of a system of particles into motion of the center of mass and motion about the center of mass:

Show dL'dt=Σri'×dp'dt
Further, show that
dL'dt=τext'
where τext' is the sum of all external torques acting on the system about the centre of mass. (Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)