Test: Angular Momentum - Civil Engineering (CE) MCQ

Acceleration of the rolling sphere,
a = gsinθ/((1+k²/R²))
Velocity of the sphere at the bottom of the inclined plane,
v = √(2gh/((1+k²/R²)))
The sphere will reach the bottom with the same speed v because h is the same in both the cases.

The force involved in the given problem is the internal forces of the system. No external force acts on the system when the child runs So, there will be no change in the speed of the centre of mass of the (trolley+child) system.

v_cm = 20cm/s = 0.20ms
Work required to stop the hoop = Rotational kinetic energy+Traslational kinetic energy
Work required = 

Acceleration of the rolling sphere,
a = gsinθ/((1+k²/R²))
Velocity of the sphere at the bottom of the inclined plane,
v = √(2gh/((1+k²/R²)))
The sphere will take a longer time to roll down one plane than the other. It will take a larger time in case if the plane with smaller inclination because the acceleration is proportional to sinθ.

No external force is acting on the centre of mass. It remains at rest. The speed of centre of mass is zero.

The centre of mass of two particles of equal masses lies midway between them. Its position vector is the average of the position vectors of the two particles.

A circular wire behaves like a ring. By perpendicular axes theorem,
I_D + I_D = 1/2 MR²
Therefore,
1/4 MR².

Only their linear velocities are exchanged. The two spheres cannot exert torques on each other, as their surfaces are frictionless, and so that angular velocities of the sphere do not change.

When a body rolls, the force if the friction acts in the same direction as the direction of motion of the centre of mass of the body.

In a hollow sphere, the mass is distributed away from the axis of rotation. So, its moment of inertia is greater than that of a solid sphere.