Question

11.4.1. Cylinders A and B are identical. Cylinder A is allowed to roll down a ramp without slipping. Cylinder B is allowed to slide down a similar ramp that has the same incline angle, but it is frictionless the following statements concerning this situation, assuming the cylinders begin from rest at the same height, is true? . Which one of a) The sum of the translational and rotational kinetic energies of cylinder A is smaller than the translational kinetic energy of cylinder B b) The sum of the translational and rotational kinetic energies of cylinder A is equal to the translational kinetic energy of cylinder B c) The sum of the translational and rotational kinetic energies of cylinder A is larger than the translational kinetic energy of cylinder B.

Please expalins about all options

answer is B C A

Answer 1

11.4.1

Correct answer: B)

In this case both the cylinders are identical. But the cylinder A is rolling down the ramp without slipping which means there are two parts of kinetic energy. They are linear and rotational kinetic energy. In this context, the term rolling without slipping is important which means the linear velocity at the center of mass is equal to the product of angular velocity and radius of the cylinder. Now while rotating the cylinder moves from one place to another which is the example of translatory motion again there is rotation which provides the rotational kinetic energy. Considering the center of mass we get total kinetic energy= translatory kinetic energy + rotational kinetic energy.

In case of Cylinder B it is sliding down without friction that means the absence of rotation but movement from one place to another. So the only kinetic energy is translational. As both the cylinder are identical (same radius, height) therefore we can say that the total kinetic energy of A is equal to the kinetic energy of B.

11.4.2

Correct answer: C)

As the cylinder is freely rolling down the inclined plane the static friction as a part of gravitational force tries to resist the movement. That means the inclined plane helps gravity to create a linear motion for the solid cylider. Again, the friction is reason behind the rotational motion (creating the essential torque) which means its upward direction at the point of contact helps the rotational motion while joining the same with linear motion.

11.4.3

Correct answer: A)

In case of Yo-Yo the angular accleration depends on the net torque which is the product of force and distance (perpendicular) from the rotating axis. Considering the Newton's second law F= M.a we can write $\alpha =\tau$ / I. Where $\alpha$ is the angular acceleration. Similarly the rolling body on an inclined plane has moment of intertia I= nMr^2 where n is the number of rotation. Again, only rolling means the torque is due to gravitational pull so total torque $\tau$ = I $\alpha$ or the acceleration $\alpha$ = $\tau$ /I. The equations are same for Yo-Yo and rolling body on an inclined plane.