Question

an body rolls down a smooth inclined plane if its rotational ke is 50%of its translation KE the body must be​

Answer 1

Answer:

The body might be solid cylinder or circular disc.

Explanation:

Let linear velocity = v

     radius of the body = r

     angular velocity = w

     Rotational Kinetic Energy = (1/2) I w^2

     Translational Kinetic Energy = (1/2) m v^2

Given, Rotational Kinetic Energy = 50% of Translational Kinetic Energy

          (1/2) I w^2 = (1/2) * (1/2) * m v^2

          I w^2 = (1/2) * m v^2

          I (v/r)^2 = (1/2) * m v^2

          I = (m r^2)/2 which is the Inertia of solid cylinder or circular disc