Question

A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy?
(A) 3/7
(B) None of these
(C) 7/2
(D) 2/5
(E) 2/7
(F) 3/5
(G) 5/3

Answer 1

Answer:

option E

Explanation:

given,

Rotational Kinetic Energy, KE_r = $\dfrac{1}{2}I\omega^2$

Moment of inertia of the solid,

$I = \dfrac{2}{5}MR^2$

$\omega = \dfrac{V}{R}$

now,

$KE_r = \dfrac{1}{2}I\omega^2$

$KE_r = \dfrac{1}{2}\times \dfrac{2}{5}MR^2 (\dfrac{V}{R})^2$

$KE_r =\dfrac{1}{5}MV^2$......(1)

transnational kinetic energy

$KE_t =\dfrac{1}{2}MV^2$

Total kinetic energy

$KE = \dfrac{1}{2}MV^2 + \dfrac{1}{5} MV^2$

$KE = \dfrac{7}{10}MV^2$

ratio of rotational kinetic energy to the total kinetic energy

$\dfrac{KE_r}{KE_t}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{7}{10}MV^2}$

$\dfrac{KE_r}{KE_t}=\dfrac{2}{7}$

hence, the correct answer is option E