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?.

Answer 1

The ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

What is the rotational kinetic energy of a sphere?

The rotational kinetic energy of the sphere is directly proportional to the square of angular acceleration.

$K = \dfrac 12 I\omega ^2$

Where,

$I$ = rotational inertia of sphere = 2/5MR²

Where $M$ is the mass of the sphere and $R$ is the radius of the sphere,

ω = angular acceleration of sphere = V/R

Where $V$ is the speed of the sphere.

So,

K = 1/2 × 2/5MR² × V²/R²

K = MV²/5

The translational kinetic energy of the sphere,

$K' = \dfrac 12 MV^2$

The total kinetic energy of the sphere,

$K_t = K + K'$

So,

$K_t= \dfrac {MV^2}5 +\dfrac { MV^2}2\\\\K_t= \dfrac {7MV^2}{10}$

Thus the Ratio of rotational kinetic energy to total kinetic energy,

$\dfrac KK_t = \dfrac {\dfrac {MV^2}{5} }{ \dfrac {7MV^2}{10}}\\\\\dfrac KK_t = \dfrac 15 \times \dfrac { 10}7\\\\\dfrac KK_t= \dfrac 27$

Therefore, the ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

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