Question

A solid sphere with mass, M, and radius, R, rolls along a level surface without slipping with a linear speed, v. What is the ratio of rotational to linear kinetic energy

Answer 1

Answer:

$\dfrac{KE_R}{KE_L}=\dfrac{2}{5}$

Explanation:

given,

mass of solid sphere = M

radius = R

linear speed = v

now,

Linear Kinetic energy

$KE_L = \dfrac{1}{2}mv^2$...............(1)

Rotational kinetic energy

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

moment of inertia of the solid sphere

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

and v = R ω

$KE_R= \dfrac{1}{2}(\dfrac{2}{5}MR^2)(\dfrac{v}{R})^2$

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

ration of rotational and Kinetic energy

$\dfrac{KE_R}{KE_L}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{1}{2}MV^2}$

$\dfrac{KE_R}{KE_L}=\dfrac{2}{5}$

hence, ratio of rotational to linear kinetic energy is equal to 2/5.