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 :

Answer:

[tex]\dfrac{KE_R}{KE_L}=\dfrac{2}{5}[/tex]

Explanation:

given,

mass of solid sphere = M

radius = R

linear speed = v

now,

Linear Kinetic energy

[tex] KE_L = \dfrac{1}{2}mv^2[/tex]...............(1)

Rotational kinetic energy

[tex] KE_R= \dfrac{1}{2}I\omega^2[/tex]

moment of inertia of the solid sphere

[tex]I = \dfrac{2}{5}MR^2[/tex]

and v = R ω

[tex] KE_R= \dfrac{1}{2}(\dfrac{2}{5}MR^2)(\dfrac{v}{R})^2[/tex]

[tex]KE_R =\dfrac{1}{5}MV^2[/tex]............(2)

ration of rotational and Kinetic energy

[tex]\dfrac{KE_R}{KE_L}=\dfrac{\dfrac{1}{5}MV^2}{\dfrac{1}{2}MV^2}[/tex]

[tex]\dfrac{KE_R}{KE_L}=\dfrac{2}{5}[/tex]

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