Question

A cylinder with its mass concentrated toward the center has a moment of inertia of 0.1 MR2 . If this cylinder is rolling without slipping along a level surface with a linear speed v, what is the ratio of its rotational kinetic energy to its linear kinetic energy?

Answer 1

Answer:

$\dfrac{K_r}{K}=0.1$

Explanation:

M = Mass of cylinder

R = Radius

Moment of inertia is given by

$I=0.1MR^2$

Velocity is given by

$v=R\omega\\\Rightarrow \omega=\dfrac{v}{R}$

Rotational kinetic energy is given by

$K_r=\dfrac{1}{2}I\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times 0.1MR^2\dfrac{v^2}{R^2}\\\Rightarrow K_r=\dfrac{1}{2}0.1Mv^2$

The linear kinetic energy is given by

$K=\dfrac{1}{2}Mv^2$

The ratio would be

$\dfrac{K_r}{K}=\dfrac{\dfrac{1}{2}0.1Mv^2}{\dfrac{1}{2}Mv^2}\\\Rightarrow \dfrac{K_r}{K}=0.1$

The ratio of the kinetic energies is $\dfrac{K_r}{K}=0.1$

Answer 2

Answer:

• Ratio of rotational kinetic energy to its linear kinetic energy = $1:10$

Explanation:

Given

moment of inertia,

$I = 0.1 MR^2$

Rotational kinetic energy,

$Rot KE = \frac{1}{2}*Iw^2$

where,  $v = Rw$

$Rot KE = 0.5 * 0.1 MR^2 * w^2\\\\Rot KE = 0.05 *Mv^2$

$linear KE = \frac{1}{2} Mv^2\\\\linear KE = 0.5Mv^2$

therefore ratio,

$= 1:10$

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