Kevin S. Huang

2014: Problem 14

Topic: Rigid Bodies
Concepts: Kinetic energy, Rotational Newton’s 2nd law

Solution:

By Newton’s 2nd law, we have

T=Ma

\tau=TR=I\alpha

so $a=T/M$ and $\alpha=TR/I$ . The kinetic energies are given by

K_{\text{trans}}=\frac{1}{2}Mv^2=\frac{1}{2}M(at)^2

K_{\text{rot}}=\frac{1}{2}I\omega^2=\frac{1}{2}I(\alpha t)^2

Then by canceling out common factors,

\frac{K_{\text{trans}}}{K_{\text{tot}}}=\frac{K_{\text{trans}}}{K_{\text{trans}}+K_{\text{rot}}}=\frac{Ma^2}{Ma^2+I\alpha^2}=\frac{M(1/M)^2}{M(1/M)^2+I(R/I)^2}=\frac{\frac{1}{M}}{\frac{1}{M}+\frac{R^2}{I}}

Multiplying the numerator and denominator by $MI$ ,

\frac{K_{\text{trans}}}{K_{\text{tot}}}=\frac{I}{I+MR^2}

so the answer is D.