Question

A disc is rolling on horizontal surface. If disc rolls without slipping, then the ratio of translational K.E. to total K.E. will be

• A. $1/3$
• B. $2/3$
• C. $4/9$
• D. $1/2$

Answer 1

The correct option is B $2/3$


$K.E_{rotation}=\frac{1}{2}I{\omega }^2(\omega =V/R)$
$K.E_{trans}=\frac{1}{2}M{V}^2=K$
$K.E_{rotation}=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\frac{M{R}^2}{2}\left(\frac{V^2}{R^2}\right)=\frac{K}{2}$
$K.E_{total}=K.E_{trans}+K.E_{rot}$
$=K+K/2=\frac{3K}{2}$
$\frac{K.E_{trans}}{K.E_{total}}=\frac{K}{3K/2}=2/3$