A solid cylinder rolls without slipping. The ratio of translational KE to the total KE is:
Solution
$I = \frac{1}{2}MR^2$. $KE_{trans} = \frac{1}{2}Mv^2$, $KE_{rot} = \frac{1}{4}Mv^2$. Ratio = $\frac{1/2}{3/4} = 2/3$.
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Rolling Motion
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