Question:
A body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \%$ of its translational kinetic energy. The body is :
Correct Option: , 2
Solution:
$\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \times \frac{1}{2} \mathrm{mv}^{2}$
$I=\frac{1}{2} m R^{2}$
Body is solid cylinder
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