Question

Upper half of an incline plane is rough and lower half is smooth. A cylinder starts its motion from the top. What is the ratio of its translational kinetic energy and rotational kinetic energy at the bottom ?

• A. 2 : 1
• B. 3 : 1
• C. 4 : 1
• D. 5 : 1

Answer 1

The correct option is D 5 : 1

In pure rolling motion of a cylinder, ratio of its translational kinetic energy and total kinetic energy is $\frac{2}{3}$.
At half way,
$K_t=\frac{2}{3}\times \frac{mgh}{2}=\frac{1}{3}mgh$
$K_r=\frac{1}{3}\times \frac{mgh}{2}=\frac{1}{6}mgh$
At bottom,
$K_t^{{}^{\prime }}=\frac{1}{3}mgh+\frac{1}{2}mgh=\frac{5}{6}mgh$
In smooth part, rotational kinetic energy remains same.
$K_r^{{}^{\prime }}=\frac{1}{6}mgh$
Thus, $\frac{K_t^{{}^{\prime }}}{K_r^{{}^{\prime }}}=5$