Question

Portion $AB$ of the wedge shown in figure is rough and $BC$ is smooth. A solid cylinder rolls without slipping from $A$ to $B$. Find the ratio of translational kinetic energy to rotational kinetic energy when the cylinder reaches point $C$.

Answer 1

When cylinder will reach the point B, it will have both translational and rotational kinetic energy

Let $\angle DAB=\theta$
Then, according to conservation of energy:

$mgABcos\theta =\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$

So rotational energy at B, $K_B=mgABcos\theta -\frac{1}{2}m{v}^2$
$I=\frac{1}{2}m{R}^2$

We get:

$mgABcos\theta =\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)(\frac{v}{R}{)}^2$
or, $mgABcos\theta =\frac{3}{4}m{v}^2$
or, $\frac{1}{2}m{v}^2=\frac{2}{3}mgABcos\theta \to \left(1\right)$

And on reaching the point c, there will be no friction through B to C so there will be no torque so rotational energy will be same as that at point B but potential energy loss will be equal to gain in translational kinetic energy.

So, $mgBCcos\theta =\frac{1}{2}m{v}^{\prime 2}\to \left(2\right)$

Translational kinetic energy,at C, $K_C=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^{\prime 2}$

Now, $K=\frac{K_C}{K_B}$
or, $K=\frac{\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^{\prime 2}}{mgABcos\theta -\frac{1}{2}m{v}^2}\to \left(3\right)$

From equations 1,2 and 3, we get:
$K=\frac{\frac{2}{3}mgABcos\theta +mgBCcos\theta }{mgABcos\theta -\frac{2}{3}mgABcos\theta }$ [$\because AB=BC$]

So, on solving above equation, $K=5$