Question

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

• A. $\frac{3}{4}$
• B. $5$
• C. $\frac{7}{5}$
• D. $\frac{8}{3}$

Answer 1

Answer: (B)

As the cylinder rolls without slipping from $A$ to $B$ ,
$v=\omega r$
Now using principle of conservation of energy for points $A$ and $B$
$mg\frac{h}{2}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{r}^2}{2}\frac{v^2}{r^2}=\frac{3}{4}m{v}^2$
$v=\sqrt{\frac{2gh}{3}}$
As $BC$ is smooth there is no change in angular velocity from $B$ to $C$ . Using principle of conservation of energy for points $B$ and $C$
$mg\frac{h}{2}+\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_C^2$
$\frac{3}{4}m{v}^2+\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_C^2$
$v_c=\sqrt{\frac{5}{2}}v$
Now
Kinetic Energy at $C$
$KE=\frac{1}{2}m{v}_C^2=\frac{5}{4}m{v}^2$
Rotational Kinetic Energy at $C$
$RE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{m{r}^2}{2}\frac{v^2}{r^2}=\frac{1}{4}m{v}^2$
Ratio of translational kinetic energy to rotational kinetic energy
$\Rightarrow \frac{KE}{RE}=\frac{\frac{5}{4}m{v}^2}{\frac{1}{4}m{v}^2}=5$