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 $B$
$mg\frac{h}{2}=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{m r^{2}}{2}\frac{v^{2}}{r^{2}}=\frac{3}{4}mv^{2}$
$v=\sqrt{\frac{2 g h}{3}}$
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 $C$
$mg\frac{h}{2}+\frac{1}{2}mv^{2}=\frac{1}{2}mv_{C}^{2}$
$\frac{3}{4}mv^{2}+\frac{1}{2}mv^{2}=\frac{1}{2}mv_{C}^{2}$
$v_{c}=\sqrt{\frac{5}{2}}v$
Now
Kinetic Energy at $C$
$KE=\frac{1}{2}mv_{C}^{2}=\frac{5}{4}mv^{2}$
Rotational Kinetic Energy at $C$
$RE=\frac{1}{2}I\omega ^{2}=\frac{1}{2}\frac{m r^{2}}{2}\frac{v^{2}}{r^{2}}=\frac{1}{4}mv^{2}$
Ratio of translational kinetic energy to rotational kinetic energy
$\Rightarrow \frac{K E}{R E}=\frac{\frac{5}{4}mv^{2}}{\frac{1}{4}mv^{2}}=5$