Question

The cylinder shown, with mass $M$ and radius $R$, has a radially dependent density. The cylinder starts from rest and rolls without slipping down an inclined plane of height $H$.

At the bottom of the plane its translational speed is $\sqrt{\frac{8gH}{7}}$. Which of the following is the rotational inertia of the cylinder?

• A. MR^2
• B. \frac{2}{5}MR^2
• C. \frac{3}{4}MR^2
• D. \frac{1}{2}MR^2

Answer 1

Answer: (C)

The rotational inertia of the cylinder is $\frac{3}{4}MR^2$.

Answer 2

The problem can be solved using the principle of conservation of energy. The initial energy of the cylinder at height $H$ is purely potential energy, $E_i = MgH$. At the bottom of the incline, this energy is converted into translational kinetic energy and rotational kinetic energy. The total final energy is $E_f = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.

For rolling without slipping, the relationship between translational velocity $v$ and angular velocity $\omega$ is $v = R\omega$, which means $\omega = \frac{v}{R}$.

Applying conservation of energy, $E_i = E_f$: $MgH = \frac{1}{2}Mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2$

We are given that $v = \sqrt{\frac{8gH}{7}}$, so $v^2 = \frac{8gH}{7}$. Substituting this into the energy equation: $MgH = \frac{1}{2}M\left(\frac{8gH}{7}\right) + \frac{1}{2}I\left(\frac{8gH}{7R^2}\right)$ $MgH = \frac{4MgH}{7} + \frac{4IgH}{7R^2}$

Dividing the entire equation by $gH$: $M = \frac{4M}{7} + \frac{4I}{7R^2}$

To solve for $I$, rearrange the equation: $M - \frac{4M}{7} = \frac{4I}{7R^2}$ $\frac{3M}{7} = \frac{4I}{7R^2}$

Multiply both sides by $7R^2$: $3MR^2 = 4I$

Therefore, the rotational inertia $I$ is: $I = \frac{3}{4}MR^2$