Question

A solid cylinder and a hollow cylinder of the same mass and radius, both initially at rest, roll down the same inclined plane without slipping. Choose the correct options.

• A. The hollow cylinder reaches the bottom first
• B. The solid cylinder reaches the bottom first
• C. The two cylinder reach the bottom together
• D. Their kinetic energies at the bottom are equal.

Answer 1

Answer: (B), (D)

The solid cylinder reaches the bottom first
Their kinetic energies at the bottom are equal.

Their kinetic energies at the bottom will be $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}km{r}^2\frac{v^2}{r^2}=\frac{1}{2}m{v}^2(1+k)$ The velocity would be greater for the material with radius of gyration $k$ is lower of for which the moment of inertia is lower. Solid cylinder with $k$ as $1/2$ will have greater velocity as compared to hollow cylinder which has $k$ as $1$. As as they have same mass and radius their energies at both the levels would remains same including the potential energy and energy lost due to friction. Thus both option B and D are correct.