Question

A solid sphere and a solid cylinder of the same mass are rolled down two inclined planes of heights \(h_1\) and \(h_2\). If at the bottom of the planes the two objects have the same linear velocity, then the ratio \(h_1:h_2\) is:

• A. \(2:3\)
• B. \(7:5\)
• C. \(14:15\)
• D. \(15:14\)

Hint:

For rolling motion:
Always include both translational and rotational KE
Larger moment of inertia \(\Rightarrow\) more energy in rotation
Same final speed does NOT imply same height

Answer 1

The Correct Option is C

Step 1: Use conservation of mechanical energy. For a body rolling without slipping: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Step 2: Solid sphere. Moment of inertia: \[ I_{\text{sphere}} = \frac{2}{5}mr^2 \] Using \(\omega = \frac{v}{r}\): \[ mgh_1 = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mr^2\cdot\frac{v^2}{r^2} \] \[ mgh_1 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \] \[ h_1 = \frac{7v^2}{10g} \]
Step 3: Solid cylinder. Moment of inertia: \[ I_{\text{cylinder}} = \frac{1}{2}mr^2 \] \[ mgh_2 = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{1}{2}mr^2\cdot\frac{v^2}{r^2} \] \[ mgh_2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \] \[ h_2 = \frac{3v^2}{4g} \]
Step 4: Find the ratio \(h_1:h_2\). \[ \frac{h_1}{h_2} = \frac{\tfrac{7}{10}}{\tfrac{3}{4}} = \frac{7}{10}\times\frac{4}{3} = \frac{14}{15} \] Final Answer: \[ \boxed{h_1:h_2 = 14:15} \]