Question

A solid sphere and a hollow cylinder roll up without slipping on the same inclined plane with the same initial speed \( v \). The sphere and the cylinder reach up to maximum heights \( h_1 \) and \( h_2 \), respectively, above the initial level. The ratio \( h_1 : h_2 \) is \( \frac{n}{10} \). The value of \( n \) is ______.

Answer 1

Correct Answer: 7

Conservation of Energy:
When the solid sphere and hollow cylinder roll up the incline, their initial kinetic energy (both translational and rotational) is converted into gravitational potential energy at the maximum height.

Total Initial Kinetic Energy for Each Object:
 For the solid sphere:
\[ \text{Total KE} = \frac{1}{2}mv^2 + \frac{1}{2}I_{\text{sphere}}\omega^2 \]
where \( I_{\text{sphere}} = \frac{2}{5}mR^2 \) and \( \omega = \frac{v}{R} \).

\[ \text{Total KE}_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2} \times \frac{2}{5}mR^2 \times \frac{v^2}{R^2} \]

\[ = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]

For the hollow cylinder:
\[ \text{Total KE} = \frac{1}{2}mv^2 + \frac{1}{2}I_{\text{cylinder}}\omega^2 \]
where \( I_{\text{cylinder}} = mR^2 \).

\[ \text{Total KE}_{\text{cylinder}} = \frac{1}{2}mv^2 + \frac{1}{2} \times mR^2 \times \frac{v^2}{R^2} \]

\[ = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2 \]

Potential Energy at Maximum Height:
At maximum height, all kinetic energy is converted to potential energy:

For the solid sphere:
\[ mgh_1 = \frac{7}{10}mv^2 \] \[ h_1 = \frac{7v^2}{10g} \]

For the hollow cylinder:
\[ mgh_2 = mv^2 \] \[ h_2 = \frac{v^2}{g} \]

Calculate the Ratio \( \frac{h_1}{h_2} \):
\[ \frac{h_1}{h_2} = \frac{\frac{7v^2}{10g}}{\frac{v^2}{g}} = \frac{7}{10} \]

Conclusion:
The ratio \( h_1 : h_2 \) is \( \frac{7}{10} \), so the value of \( n \) is \( 7 \).