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 \( 7 \).

Answer 2

Step 1: Given information.
A solid sphere and a hollow cylinder roll up the same inclined plane without slipping, both having the same initial speed \( v \).
The solid sphere reaches a maximum height \( h_1 \), and the hollow cylinder reaches a maximum height \( h_2 \).
We need to find the ratio \( h_1 : h_2 = \frac{n}{10} \).

Step 2: Concept – Energy conservation.
When the body rolls up the incline without slipping, total mechanical energy is conserved.
At the base (initial point):
\[ \text{Total energy} = \text{Translational KE} + \text{Rotational KE} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] At the maximum height, the body momentarily comes to rest, so all kinetic energy converts into potential energy:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Step 3: Relation between linear and angular speed.
Since rolling without slipping, \( v = \omega R \).
Substitute \( \omega = \frac{v}{R} \):
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v^2}{R^2}\right) \] \[ h = \frac{v^2}{2g}\left(1 + \frac{I}{mR^2}\right) \]
Step 4: Moment of inertia for each body.
For a solid sphere: \( I = \frac{2}{5}mR^2 \)
\[ h_1 = \frac{v^2}{2g}\left(1 + \frac{2}{5}\right) = \frac{v^2}{2g} \times \frac{7}{5} \] For a hollow cylinder: \( I = mR^2 \)
\[ h_2 = \frac{v^2}{2g}\left(1 + 1\right) = \frac{v^2}{2g} \times 2 \]
Step 5: Ratio of heights.
\[ \frac{h_1}{h_2} = \frac{\frac{7}{5}}{2} = \frac{7}{10} \] Hence, \( h_1 : h_2 = \frac{7}{10} \).

Step 6: Final Answer.
\[ \boxed{n = 7} \]