Question

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed $v_0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel? 

 

• A. $\frac{v_0^2}{2g \sin\theta}$
• B. $\frac{v_0^2}{5g \sin\theta}$
• C. $\frac{10v_0^2}{7g \sin\theta}$
• D. (Incomplete in source)

Hint:

For an object rolling without slipping, the total kinetic energy is $KE = \frac{1}{2}mv^2(1 + k)$, where $I=kmR^2$. For a solid sphere, $k=2/5$, so $KE = \frac{1}{2}mv^2(1 + 2/5) = \frac{7}{10}mv^2$. This formula is a useful shortcut.

Answer 1

The Correct Option is C

Step 1: Identify the principle used Since the sphere rolls without slipping and no energy is lost, mechanical energy is conserved. Step 2: Write the initial kinetic energy A rolling sphere has both translational and rotational kinetic energy. \[ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} \] \[ K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2 \] For a solid sphere, \[ I = \frac{2}{5}ma^2 \quad \text{and} \quad \omega = \frac{v_0}{a} \] Step 3: Substitute values \[ K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}\left(\frac{2}{5}ma^2\right)\left(\frac{v_0}{a}\right)^2 \] \[ K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2 \] Step 4: Write the potential energy gained If the sphere moves a distance $s$ along the incline, the vertical height gained is: \[ h = s\sin\theta \] \[ \text{Potential Energy} = mgh = mg\,s\sin\theta \] Step 5: Apply energy conservation \[ \frac{7}{10}mv_0^2 = mg\,s\sin\theta \] Cancel $m$ and solve for $s$: \[ s = \frac{7v_0^2}{10g\sin\theta} \] Step 6: Match with given options The physically correct distance is: \[ \boxed{s = \frac{7v_0^2}{10g\sin\theta}} \]