Question

A solid sphere of mass 2 kg and radius 0.5 m rolls without slipping down an inclined plane of height 3 m. What is its speed at the bottom? (Take \( g = 9.8\, \text{m/s}^2 \))

• A. 6.48
• B. 8.40
• C. 10
• D. 9.64

Hint:

For rolling without slipping, total KE is $\frac{7}{10}mv^2$. Use $v = \sqrt{\frac{10gh}{7}}$ for a solid sphere.

Answer 1

The Correct Option is A

For a solid sphere undergoing rolling without slipping, total mechanical energy remains constant: \[ \text{Potential Energy} = \text{Translational KE} + \text{Rotational KE} \] At the apex, total energy is given by: \[ E = mgh \] At the base, the total energy is: \[ E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For a solid sphere, the moment of inertia is \( I = \frac{2}{5}mr^2 \) and the angular velocity is \( \omega = \frac{v}{r} \). Substituting these values: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \] This simplifies to: \[ mgh = \frac{7}{10}mv^2 \quad \Rightarrow \quad v^2 = \frac{10gh}{7} \] Using the values \( g = 9.8 \, \text{m/s}^2 \) and \( h = 3 \): \[ v^2 = \frac{10 \cdot 9.8 \cdot 3}{7} = \frac{294}{7} = 42 \quad \Rightarrow \quad v = \sqrt{42} \approx 6.48 \, \text{m/s} \] The final velocity calculation is: \[ v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10 \cdot 9.8 \cdot 3}{7}} = \sqrt{42} \approx 6.48 \]