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 rolling without slipping, total mechanical energy is conserved: \[ \text{Potential Energy} = \text{Translational KE} + \text{Rotational KE} \] Total energy at top: \[ E = mgh \] At the bottom: \[ E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For a solid sphere: \[ I = \frac{2}{5}mr^2, \quad \omega = \frac{v}{r} \] Substitute values: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \] \[ mgh = \frac{7}{10}mv^2 \quad \Rightarrow \quad v^2 = \frac{10gh}{7} \] Substitute \( g = 9.8 \, \text{m/s}^2, h = 3 \): \[ v^2 = \frac{10 \cdot 9.8 \cdot 3}{7} = \frac{294}{7} = 42 \quad \Rightarrow \quad v = \sqrt{42} \approx 6.48 \, \text{m/s} \] \[ v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10 \cdot 9.8 \cdot 3}{7}} = \sqrt{42} \approx 6.48 \]

Answer 2

To solve the problem, we need to find the speed of a solid sphere rolling without slipping down an inclined plane of height 3 m.

1. Given Data: 
Mass of sphere, $m = 2\, kg$
Radius of sphere, $r = 0.5\, m$
Height of incline, $h = 3\, m$
Acceleration due to gravity, $g = 9.8\, m/s^2$

2. Using Energy Conservation Principle:
Potential energy at top = Kinetic energy at bottom (translational + rotational)
$ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 $

3. Moment of Inertia and Rolling Condition:
For a solid sphere, $I = \frac{2}{5} mr^2$
Rolling without slipping: $v = \omega r \Rightarrow \omega = \frac{v}{r}$

4. Substitute and Simplify:
$ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} mr^2 \times \left(\frac{v}{r}\right)^2 $
$ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 $
$ mgh = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 = \frac{7}{10} mv^2 $

5. Solve for $v$:
$ gh = \frac{7}{10} v^2 \Rightarrow v^2 = \frac{10}{7} gh $
$ v = \sqrt{\frac{10}{7} \times 9.8 \times 3} $

6. Calculate Numerical Value:
$ v = \sqrt{\frac{10}{7} \times 29.4} = \sqrt{42} = 6.48\, m/s $

Final Answer:
The speed of the sphere at the bottom is $ {6.48\, m/s} $.