Question

A solid sphere rolls down without slipping from the top of an inclined plane of height 28 m and angle of inclination 30°. The velocity of the sphere when it reaches the bottom of the plane is (Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \))

• A. 20 m/s
• B. 28 m/s
• C. 10 m/s
• D. 14 m/s

Hint:

For rolling motion, use energy conservation considering both translational and rotational kinetic energy.

Answer 1

The Correct Option is A

Let's solve the problem step by step. Given: Height of the inclined plane, $h = 28$ m Angle of inclination, $\theta = 30^\circ$ Acceleration due to gravity, $g = 10$ m/s$^2$ The sphere rolls without slipping. The velocity of a rolling solid sphere at the bottom of an inclined plane is given by: $v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^2}}}$ where $I$ is the moment of inertia of the solid sphere, $m$ is its mass, and $r$ is its radius. For a solid sphere, the moment of inertia is $I = \frac{2}{5} mr^2$. So, $\frac{I}{mr^2} = \frac{\frac{2}{5} mr^2}{mr^2} = \frac{2}{5}$. Now, substitute the values into the velocity equation: $v = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{2gh}{\frac{7}{5}}} = \sqrt{\frac{10gh}{7}}$ Plug in the given values: $v = \sqrt{\frac{10 \times 10 \times 28}{7}} = \sqrt{\frac{2800}{7}} = \sqrt{400} = 20$ m/s Therefore, the velocity of the sphere when it reaches the bottom of the plane is 20 m/s. Final Answer: The final answer is $\boxed{(1)}$