Question

A body of mass \( m \) and radius \( r \) rolling horizontally with velocity \( V \), rolls up an inclined plane to a vertical height \( \frac{V^2}{g} \). The body is: 

• A. a sphere
• B. a circular disc
• C. a circular ring
• D. a solid cylinder

Hint:

For rolling motion on an incline, use conservation of energy and the moment of inertia formula to determine how height relates to velocity.

Answer 1

The Correct Option is C

Step 1: Apply Energy Conservation Using the conservation of mechanical energy: \[ \text{Initial Energy} = \text{Final Energy} \] \[ \frac{1}{2} m V^2 + \frac{1}{2} I \omega^2 = mg h \] Since the body is rolling without slipping: \[ \omega = \frac{V}{r} \] The moment of inertia of a rolling object determines how much kinetic energy is converted into potential energy.  

Step 2: Finding the Height \[ \frac{1}{2} m V^2 + \frac{1}{2} \left( I \frac{V^2}{r^2} \right) = mg h \] \[ \frac{1}{2} m V^2 \left( 1 + \frac{I}{m r^2} \right) = mg h \] Solving for \( h \): \[ h = \frac{V^2}{g} \times \frac{1}{1 + \frac{I}{m r^2}} \] For a circular ring, \( I = m r^2 \), so: \[ h = \frac{V^2}{g} \times \frac{1}{1 + 1} = \frac{V^2}{2g} \] The problem states that \( h = \frac{V^2}{g} \), which matches the condition for a circular ring. Thus, the body is a circular ring.