Question

If the rolling object reaches a maximum height \(\frac{7v_0^2}{10g}\), what may be the object? 

• A. Solid sphere
• B. Ring
• C. Disc
• D. Hollow sphere

Hint:

Common moments of inertia for rolling objects: \[ \text{Ring} = mR^2 \] \[ \text{Disc} = \frac12 mR^2 \] \[ \text{Solid sphere} = \frac{2}{5}mR^2 \] \[ \text{Hollow sphere} = \frac{2}{3}mR^2 \]

Answer 1

The Correct Option is A

Concept: For rolling motion without slipping, total kinetic energy is \[ K = \frac12 mv^2 + \frac12 I_{cm}\omega^2 \] Using rolling condition \[ \omega = \frac{v}{R} \] At maximum height all kinetic energy converts into potential energy. \[ \frac12 mv_0^2 + \frac12 I_{cm}\left(\frac{v_0}{R}\right)^2 = mgh \]
Step 1: Substitute given maximum height. \[ h = \frac{7v_0^2}{10g} \] Thus, \[ \frac12 mv_0^2 + \frac12 I_{cm}\frac{v_0^2}{R^2} = mg\frac{7v_0^2}{10g} \] \[ \frac12 m + \frac12 \frac{I_{cm}}{R^2} = \frac{7}{10}m \]
Step 2: Solve for moment of inertia. \[ \frac12 m + \frac12\frac{I_{cm}}{R^2} = 0.7m \] \[ \frac12\frac{I_{cm}}{R^2} = 0.2m \] \[ \frac{I_{cm}}{R^2} = 0.4m \] \[ I_{cm} = \frac{2}{5}mR^2 \]
Step 3: Identify the object. Moment of inertia \[ I_{cm} = \frac{2}{5}mR^2 \] This corresponds to a solid sphere. \[ \boxed{\text{Object = Solid Sphere}} \]