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

Step 1: Understanding the Concept:
When an object rolls up an incline without slipping, its total initial kinetic energy is converted into gravitational potential energy at the maximum height.
The total kinetic energy consists of both translational and rotational components.
Step 2: Key Formula or Approach:
Using the conservation of mechanical energy:
\[ KE_{\text{trans}} + KE_{\text{rot}} = PE_{\text{final}} \]
\[ \frac{1}{2} m v_0^2 + \frac{1}{2} I_{\text{cm}} \omega^2 = mgh \]
For pure rolling, we use the constraint $v_0 = R\omega \implies \omega = \frac{v_0}{R}$.
Step 3: Detailed Explanation:
Given the maximum height is $h = \frac{7v_0^2}{10g}$.
Substitute $h$ and $\omega$ into the energy equation:
\[ \frac{1}{2} m v_0^2 + \frac{1}{2} I_{\text{cm}} \left(\frac{v_0}{R}\right)^2 = mg \left( \frac{7v_0^2}{10g} \right) \]
Cancel out the common terms to simplify:
\[ \frac{1}{2} m v_0^2 + \frac{I_{\text{cm}} v_0^2}{2 R^2} = \frac{7}{10} m v_0^2 \]
Divide the entire equation by $v_0^2$:
\[ \frac{1}{2} m + \frac{I_{\text{cm}}}{2 R^2} = \frac{7}{10} m \]
Rearrange to solve for $I_{\text{cm}}$:
\[ \frac{I_{\text{cm}}}{2 R^2} = \frac{7}{10} m - \frac{1}{2} m = \frac{7}{10} m - \frac{5}{10} m = \frac{2}{10} m = \frac{1}{5} m \]
Multiply by $2R^2$:
\[ I_{\text{cm}} = 2 \times \frac{1}{5} m R^2 = \frac{2}{5} m R^2 \]
The moment of inertia $I_{\text{cm}} = \frac{2}{5}mR^2$ corresponds exclusively to a uniform solid sphere.
Step 4: Final Answer:
The rolling object is a solid sphere.