Question

A small object of uniform density rolls up a curved surface with an initial velocity \( v \). It reaches upto a maximum height of \( \frac{7v^2}{10g} \) with respect to initial position. Then the object is

• A. ring
• B. solid sphere
• C. hollow sphere
• D. disc

Hint:

Always remember standard \( \frac{I}{mR^2} \) values: Ring = 1, Disc = \( \frac{1}{2} \), Solid sphere = \( \frac{2}{5} \), Hollow sphere = \( \frac{2}{3} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When an object rolls without slipping, it possesses both translational and rotational kinetic energy. As it climbs a height, this total kinetic energy is converted into gravitational potential energy. The specific type of object determines the ratio between these two types of kinetic energy.
Step 2: Key Formula or Approach:
1. Total Kinetic Energy: \( KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \).
2. For rolling without slipping: \( \omega = \frac{v}{R} \) and \( I = kmR^2 \).
3. Conservation of Energy: \( KE_{total} = mgh \).
Step 3: Detailed Explanation:
Let the moment of inertia be \( I = \beta mR^2 \).
Total initial energy \( E_i = \frac{1}{2}mv^2 + \frac{1}{2}(\beta mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \beta) \).
At maximum height \( h \), potential energy \( E_f = mgh \).
Equating \( E_i = E_f \):
\[ \frac{1}{2}mv^2(1 + \beta) = mg \left( \frac{7v^2}{10g} \right) \]
\[ \frac{1}{2}(1 + \beta) = \frac{7}{10} \]
\[ 1 + \beta = \frac{14}{10} = 1.4 \]
\[ \beta = 0.4 = \frac{2}{5} \]
The moment of inertia of the object is \( I = \frac{2}{5}mR^2 \). This formula corresponds to a solid sphere.
Step 4: Final Answer:
The object is a solid sphere.