Question

An object of uniform density rolls up the curved path with the initial velocity $v_o$ as shown in the figure. If the maximum height attained by an object is $\frac{7v_o^2}{10 g}$ ($g=$ acceleration due to gravity), the object is a _______

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

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves the conservation of mechanical energy for a rolling object. As the object rolls up, its initial total kinetic energy (translational + rotational) is converted into gravitational potential energy at the maximum height.
: Key Formula or Approach:
Total Kinetic Energy $K_{total} = \frac{1}{2}mv_o^2 + \frac{1}{2}I\omega^2$.
For rolling without slipping, $\omega = v_o/R$ and $I = kmR^2$ (where $k$ is the shape factor).
By conservation of energy: $K_{total} = mgh_{max}$.
Step 2: Detailed Explanation:
Initial Total Kinetic Energy:
\[ K = \frac{1}{2}mv_o^2 + \frac{1}{2}(kmR^2)\left(\frac{v_o}{R}\right)^2 = \frac{1}{2}mv_o^2(1+k) \]
At maximum height $h$, Potential Energy $U = mgh$.
Equating $K = U$:
\[ \frac{1}{2}mv_o^2(1+k) = mg\left(\frac{7v_o^2}{10g}\right) \]
\[ \frac{1}{2}(1+k) = \frac{7}{10} \]
\[ 1+k = \frac{14}{10} = 1.4 \]
\[ k = 0.4 = \frac{2}{5} \]
For a solid sphere, the moment of inertia is $I = \frac{2}{5}mR^2$, so $k = \frac{2}{5}$.
Step 3: Final Answer:
The object is a solid sphere.