Identify Rolling Object Given Maximum Height Attained

To determine the identity of the rolling object, we use the principle of conservation of mechanical energy.

Assuming the object rolls without slipping, its initial total kinetic energy ($K$) consists of both translational kinetic energy ($K_t$) and rotational kinetic energy ($K_r$): $K = K_t + K_r = \frac{1}{2} m v_0^2 + \frac{1}{2} I \omega_0^2$

where:

• $m$ is the mass of the object,
• $v_0$ is the initial linear velocity of the center of mass,
• $I$ is the moment of inertia about the axis of rotation, which can be expressed as $I = k m R^2$ (with $k$ being a shape-dependent constant and $R$ being the radius),
• $\omega_0$ is the initial angular velocity.

For rolling without slipping, $\omega_0 = \frac{v_0}{R}$. Substituting $I = k m R^2$ and $\omega_0 = \frac{v_0}{R}$ into the kinetic energy equation gives: $K = \frac{1}{2} m v_0^2 + \frac{1}{2} (k m R^2) \left(\frac{v_0}{R}\right)^2 = \frac{1}{2} m v_0^2 (1 + k)$

At the maximum height $h$ attained by the object, all of its initial kinetic energy is converted into gravitational potential energy ($U = mgh$): $m g h = \frac{1}{2} m v_0^2 (1 + k)$

Solving for the maximum height $h$: $h = \frac{(1 + k) v_0^2}{2g}$

We are given that the maximum height attained is: $h = \frac{7 v_0^2}{10 g}$

Equating the two expressions for $h$: $\frac{1 + k}{2} = \frac{7}{10}$

$1 + k = \frac{14}{10} = \frac{7}{5}$

$k = \frac{7}{5} - 1 = \frac{2}{5}$

The constant $k = \frac{2}{5}$ corresponds to a moment of inertia $I = \frac{2}{5} m R^2$, which is characteristic of a solid sphere.

Correct Answer: D (solid sphere)