Identification of Rolling Object from Maximum Height Reached

To identify the object, we use the principle of conservation of mechanical energy for pure rolling motion.

Let:

• $m$ be the mass of the object,
• $R$ be the radius of the object,
• $I = k m R^2$ be the moment of inertia about its axis of rotation, where $k$ is a dimensionless geometric constant.

Since the object is rolling purely without slipping, the relationship between its linear velocity $v$ and angular velocity $\omega$ is:
$\omega = \frac{v}{R}$

The total initial kinetic energy ($K_i$) consists of both translational and rotational kinetic energy:
$K_i = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Substituting $I = kmR^2$ and $\omega = \frac{v}{R}$:
$K_i = \frac{1}{2}mv^2 + \frac{1}{2}(kmR^2)\left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2(1+k)$

At the maximum height $h$, the object momentarily comes to rest, so its final kinetic energy is zero, and its energy is purely gravitational potential energy:
$U_f = mgh$

Given that the maximum height reached is $h = \frac{3v^2}{4g}$, we apply the law of conservation of mechanical energy ($K_i = U_f$):
$\frac{1}{2}mv^2(1+k) = mg\left(\frac{3v^2}{4g}\right)$

Simplifying both sides:
$\frac{1}{2}(1+k) = \frac{3}{4}$

$1 + k = \frac{3}{2}$

$k = \frac{1}{2}$

Comparing the value of $k$ with the values for standard bodies:

• For a ring: $k = 1$
• For a solid sphere: $k = \frac{2}{5}$
• For a hollow sphere: $k = \frac{2}{3}$
• For a disc: $k = \frac{1}{2}$

Therefore, the object is a disc.

Correct Answer: D (disc)