A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is

Step 1: Express the initial kinetic energy of the rolling object.

The total kinetic energy is the sum of translational and rotational kinetic energies.

$$KE_{i} = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$$

Step 2: Relate the angular velocity to the linear velocity.

For rolling without slipping, the angular velocity is determined by the linear velocity and the radius $R$.

$$\omega = \frac{v}{R}$$

Step 3: Simplify the kinetic energy expression.

Define the shape parameter $k = \frac{I}{mR^{2}}$.

$$KE_{i} = \frac{1}{2}mv^{2}\left(1 + \frac{I}{mR^{2}}\right)$$ $$KE_{i} = \frac{1}{2}mv^{2}(1+k)$$

Step 4: Apply the conservation of energy.

At the maximum height $H$, the object momentarily stops and all kinetic energy is converted into gravitational potential energy.

$$KE_{i} = mgH$$ $$\frac{1}{2}mv^{2}(1+k) = mgH$$ $$H = \frac{v^{2}(1+k)}{2g}$$

Step 5: Solve for the shape parameter $k$.

Equate the expression to the given maximum height $H = \frac{3v^{2}}{4g}$.

$$\frac{3v^{2}}{4g} = \frac{v^{2}(1+k)}{2g}$$ $$\frac{3}{4} = \frac{1+k}{2}$$ $$1+k = \frac{3}{2}$$ $$k = \frac{1}{2}$$

Step 6: Identify the object.

Compare the calculated value of $k = \frac{I}{mR^{2}}$ with standard bodies of uniform density.

• Ring: $k = 1$
• Disc: $k = \frac{1}{2}$
• Solid sphere: $k = \frac{2}{5}$
• Hollow sphere: $k = \frac{2}{3}$

Answer: Disc