Question

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 a

Answer 1

Step1: Given data

1. The initial velocity of the body is $u=v$.
2. The maximum height of the body on the curved surface is $H=\frac{3v^2}{4g}$.

Diagram

Step2: Conservation of energy

1. From the concept of conservation of energy, the sum of kinetic and potential energy is conserved in every point of motion. i.e., $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$
2. The kinetic energy of a body moving with a velocity is $K_t=\frac{1}{2}m{v}^2$ and the kinetic energy due to rotation is $\frac{1}{2}I{\omega }^2$, where, m is the mass of the body, I is the moment of inertia, and $\omega$ is the angular velocity of the body.
3. The potential energy of a body at height h is $mgh$, and g is the acceleration due to gravity.

Step4: Finding the moment of inertia

We know from the conservation of kinetic energy, that the sum of kinetic and potential energy is conserved at $h=0$ and $h=h$.

So, $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$.

$$\begin{gathered} or\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh. \\ or\frac{1}{2}m{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=mg\left(\frac{3v^2}{4g}\right)\left(\sin ce,\omega =\frac{v}{R}andh=\frac{3v^2}{4g}\right) \\ or\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=mg\left(\frac{3v^2}{4g}\right)-\frac{1}{2}m{v}^2 \\ or\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{3m{v}^2}{4}-\frac{1}{2}m{v}^2 \\ or\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=\frac{1}{4}m{v}^2 \\ orI=\frac{\frac{1}{4}m{v}^2}{{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{v}{R}}\right)}^2}=\frac{1}{2}\frac{m{v}^2.R^2}{v^2}=\frac{1}{2}m{R}^2 \\ orI=\frac{1}{2}m{R}^2. \end{gathered}$$

Therefore, the moment of inertia of the body is $I=\frac{1}{2}m{R}^2$, this is the moment of inertia of a disc about an axis passing through the center perpendicular to the plane. So, the object is disc.