19.12 Work and energy in rolling (check your understnading)  (Page 2/2)

The hollow spherical ball is at rest in the beginning. Thus, hollow spherical ball has no kinetic energy, but only gravitational potential energy due to its place at a height from the ground.

There is no loss of energy due to friction and hence, mechanical energy of the ball is conserved. The hollow spherical ball acquires the kinetic energy, which is equal to the change in gravitational potential energy.

$$\begin{array}{l}K=mg(\mathrm{AB}-\mathrm{CD})=mg(10-4)=6mg\end{array}$$

In order to know the speed at the end of the track, we need to know the kinetic energy due to rolling. The kinetic energy of rolling, in terms, of linear velocity is given as :

$$\begin{array}{l}K=(I+m{r}^2)x\frac{{v_C}^2}{2r^2}\end{array}$$

Solving for ${v_C}^2$ and putting value of kinetic energy as derived earlier, we have :

$$\begin{array}{l}\Rightarrow {v_C}^2=\frac{2K{r}^2}{(I+m{r}^2)}\\ \Rightarrow {v_C}^2=\frac{2x6mg{r}^2}{(\frac{2m{r}^2}{3}+m{r}^2)}\end{array}$$

$$\begin{array}{l}\Rightarrow v_C=6\surd 2m/s\end{array}$$

Hence, option (c) is correct.

The mechanical energy of the rolling body is conserved in both cases. As such, the rolling of spheres along two inclines through same vertical displacement means that final kinetic energies of the solid spheres are same.

$$\begin{array}{l}K=Mgh\end{array}$$

We also know that distribution of kinetic energy between translation and rotation is fixed for a given geometry and mass of the body.

$$\begin{array}{l}\frac{K_T}{K_R}=\frac{M{R}^2}{I}\end{array}$$

This means that translational kinetic energy of the solid sphere will be same. In turn, it means that final speeds of the solid sphere in two cases will be same.

However, as the linear distance is less in the case of steeper incline, the sphere will take lesser time.

Hence, option (c) is correct.

The kinetic energy of a rolling body, in terms of linear velocity of center of mass, is given as :

$$\begin{array}{l}K=(I+M{R}^2)x\frac{{V_C}^2}{2R^2}\end{array}$$

$$\begin{array}{l}\Rightarrow {V_C}^2=\frac{K{R}^2}{I+M{R}^2}\end{array}$$

For ring, m = 0.3 kg and $I=M{R}^2$

$$\begin{array}{l}\Rightarrow {V_C}^2=\frac{K{R}^2}{M{R}^2+M{R}^2}=\frac{K}{M}=\frac{K}{0.3}=\frac{10K}{3}\end{array}$$

For disk, m = 0.4 kg and $I=\frac{M{R}^2}{2}$ .

$$\begin{array}{l}\Rightarrow {V_C}^2=\frac{K{R}^2}{\frac{M{R}^2}{2}+M{R}^2}=\frac{4K}{3m}=\frac{4K}{3x\mathrm{0.4}}=\frac{10K}{3}\end{array}$$

Thus, velocities of centers of mass for both ring and disk are same.

Hence, option (d) is correct

Since mechanical energy is conserved in rolling, the total kinetic energy for a circular body is a constant. Now, ratio of translational and rotational kinetic energy is :

$$\begin{array}{l}\frac{K_T}{K_R}=\frac{M{R}^2}{I}\end{array}$$

This ratio is independent of the position of the rolling body. Thus, rotational kinetic energy (and hence angular velocity) is maximum for the circular body having maximum moment of inertia. Here, ring for same mass and radius has the maximum moment of inertia. As such, angular velocity is maximum for the ring.

Hence, option (a) is correct.

We know that the minimum speed of the ball at the top (D) of loop, so that it does not leave the highest point, is :

$$\begin{array}{l}{v}_D=\surd \left(gR\right)\end{array}$$

Corresponding to this requirement, the velocity required at the bottom of the loop is :

$$\begin{array}{l}{v}_B=\surd \left(5gR\right)\end{array}$$

Now, the kinetic energy at the bottom of the loop, in terms of linear velocity, is :

$$\begin{array}{l}K=(I+M{R}^2)x\frac{{V_C}^2}{2r^2}\end{array}$$

$$\begin{array}{l}\Rightarrow K=\frac{(I+m{r}^2)x5gR}{2r^2}\end{array}$$

Let “h” be the height, then (assuming r<<R) :

$$\begin{array}{l}\Rightarrow \frac{(I+m{r}^2)x5gR}{2r^2}=mgh\end{array}$$

$$\begin{array}{l}\Rightarrow h=\frac{(I+m{r}^2)x5gR}{2r^{2}mg}\end{array}$$

For solid sphere,

$$\begin{array}{l}\Rightarrow h=\frac{(\frac{2m{r}^2}{5}+m{r}^2)x5gR}{2r^{2}mg}\\ \Rightarrow h=\frac{\frac{7m{r}^2}{5}x5gR}{2r^{2}mg}\end{array}$$

$$\begin{array}{l}\Rightarrow h=\frac{7R}{10}\end{array}$$

Hence, option (c) is correct.