20.7 Conservation of angular momentum (check your understanding)

The gravitational pull of the Sun passes through COM of the Earth, providing centripetal force required for the rotation of the Earth about it. Thus, there is no external torque on the Earth. It means that the angular momentum of the Earth remains constant.

The melting of ice cap will result in the rise of sea level. From the point of view of MI of the Earth, it means redistribution of mass. The water equivalent of ice moves away from the axis of rotation, which passes through the poles. This results in an increase in the MI of the Earth.

Now, applying law of conservation of angular momentum :

$$\begin{array}{l}{L}_i=L_f\\ I_i{\omega }_i=I_f{\omega }_f\end{array}$$

$$\begin{array}{l}\Rightarrow {\omega }_f=\frac{I_i{\omega }_i}{I_f}\end{array}$$

Hence, increase in the MI of the Earth, due to melting of ice, will decrease angular velocity. On the other hand, the duration of a day on the Earth is equal to its time period of rotation, which is given as :

$$\begin{array}{l}\Rightarrow T=\frac{2\pi }{{\omega }_f}\end{array}$$

As the Earth rotate slowly (lesser angular speed, ${\omega }_f$ ), the duration of a day on the Earth will increase.

Hence, options (b) and (d) are correct.

Since no external torque is operating on the disk – objects system, the angular momentum of the system is conserved.

Let “ ${\omega }_f$ ” be the common angular velocity of the composite system after objects are placed on the disk. Let subscripts “D”, “O” and “C” represent disk, objects and composite system respectively , then according to conservation of angular momentum,

$$\begin{array}{l}{I}_{\mathrm{Di}}{\omega }_{\mathrm{Di}}+I_{\mathrm{Oi}}{\omega }_{\mathrm{Oi}}=I_{\mathrm{Ci}}{\omega }_f\end{array}$$

Here,

$$\begin{array}{lllllll}{I}_{\mathrm{Di}}=\frac{M{R}^2}{2}& ;& {\omega }_{\mathrm{Di}}=\omega & ;& I_{\mathrm{Oi}}=0& ;& {\omega }_{\mathrm{Oi}}=0\end{array}$$

Also, the MI of the composite system is :

$$\begin{array}{l}{I}_{\mathrm{Cf}}=(\frac{M{R}^2}{2}+2m{R}^2)=(\frac{M}{2}+2m)R^2\end{array}$$

Putting in the equation of conservation law,

$$\begin{array}{l}\frac{M{R}^2\omega }{2}=(\frac{M}{2}+2m)R^2{\omega }_f\end{array}$$

$$\begin{array}{l}{\omega }_f=\frac{M\omega }{2(\frac{M}{2}+2m)}=\frac{M\omega }{(M+4m)}\end{array}$$

Hence, options (a) is correct.

The Earth rotates around the Sun as gravitational pull provides the necessary centripetal force. This force passes through the center of mass of the Earth. As such, gravitational pull does not constitute torque on the Earth. Therefore, we can consider that angular momentum of the Earth remains unchanged. Now, let us use "i" and "f" subscripts to denote initial and final values, then according to law of conservation of angular momentum :

$$\begin{array}{l}{L}_i=L_f\\ I_i{\omega }_i=I_f{\omega }_f\end{array}$$

$$\begin{array}{l}\Rightarrow {\omega }_f=\frac{I_i{\omega }_i}{I_f}\end{array}$$

We know that time period of the Earth is :

$$\begin{array}{l}{T}_i=24\mathrm{hrs}\end{array}$$

Hence, its angular velocity is :

$$\begin{array}{l}\Rightarrow {\omega }_i=\frac{2\pi }{T_i}=\frac{2\pi }{24}=\frac{\pi }{12}\mathrm{rad}/s\end{array}$$

Since data on mass and radius are missing, it is not possible to calculate MIs in two cases separately. However, we can find the ratio of MIs :

$$\begin{array}{l}\frac{I_i}{I_f}=\frac{\frac{2M{R}^2}{5}}{\frac{2}{5}x\frac{2M}{3}x{\left(\frac{R}{2}\right)}^2}=\frac{1x3x4}{2}=6\end{array}$$

Putting in the equation,

$$\begin{array}{l}{\omega }_f=6{\omega }_i=6x\frac{\pi }{12}=\frac{\pi }{2}\mathrm{rad}/s\end{array}$$

The new time period of rotation is :

$$\begin{array}{l}{T}_f=\frac{2\pi }{{\omega }_f}=\frac{2\pi x2}{\pi }=4\mathrm{hrs}\end{array}$$

Hence, options (b) is correct.