20.7 Conservation of angular momentum (check your understanding)  (Page 2/2)

The bodies acquire equal angular velocity due to friction operating between the surfaces in contact. However, friction here is internal to the system of two rotating disks. Thus, there is no external torque on the system and we can employ law of conservation of angular momentum :

Two rotating disks

$$\begin{array}{l}{L}_i=L_f\end{array}$$

Here,

$$\begin{array}{l}{L}_i=I_1{\omega }_1+I_2{\omega }_2\end{array}$$

When disks come in contact and rotate about a common axis with equal angular velocity, they acquire common angular velocity, say ”ω”. Since, two disks are rotating about a common axis of rotation, the MI of the combination is arithmetic sum of individual MIs. The moment of inertia of the composite system, “ $I_{\mathrm{C}}$ ” is given as :

$$\begin{array}{l}{I}_{\mathrm{C}}=I_1+I_2\end{array}$$

Thus, final angular momentum of the system is :

$$\begin{array}{l}{L}_{\mathrm{f}}=(I_1+I_2)\omega \end{array}$$

Putting values in the equation of conservation of angular momentum, we have :

$$\begin{array}{l}\Rightarrow I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega \end{array}$$

$$\begin{array}{l}\Rightarrow \omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}\end{array}$$

Hence, option (c) is correct.

Applying law of conservation of angular momentum, the expression of common angular velocity is given as (as derived in the earlier question) :

Two rotating disks

$$\begin{array}{l}\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}\end{array}$$

The energy dissipated due to the friction is equal to the change in the angular kinetic energy. Also as there is dissipation of energy, final angular kinetic energy is less than initial angular kinetic energy of the system. The change in angular kinetic energy, therefore, is :

$$\begin{array}{l}\Delta K=\frac{1}{2}x{I}_i{\omega }_i^2-\frac{1}{2}x{I}_f{\omega }_f^2\end{array}$$

Here, final common angular speed and moment of inertia of the combined system are :

$$\begin{array}{lll}{\omega }_f=\omega & ;& I_f=I_1+I_2\end{array}$$

Putting values, we have :

$$\begin{array}{l}\Rightarrow \Delta K=\frac{1}{2}x{I}_1{\omega }_1^2-\frac{1}{2}x{I}_2{\omega }_2^2=\frac{1}{2}x(I_1+I_2){\omega }^2\end{array}$$

Substituting value of common angular velocity and solving, we have :

$$\begin{array}{l}\Rightarrow \Delta K=\frac{I_1{I}_2{({\omega }_1-{\omega }_2)}^2}{2(I_1+I_2)}\end{array}$$

Hence, option (b) is correct.

The line of action of the force on the block due to obstruction passes through the axis of rotation about which block suddenly rotates i.e. about the obstruction for a brief period. These forces (operating at the time of collision) do not constitute torque. We can, therefore, conclude that there is no torque on the block. According to law of conservation of angular momentum,

$$\begin{array}{l}{L}_i=L_f\end{array}$$

The angular momentum of the block, treating it as a particle like body in translation, about the point of obstruction,"P", is :

$$\begin{array}{l}{L}_i=-Mv{r}_{\perp }=-\frac{MvL}{2}\end{array}$$

Negative sign indicates that the angular momentum of the block about "P" is clockwise. The angular momentum of the block after collision, now treating it as a rigid body rotating about the axis perpendicular to the plane of motion and passing through the point of obstruction, is :

A block hitting an obstruction

$$\begin{array}{l}{L}_f=I\omega \end{array}$$

where “ω” is the angular velocity of the block about the axis of rotation.

Thus,

$$\begin{array}{l}\Rightarrow -\frac{MvL}{2}=I\omega \end{array}$$

For moment of inertia of the block, it can be treated like a square plate. Its MI about the axis passing through COM is,

$$\begin{array}{l}\Rightarrow I_{\mathrm{COM}}=\frac{(L^2+L^2)}{12}=\frac{M{L}^2}{6}\end{array}$$

Axis of rotation

Recall that the formula used for MI here is that of plate about an axis that passes through COM but which is in the plane of plate. It is a valid assumption as cube can be considered to be a cylinder with square face. Since MI of the plate and its corresponding cylinder are same, MI of the cube is same as that of corresponding square plate.

Here, the axis can be considered to lie in the central plane consisting of its COM. A central plane with perpendicular axis is shown for an enlarged cube. Since cube is symmetric about any such axis as seen from any of the six faces, we can conclude that MI as calculated above is about an axis, which is perpendicular to the plane of motion and passing through its COM. Now, using theorem of parallel axes, we have MI about the point of obstruction :

$$\begin{array}{l}\Rightarrow I=I_{\mathrm{COM}}+M{r}^2\end{array}$$

From the geometry,

$$\begin{array}{l}\mathrm{OP}=r=\frac{L}{\surd 2}\end{array}$$

$$\begin{array}{l}\Rightarrow r^2=\frac{L^2}{2}\end{array}$$

Putting in the expression,

$$\begin{array}{l}\Rightarrow I=\frac{M{L}^2}{6}+\frac{M{L}^2}{2}=\frac{2M{L}^2}{3}\end{array}$$

Now, putting in the equation of conservation of angular momentum, we have :

$$\begin{array}{l}\Rightarrow -\frac{MvL}{2}=\frac{2M{L}^2}{3}x\omega \end{array}$$

$$\begin{array}{l}\Rightarrow \omega =-\frac{3v}{4L}\end{array}$$

The negative sign indicates that the cubical block overturns in clockwise rotation about point "P". The angular speed of the block about obstruction, therefore, is equal to the magnitude of angular velocity.

Hence, option (a) is correct.