20.4 Law of motion in angular form for a system of particles (check

The angular momentum of a rigid body is given by the equation,

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

In order to find MI about diameter, we first calculate moment of inertia, " $I_C$ ", of the disk about a perpendicular axis passing through the center as :

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

Applying theorem of perpendicular axes, the moment of inertia about the axis of rotation (i.e. about one of the diameters) is :

$$\begin{array}{l}I=\frac{I_C}{2}=\frac{M{R}^2}{4}\end{array}$$

Therefore, the angular momentum of the disk about one of its diameter is :

$$\begin{array}{l}L=I\omega =\frac{M{R}^2\omega }{4}\end{array}$$

Hence, option (d) is correct.

The angular momentum of a rigid body is given by the equation,

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

Here, the rod rotates about a perpendicular axis at its end. Thus, we need to find MI of the rod, "I", about this axis. Using theorem of parallel axes, the MI about the axis of rotation is :"

$$\begin{array}{l}I=I_{\mathrm{C}}+Mx{\left(\frac{x}{2}\right)}^2\\ \Rightarrow I=\frac{M{x}^2}{12}+\frac{M{x}^2}{4}=\frac{M{x}^2}{3}\end{array}$$

Putting in the expression of angular momentum :

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

The angular velocity is expressed in terms of tangential velocity as :

$$\begin{array}{l}\omega =\frac{v}{x}\end{array}$$

Substituting this value in the expression of angular momentum, we have

$$\begin{array}{l}\Rightarrow L=\frac{Mvx}{3}\end{array}$$

Hence, option (b) is correct.

We know that angular momentum of a rigid body is given by :

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

We need to calculate the MI of the rod about the axis of rotation. Now, the MI of the rod about central axis is given by :

$$\begin{array}{l}\Rightarrow I_C=\frac{M{\left(3x\right)}^2}{12}=\frac{3M{x}^2}{4}\end{array}$$

Applying theorem of parallel axis, the MI about the axis of rotation passing through point “O” is :

$$\begin{array}{l}\Rightarrow I_O=I_C+M{d}^2\end{array}$$

Here,

$$\begin{array}{l}d=\frac{3x}{2}-x=\frac{x}{2}\end{array}$$

$$\begin{array}{l}\Rightarrow I_O=\frac{3M{x}^2}{4}+\frac{M{x}^2}{4}=M{x}^2\end{array}$$

Therefore, angular momentum about the axis of rotation is :

$$\begin{array}{l}\Rightarrow L=I\omega =M{x}^2\omega \end{array}$$

Hence, option (a) is correct.

Note 1 : Alternatively, we can also calculate the angular momentum of the rod in terms of angular momentum of the parts on either side of the axis of rotation. Since each part rotates about the same axis of rotation, the net angular momentum of the rod about the axis of rotation is algebraic sum of the angular momentum of the parts.

The MI of the left rod, about a perpendicular axis at one of the ends, is :

$$\begin{array}{l}{I}_{\mathrm{OL}}=\frac{\left(\frac{M}{3}\right)x^2}{3}=\frac{M{x}^2}{9}\end{array}$$

$$\begin{array}{l}\Rightarrow L_{\mathrm{OL}}=I\omega =\frac{M{x}^2\omega }{9}\end{array}$$

Similarly, the MI of the right rod is :

$$\begin{array}{l}{I}_{\mathrm{OR}}=\frac{\left(\frac{2M}{3}\right){\left(2x\right)}^2}{3}=\frac{8M{x}^2}{9}\end{array}$$

$$\begin{array}{l}\Rightarrow L_{\mathrm{OR}}=I\omega =\frac{8M{x}^2\omega }{9}\end{array}$$

$$\begin{array}{l}\Rightarrow L=L_{\mathrm{OL}}+L_{\mathrm{OR}}\\ \Rightarrow L=\frac{M{x}^2\omega }{9}+\frac{8M{x}^2\omega }{9}=M{x}^2\omega \end{array}$$

Note 2 : It must be understood here that we can not treat the rod as a particle at its center of mass and then find the angular momentum. For example, in this case, the center of mass is at distance of "3x/2" from either end of the rod and at a distance of "x/2" from the point "O" through which axis of rotation passes. The MI of this particle equivalent rod about the axis of rotation is :

$$\begin{array}{l}\Rightarrow I_O=\frac{M{x}^2}{4}\end{array}$$

and

$$\begin{array}{l}\Rightarrow L=I\omega =\frac{M{x}^2\omega }{4}\end{array}$$

Obviously, this approach is conceptually wrong. The concept of center of mass by definition is valid for translation - not rotation.

Here, the point of contact and center of mass are in the plane of motion. Hence,

$$\begin{array}{l}\mathbf{L}=M{\mathbf{r}}_C\mathbf{x}{\mathbf{v}}_C)+I\mathbf{\omega }\end{array}$$

Now, the magnitude of angular momentum of the body, considering it as a particle at center of mass, with respect to point of contact is :

$$\begin{array}{l}\mathbf{L}=M{r}_C{v}_C=MRv\end{array}$$

This angular momentum is clockwise and into the page.

The magnitude of angular momentum about center of mass is same as that about its central axis i.e. perpendicular axis passing through center of mass :

$$\begin{array}{l}I\omega =M{R}^2\omega \end{array}$$

For rolling motion, $\omega =\frac{v}{R}$

$$\begin{array}{l}I\omega =M{R}^{2}x\frac{v}{R}=MRv\end{array}$$

This angular momentum is also clockwise and into the page. Therefore, the magnitude of net angular momentum is :

$$\begin{array}{l}L=2MRv\end{array}$$

Hence, option (b) is correct.

This question is similar to the previous one - with one exception that sphere is a three dimensional body as against the disk, which is two dimensional. However, the situation is same as far as the point about angular momentum is calculated and the center of mass. The point of contact and center of mass are in the same middle section of the sphere, along which motion takes place. Thus, the point of contact and center of mass of the solid sphere are in the plane of motion. Hence,

$$\begin{array}{l}\mathbf{L}=M({\mathbf{r}}_C\mathbf{x}{\mathbf{v}}_C)+I\mathbf{\omega }\end{array}$$

Now, the magnitude of angular momentum of the body as particle at center of mass with respect to point of contact is :

$$\begin{array}{l}M\left(r_C{v}_C\right)=MRv\end{array}$$

This angular momentum is clockwise and into the page.

The magnitude of angular momentum about center of mass is same as that about its central axis i.e. perpendicular axis passing through center of mass :

$$\begin{array}{l}I\omega =\frac{2M{R}^2\omega }{5}\end{array}$$

For rolling motion, $\omega =\frac{v}{R}$

$$\begin{array}{l}I\omega =\frac{2M{R}^{2}v}{5R}=\frac{2MRv}{5}\end{array}$$

This angular momentum is also clockwise and into the page. The magnitude of net angular momentum, therefore, is :

$$\begin{array}{l}L=MRv+\frac{2M{R}^{2}v}{5}=\frac{7M{R}^{2}v}{5}\end{array}$$

Hence, option (d) is correct.

The pulley and string are both considered “mass-less”. Therefore, these elements do not contribute to the angular momentum of the system. Further, we also observe that we are required to find angular momentum about an axis - not about a point. We need to know the linear velocity of the blocks in order to calculate their angular momentum about the axis of rotation of the pulley. This, in turn, requires us to find the acceleration of the blocks in vertical direction.

Here, pulley is “mass-less”. Thus, tension in the string on either side of the pulley is same. From force analysis of each block :

$$\begin{array}{l}{m}_{2}g-T=m_{2}a\\ T-m_{1}g=m_{1}a\end{array}$$

$$\begin{array}{l}\Rightarrow a=\frac{(m_2-m_1)}{(m_1+m_2)}g\end{array}$$

For vertical motion of block " $m_2$ ", we have :

$$\begin{array}{l}{v}^2=u^2+2ay=0+2ay=2ay\end{array}$$

$$\begin{array}{l}\Rightarrow v^2=2ay=\frac{2(m_2-m_1)gy}{(m_1+m_2)}\end{array}$$

Now, angular momentums of the blocks are both clockwise. Hence, we can get magnitude of net angular momentum as arithmetic sum :

$$\begin{array}{l}L=m_{1}vr+m_{2}vr=(m_1+m_2)vr\end{array}$$

Substituting the value of the speed,

$$\begin{array}{l}\Rightarrow L=(m_1+m_2)xrx\surd \frac{2(m_2-m_1)gy}{(m_1+m_2)}\end{array}$$

$$\begin{array}{l}\Rightarrow L=\surd \frac{2({m_2}^2-{m_1}^2)g{r}^{2}y}{(m_1+m_2)}\end{array}$$

Hence, option (a) is correct.