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1 Million Means: 1 Million in Rupees, Lakhs and Crores

September 13, 2023The term **angular momentum** can be associated with any object that performs rotation about a point or axis. Just as Moment of force (Torque) in rotational motion is analogous to force in linear motion, the angular momentum in rotational motion is similar to linear momentum in linear motion. The **angular momentum formula **is articulated as \(L = mvr\).

Furthermore, one can find the direction of angular momentum by utilising the right-hand thumb rule. The order of angular momentum will be perpendicular to the plane of rotation. The angular momentum formula for a rigid body is \(L = I\omega \). This article will provide detailed information on the angular momentum formula. Scroll down to learn more!

It is defined as the cross product of the position vector of rotating mass concerning the point of rotation and linear momentum of the mass. To understand the formula of Angular momentum in different cases, we will classify the angular momentum into three different types.

**1. Angular momentum of a point mass about some point: **Let a point mass of mass \(m\) is moving with velocity \(\vec V\) at point \(A\). The linear momentum of the mass will be, \(\vec P = m\vec V\). The position vector of mass is \(\vec r\). Now the angular momentum of the mass about point** **\(O\) will be given by,

\(\vec L = \vec r \times \vec P = \vec r \times (m\vec V) = {m}(\vec r \times \vec V)\)

The angular momentum can also be written as:

\(L = mvr\,\sin \,\theta = mv{r_⏊}\)

Here, \({r_⏊} = r\,\sin \,\theta \) is the perpendicular distance of the line of the action of velocity \(\vec V\) from point \(O\).

The direction of angular momentum is found by using the Right-hand thumb rule. To get the direction, curl your fingers of your right hand; the direction of thumb will give the direction of angular momentum. The direction of angular momentum will be perpendicular to both the velocity vector and position vector.

**NCERT Solutions for 11th Physics Chapter 7**

**2. Angular momentum of a rigid body rotating about a fixed axis:** Assume a rigid body rotating with angular speed \(\omega \) about a fixed axis. Suppose any small mass \({m_i}\)**_{ }**at distance \({r_i}\)

\({L_i} = m(\vec r \times \vec V) = {m_i}{r_i}{v_i}\)

We know that, \(V = r\omega \)

\(\therefore \,{L_i} = {m_i}{r_i}{v_i} = {m_i}r_i^2\omega \)

Now the angular momentum of the whole rigid body about the axis will be equal to the sum of angular momentum of individual particles of the body.

So, total angular momentum, \(L = \sum {{L_i} = } \sum {{m_i}{r_i}^2\omega } \)

\( \Rightarrow L = \omega \sum {{m_i}} {r_i}^2\)

\( \Rightarrow L = I\omega \)

Where \(I\) is the Moment of inertia of the rigid body about the rotation axis.

In this case, the right-hand thumb rule gives the direction of net angular momentum.

**3. Angular momentum of a rigid body in combined rotation and translation:- **Take a rigid body having both angular velocity \(\omega \) and translational velocity \(\overrightarrow {{V_0}} \)**.**

Let us assume that point \(O\) is a fixed point in an inertial frame of reference ( Not an accelerated frame). Then the angular momentum of any small mass \({m_i}\)_{ }at distance \({r_i}\)_{ }_{ }from point \(O\)** **will be given by,

\(\overrightarrow L = \sum {{m_i}\left( {\overrightarrow {{r_i}} \times {{\overrightarrow V }_i}} \right)} \)

\( = \sum {{m_i}} \left( {{{\vec r}_{i,\,CM}} + {{\vec r}_0}} \right) \times \left( {{{\vec V}_{i,\,CM}} + {{\vec V}_0}} \right)\)

Here \({{{\vec r}_o}}\)** **is the position vector of the center of mass and** **\({{{\vec r}_{i,\,CM}}}\)** **is the position vector of small mass \({{m_i}}\) with respect to the center of mass. Similarly \({{{\vec V}_{i,\,CM}}}\)** **is the velocity of the small mass with respect to the center of mass. After simplification of the above equation, we will get,

\(\overrightarrow L = \sum {{m_i}} \left( {{{\overrightarrow r }_{i,\,CM}} \times {{\overrightarrow V }_{i,\,CM}}} \right) + \left\{ {\sum {{m_i}} {{\overrightarrow r }_{i,\,CM}}} \right\} \times {\overrightarrow V _0} + {\overrightarrow r _0} \times \left\{ {\sum {{m_i}} {{\overrightarrow V }_{i,\,CM}}} \right\} +\)

\(\left( {\sum {{m_i}} } \right){\overrightarrow r _0} \times {\overrightarrow V _0}\)

Now,

\(\sum {{m_i}} {\vec r_{i,CM}} = M{\vec r_{CM,CM}} = 0\)** ( **Here** **\(M\) is the total mass )

Similarly,

\(\sum {{m_i}} {\vec V_{i,CM}} = M{\vec V_{CM,CM}} = 0\)

Thus we will get,

\(\overrightarrow L = \sum {{m_i}\left( {{{\vec r}_{i,CM}} \times {{\vec V}_{i,CM}}} \right) + } \left( {\sum {{m_i}} } \right){\vec r_0} \times {\vec V_0}\)

\( = {\overrightarrow L _{CM}} + M\left( {{{\overrightarrow r }_0} \times {{\overrightarrow V }_0}} \right)\)

In the above equation the term \({\overrightarrow L _{CM}}\)** **represents the angular momentum of the rigid body as observed from the center of mass. The second term represents the angular momentum of the center of mass about point \(O\)*.*

**Q.1. A body of mass \(m\) is moving along the line \(y = b, z = 0\) with a constant speed \(v\). Does the body’s angular momentum about the origin is increasing, decreasing, or constant?**Ans: As the value of the \(z\) coordinate is zero. So, we can say the motion is in the \(x-y\) plane. It is also given that the body’s motion is in a straight line, \(y = b\).

Now the angular momentum of the body about the origin will be given by,

\(L = mvr\,\sin \,\theta = mv{r_ \bot }\)

Since the body is moving parallel to the \(x\)-axis, the perpendicular distance from the origin will not change and will be equal to \(b\)**. **So, the magnitude of angular momentum,

\(L = mvb\)

Its value will remain constant.

**Q.2. A circular disc of mass \(m\) and radius \(R\) is set into motion on a horizontal floor with a linear speed \(v\) in the forward direction and an angular speed \(\omega = \frac{v}{R}\) in the clockwise direction. Find the magnitude of total angular momentum of the disc about the bottommost point of the disc. Ans:** We know that the angular momentum of a body having both rotational motion and translational motion is given by,

\(\overrightarrow L = {\overrightarrow L _{CM}} + M\left( {{{\overrightarrow r }_0} \times {{\overrightarrow V }_0}} \right)\)

Here \({\overrightarrow L _{CM}}\)** **represents the angular momentum of the rigid body as observed from the center of mass. Its value is given by,

\(\left| {{{\vec L}_{CM}}} \right| = I\omega \)

The moment of inertia of disc about the center of mass will be, \(I = \frac{{m{R^2}}}{2}\)**. **By putting this value in the above equation,

\(\left| {{{\vec L}_{CM}}} \right| = \frac{{m{R^2}}}{2} \times \frac{v}{R} = \frac{{mvR}}{2}\)** **(Along the axis of the disc)

Now the angular momentum of the center of mass about the point the bottommost point of the disc will be,

\(\left| {m\left( {{{\vec r}_0} \times {{\vec V}_0}} \right)} \right| = mvR\) (Along the axis of the disc)

So, the magnitude of total angular momentum of the disc about the bottommost point of the disc will be,

\(\left| {\overrightarrow L } \right| = \left| {{{\overrightarrow L }_{CM}}} \right| + \left| {m\left( {{{\overrightarrow r }_0} \times {{\overrightarrow V }_0}} \right)} \right|\)

\( = mvR + \frac{{mvR}}{2} = \frac{{3mvR}}{2}\)

Angular momentum is a vector quantity. It is analogous to linear momentum in linear motion. It is given by the cross product of the position vector of rotating mass with respect to the point of rotation and linear momentum of the mass. For point mass, the angular momentum is given by, \(L = mvr\,\sin \,\theta = mv{r_ \bot }\)

The right-hand thumb rule gives the direction of angular momentum. The direction of angular momentum will be perpendicular to the plane of rotation. Angular momentum of a rigid body is given by, \(L = I\omega \)

When the rigid body has both rotational motion and translational motion, then the angular momentum is given by, \(\vec L = {\vec L_{CM}} + M\left( {{{\vec r}_0} \times {{\vec V}_0}} \right)\).

**Q.1. When the Moment of inertia of a rotating body is halved, then what will be the effect on angular velocity? Ans:** As there is no external torque is acting on the body, so by the law of conservation of angular momentum, we can say that initial angular momentum will be equal to final angular momentum.

\({I_1}{\omega _1} = {I_2}{\omega _2}\)

\( \Rightarrow {\omega _2} = \frac{{{I_1}}}{{{I_2}}}{\omega _1} = 2{\omega _1}\)

So, the angular velocity will be doubled.

**Q.2. What is the dimensional formula of angular momentum? Ans:** Angular momentum, \(L = Mvr\)

\(\left[ L \right] = \left[ m \right]\left[ v \right]\left[ r \right] = ML{T^{ – 1}}L\)

\([L] = M{L^2}{T^{ – 1}}\)

**Q.3. For an isolated rotating body, what is the relation between angular velocity and radius? Ans:** For an isolated rotating body, the angular momentum will be conserved.

\(Mvr = \)** **Constant

\( \Rightarrow v \propto \frac{1}{r}\)

Also, we know that, \(v = r\omega \).

\( \Rightarrow r\omega \propto \frac{1}{r}\)

\( \Rightarrow \omega \propto \frac{1}{{{r^2}}}\)

So, the angular velocity will be inversely proportional to the square of the radius.

**Q.4. Suppose the angular momentum of a body is zero at about some point. Is it necessary that it will be zero at a different point? Ans:** To get the answer to the above question, let us take an example. Assume a body travelLing along the \(x-\)axis with some velocity. Now the angular momentum of the body about the origin will be zero \(\left( {{r_ \bot } = 0} \right)\) It will be zero about every point lying on the \(x-\)axis. But the value of angular momentum will be non-zero about any other point having non zero y-coordinate value. So, it is not necessary that angular momentum will be zero about all the points if it is zero about any point.

**Q.5. A body of mass \(m\) moving with velocity \(v\) along \(y = 4x\). What will be the value of the angular momentum of the body about the origin? Ans:** Since the line, \(y = 4x\) passes through the origin, the perpendicular distance will be zero from the origin. We know that, angular momentum, \(L = mvr\,\sin \,\theta = mv{r_\bot}\). So, the value of angular momentum will be zero.