Informative line

Angular Momentum Of Point

Learn formula and direction of angular momentum and Torque. Practice to calculate angular momentum of a particle and vector, linear momentum, axis of rotation, translational motion and magnitude of angular momentum.

Identification of Motion of a Body

Translational Motion

If the velocity of all the particles of a body is same, the body is said to be in translational motion.

The path of the body does not need to be linear.

Rotational Motion

If all the particles of a body move in circular motion such that their centers lie on a straight line, it is called rotational motion.

Axis of Rotation

The straight line perpendicular to the circular path passing through all the centers is called the axis of rotation.

If a body is in translational motion i.e., all the particles have same velocity \(\overrightarrow{v}\), then we can replace it by a point at its center of mass moving with \(\overrightarrow{v}\), as shown in figure.

 

If a body is rotating about a fixed axis, its moment of inertia about the fixed axis \(I_{axis}\) is always involved while writing the formulae.

Illustration Questions

Which one of the following bodies is performing translational motion?

A

B

C

D

×

Option (C) is correct, because all the points on the block move with same velocity.

image

Option (A) is incorrect, because the body is in rotational motion about axis AB.

image

Option (B) is incorrect because all the particles on the rod perform circular motion about an axis passing through H and is perpendicular to the plane of motion. Thus, rod is in rotational motion.

image

Option (D) is incorrect because all the particles on the disk perform circular motion with centers at O. Thus, they perform rotational motion with AB as axis of rotation.

image

Which one of the following bodies is performing translational motion?

A image
B image
C image
D image

Option C is Correct

Calculation of Perpendicular Distance of a Point from Line of Velocity

To calculate the perpendicular distance of a point O from line of velocity, we extend the line of velocity and drop perpendicular from point O on the line of velocity, as shown in figure.

\(r_{\perp}=\ell\,sin\,\theta\)

where \(\theta\) is the angle between \(\overrightarrow{v}\) and \(\overrightarrow{r}\) when these are joined tail to tail.

Also, \(\overrightarrow{r}\) is the position vector of particle P with respect to point O.

Illustration Questions

Find the perpendicular distance of point of projection from line of velocity of a projectile (at its maximum height).

A \(H_{max}\)

B Range, R

C \(\dfrac{R}{2}\)

D \(2\,H_{max}\)

×

At maximum height, the velocity of particle is horizontal.

image

Extending the line of velocity and dropping a perpendicular on it from point of projection.

image

\(\therefore r_{\perp}=H_{max}\)

Find the perpendicular distance of point of projection from line of velocity of a projectile (at its maximum height).

A

\(H_{max}\)

.

B

Range, R

C

\(\dfrac{R}{2}\)

D

\(2\,H_{max}\)

Option A is Correct

Direction of Angular Momentum

Angular momentum is the rotational equivalent of linear momentum.

Direction of Torque

The direction of torque of a force about a point is determined by its tendency to rotate the body clockwise or anticlockwise.

Direction of angular momentum

The direction of angular momentum is determined by considering the sense of rotation of the particle with respect to the given point (clockwise or anticlockwise).

For simplicity, assume a handycam is placed at O to capture the moving particle P. The required sense of rotation of handycam to shoot the particle is the sense of rotation of particle.

Illustration Questions

In which of the following cases, the direction of angular momentum with respect to O, is wrong?

A

B

C

D

×

Option (A) is incorrect, because the direction of angular momentum is clockwise.

image

Option (B) is incorrect, because the direction of angular momentum is anticlockwise.

image

Option (C) is incorrect, because the direction of angular momentum is clockwise.

image

Option (D) is correct, because there is no rotation. Since, initial and final lines are same.

image

In which of the following cases, the direction of angular momentum with respect to O, is wrong?

A image
B image
C image
D image

Option D is Correct

Calculation of Angular Momentum of Point Mass 

Consider a particle P of mass m, moving with velocity \(\overrightarrow{v}\), as shown in figure.

Calculation of \(r_{\perp}\)

We find perpendicular distance (\(r_{\perp}\)) of point O from line of velocity.

\(\theta\) is the angle between vectors \(\overrightarrow{v}\) and \(\overrightarrow{r}\) when joined from tail to tail, where \(\overrightarrow{r}\) is the position vector of P with respect to O.

\(r_{\perp}=\ell \;sin\,\theta\)

Calculation of angular momentum

Angular Momentum of a particle of mass m, moving with velocity v with respect to O, is given by,

\(L=mvr_{\perp}\)

where \(r_{\perp}\) is the perpendicular distance of point O from line of velocity.

\(\because r_{\perp}=\ell \;sin\,\theta\)

\(\therefore L=mv\ell \;sin\,\theta\)

Direction of angular momentum

Direction of angular momentum is same as the sense of rotation of line of sight of the observer at O to see the particle P. 

Clockwise in this case.

Illustration Questions

Find the angular momentum of a particle P, having mass \((m)\,2\,kg\), moving with velocity \(\overrightarrow{v}\) about point O, as shown in figure. \(|\overrightarrow{r}|=\ell=10\,m\) \(\overrightarrow{v}=10\,m/s\) \(\theta=\dfrac{\pi}{4}\)

A \(30\sqrt2\;kg\,m^2/s\)

B \(50\;kg\,m^2/s\)

C \(100\sqrt2\;kg\,m^2/s\)

D \(100\;kg\,m^2/s\)

×

Given:

Magnitude of \(\overrightarrow{r}=\ell=10\,m\)

Velocity of particle P, \(\overrightarrow{v}=10\,m/s\)

Angle, \(\theta=\dfrac{\pi}{4}\)

Mass of particle P, \(m=2\,kg\)

image

Calculation of \(r_{\perp}\):

\(r_{\perp}=\ell\; sin\,\theta\)

\(r_{\perp}=10\; sin\left(\dfrac{\pi}{4}\right)\)

\(=\dfrac{10}{\sqrt{2}}\)

\(=5\sqrt{2}\,m\)

image

Calculation of angular momentum :

\(L=mvr_{\perp}\)

\(=2×10×5\sqrt{2}\)

\(=100\sqrt{2}\;kg\;m^2/s\)

Sense of rotation is the direction of angular momentum.

Direction of angular momentum is clockwise.

image

Find the angular momentum of a particle P, having mass \((m)\,2\,kg\), moving with velocity \(\overrightarrow{v}\) about point O, as shown in figure. \(|\overrightarrow{r}|=\ell=10\,m\) \(\overrightarrow{v}=10\,m/s\) \(\theta=\dfrac{\pi}{4}\)

image
A

\(30\sqrt2\;kg\,m^2/s\)

.

B

\(50\;kg\,m^2/s\)

C

\(100\sqrt2\;kg\,m^2/s\)

D

\(100\;kg\,m^2/s\)

Option C is Correct

Calculation of Angular Momentum by Cross-Product

Angular Momentum

The angular momentum of a particle P of mass m, moving with velocity \(\overrightarrow{v}\) with respect to O, is given by

If \(\overrightarrow{P}\) is the linear momentum of particle then

\(\overrightarrow{P}=m\overrightarrow{v}\)

Angular momentum = \(\overrightarrow{r}\) × linear momentum

\(\overrightarrow{L}=\overrightarrow{r}×(m\overrightarrow{v})\)

\(=m(\overrightarrow{r}×\overrightarrow{v})\)

where \(\overrightarrow{r}\) is the position vector of P with respect to O.

Magnitude of Angular Momentum

\(|\overrightarrow{L}|=mvr\;sin\,\theta\)

where \(\theta\) is the angle between \(\overrightarrow{r}\) and \(\overrightarrow{v}\)

Direction of Angular Momentum

Direction of angular momentum is determined by right hand thumb rule.

Here, angular momentum is perpendicular and inside the plane.

The angular momentum vector is perpendicular to \(\overrightarrow{r}\) and \(\overrightarrow{v}\).

If point O is in the plane of motion, the angular momentum vector is perpendicular to the plane of motion.

Illustration Questions

Find the angular momentum of a particle P (having mass 2 kg and moving with velocity \(\overrightarrow{v}=10\,m/s\)) with respect to point Q (2,1), as shown in figure.

A \(80\hat k\;kg\;m^2/sec\)

B \(10\hat k\;kg\;m^2/sec\)

C \(20\hat k\;kg\;m^2/sec\)

D \((-40\hat k)\;kg\;m^2/sec\)

×

Given :

Mass of particle P, \(m=2\,kg\)

Velocity of particle P, \(\overrightarrow{v}=10 \hat i\,m/s\)

Coordinates of P (4,3) and Q (2,1)

image

Calculation of \(\overrightarrow{r}\) :

\(\overrightarrow{r}_{P/Q}=(2\hat i+2\hat j)\,m\)

image

Calculation of angular momentum :

\(\overrightarrow{L}=(\overrightarrow{r}×m\overrightarrow{v})\)

\(\overrightarrow{L}=m(\overrightarrow{r}×\overrightarrow{v})\)

\(\overrightarrow{L}=2\,(2\hat i+2\hat j)×(10\hat i)\)

\(\overrightarrow{L}=2[-20\hat k]\)

\(\overrightarrow{L}=(-40\hat k)\)

\(\vec L=(-40\hat k)\;kg\;m^2/sec\)

image

Find the angular momentum of a particle P (having mass 2 kg and moving with velocity \(\overrightarrow{v}=10\,m/s\)) with respect to point Q (2,1), as shown in figure.

image
A

\(80\hat k\;kg\;m^2/sec\)

.

B

\(10\hat k\;kg\;m^2/sec\)

C

\(20\hat k\;kg\;m^2/sec\)

D

\((-40\hat k)\;kg\;m^2/sec\)

Option D is Correct

Calculation of Angular Momentum of a Point Mass about Point of Suspension in a Conical Pendulum

  • Consider two particles A and B of mass m each and velocities \(v_A\) and \(v_B\) respectively. The length of string by which these are attached to the point of suspension (O) is \(\ell\).

Here

\(\overrightarrow{v}_A\) is out of the plane of paper

\(\overrightarrow{v}_B\) is into the plane of paper.

Angular Momentum of Point A

Here, \(\overrightarrow{r}\) is \(\overrightarrow{OA}\).

\(\overrightarrow{v}_A\) is outside and perpendicular to the plane.

(i) Direction

  • Direction of angular momentum is determined by right hand thumb rule.

(ii) Magnitude

\(|\overrightarrow{L}_A|=mv\ell \)

Angular Momentum at Point B

Here, \(\overrightarrow{r}\) is \(\overrightarrow{OB}\)

\(\overrightarrow{v}\) is inside and perpendicular to the plane of paper.

(i) Direction

  • Direction of angular momentum is determined by right hand thumb rule.

(ii) Magnitude

\(|\overrightarrow{L}_B|=mv\ell \)

  •  The plane of screen is \(x-y\) plane and z-axis is perpendicular outside the plane.

(i) Angular Momentum at Point A

\(\overrightarrow{v}_A=v \hat k\)

\(\overrightarrow{r}_A=-\ell \;sin\, \theta \;\hat i -\ell \;cos \,\theta \;\hat j\)

\(\overrightarrow{L}_A=m(\overrightarrow{r}×\overrightarrow{v}_A)\)

\(\overrightarrow{L}_A=m(-\ell \;sin\, \theta \;\hat i -\ell \;cos \,\theta \;\hat j)× v \hat k\)

\(\overrightarrow{L}_A=m[-\ell \;sin\, \theta \;v(-\hat j) -\ell \;cos \,\theta \;v \hat i ]\)

\(\overrightarrow{L}_A=m(\ell v\;sin \,\theta \; \hat j -\ell v \;cos\, \theta \; \hat i )\)

\(\overrightarrow{L}_A=(-mv\ell \;cos\, \theta)\hat i+(mv\ell \;sin \,\theta)\hat j\)

(ii) Angular Momentum at Point B

\(\overrightarrow{v}_B=-v \hat k\)

\(\overrightarrow{r}_B=\ell \;sin \,\theta \;\hat i -\ell \;cos \,\theta \;\hat j\)

\(\overrightarrow{L}_B=m(\overrightarrow{r}×\overrightarrow{v})\)

\(\overrightarrow{L}_B=m(\ell \;sin\, \theta \;\hat i -\ell \;cos \,\theta \;\hat j)(-v \hat k)\)

\(\overrightarrow{L}_B=m[-\ell \;sin \,\theta \;v(-\hat j) +\ell \;cos\, \theta \;v \hat i ]\)

\(\overrightarrow{L}_B=m[v\ell \;cos \,\theta \; \hat i +v\ell \;sin\, \theta \; \hat j ]\)

\(\overrightarrow{L}_B=mv\ell \;cos \,\theta \;\hat i+ mv\ell \;sin \,\theta\;\hat j\)

Illustration Questions

Which of the following vectors represent angular momentum \((\overrightarrow{L})\) of the particle in conical pendulum at point B, about O?

A \(|\overrightarrow{L}|=mv\ell\)

B \(|\overrightarrow{L}|=mv\ell\)

C \(|\overrightarrow{L}|=mv\ell\;cos\,\theta\)

D \(|\overrightarrow{L}|=mv\ell\;sin\,\theta\)

×

Angular momentum \(\overrightarrow{L}\) is perpendicular to \(\overrightarrow{r}\) as well as \(\overrightarrow{v}\).

image

Direction :

Direction of angular momentum is determined by right hand thumb rule.

image

Magnitude : 

\(|\overrightarrow{L}|=mv\ell\)

Which of the following vectors represent angular momentum \((\overrightarrow{L})\) of the particle in conical pendulum at point B, about O?

image
A

\(|\overrightarrow{L}|=mv\ell\)

.

image
B

\(|\overrightarrow{L}|=mv\ell\)

image
C

\(|\overrightarrow{L}|=mv\ell\;cos\,\theta\)

image
D

\(|\overrightarrow{L}|=mv\ell\;sin\,\theta\)

image

Option A is Correct

Torque

The rate of change of angular momentum is Torque.

Formula of Torque :

\(\because \;\overrightarrow{F}=\dfrac{d\overrightarrow{P}}{dt}\)

\(\therefore \overrightarrow{r}×\overrightarrow{F}=\overrightarrow{r}×\dfrac{d\overrightarrow{P}}{dt}\)

\(\overrightarrow{\tau}=\dfrac{d(\overrightarrow{r}×\overrightarrow{P})}{dt}\)

\(\implies \overrightarrow{\tau}=\dfrac{d\overrightarrow{L}}{dt}\)

Illustration Questions

A particle of mass m is projected from point O with speed u at an angle \(\theta\) with the horizontal. Find the angular momentum of the particle with respect to O, at any time t.

A \(\dfrac{mg\;u\,cos\,\theta \;t^2}{2}\)

B \(\dfrac{mg\;t^2}{2\,cos\,\theta}\)

C \(mg\,u\,cos\,\theta\)

D \(\dfrac{mg\,cos\,\theta }{t^2}\)

×

Speed of particle = u

Mass of particle = m

Angle of projection = \(\theta\)

image

Torque = Rate of change of angular momentum

\(\overrightarrow{\tau}=\dfrac{d\overrightarrow{L}}{dt}\)

\(mg\;x_t=\dfrac{d\overrightarrow{L}}{dt}\)       \(\left[\because\;\dfrac{x_t}{t}=u\;cos\,\theta\right]\)

\(d\overrightarrow{L}=mg(u\;cos\,\theta×t)dt\)

\(\overrightarrow{L}=\displaystyle\int\limits_0^t mg\;u\,cos\,\theta \;t\;dt\)

\(\overrightarrow{L}=\dfrac{mg\;u\,cos\,\theta \;t^2}{2}\)

image

A particle of mass m is projected from point O with speed u at an angle \(\theta\) with the horizontal. Find the angular momentum of the particle with respect to O, at any time t.

A

\(\dfrac{mg\;u\,cos\,\theta \;t^2}{2}\)

.

B

\(\dfrac{mg\;t^2}{2\,cos\,\theta}\)

C

\(mg\,u\,cos\,\theta\)

D

\(\dfrac{mg\,cos\,\theta }{t^2}\)

Option A is Correct

Calculation of Angular Momentum about the Center of a Circle in a Conical Pendulum

Consider a point mass at P, of mass m and velocity \(\overrightarrow{v}\). The point mass makes angle \(\theta\) with the x-axis, as shown in figure.

2-Dimensional View of x-z Plane (Plane of Circle)

Here, \(\overrightarrow{r}=r\;cos\,\theta\;\hat i + r\;sin\,\theta \;\hat k\)

\(\overrightarrow{v}\) is perpendicular to \(\overrightarrow{r}\) in circular motion :

\(\overrightarrow{v}=v\;sin\,\theta\;\hat i - v\;cos\,\theta \;\hat k\)

Angular Momentum

\(\overrightarrow{L}=m(\overrightarrow{r}×\overrightarrow{v})\)

\(=m(r\;cos\,\theta\;\hat i + r\;sin\,\theta \;\hat k)×(v\;sin\,\theta\;\hat i - v\;cos\,\theta \;\hat k)\)

\(=m(rv\;cos^2\,\theta\;\hat j + rv\;sin^2\,\theta \;\hat j)\)

\(=mrv(sin^2\,\theta +cos^2\,\theta)\hat j\)

\(=mrv\;\hat j\)

\(\because \overrightarrow{L}\) is independent of \(\theta\) and is constant.

\(\therefore \dfrac{d\overrightarrow L}{dt}=0\)

Torque on the particle about point C

Net force perpendicular to the plane of motion is zero.

\(\therefore Tcos\,\theta_o=mg\)

\(\therefore\) their torque gets canceled out.

Also, \(Tsin\,\theta_o\) passes through C.

\(\therefore \tau_c=0\)

\(\implies \overrightarrow{\tau}_c=\dfrac{d\overrightarrow{L}_c}{dt}=0\)

Illustration Questions

Find the angular momentum and torque of a particle moving with uniform speed \(\overrightarrow{v}\) in conical pendulum about C, when it is at position A, as shown in figure.

A \(\overrightarrow{L}=mvr\; \hat j,\overrightarrow{\tau}=0\)

B \(\overrightarrow{L}=mvr\;sin\,\theta\; \hat j,\overrightarrow{\tau}=0\)

C \(\overrightarrow{L}=mvr\; \hat k,\overrightarrow{\tau}=0\)

D \(\overrightarrow{L}=mvr\;sin\,\theta\; \hat k,\overrightarrow{\tau}=0\)

×

Position vector, \(\overrightarrow{r}=r\;\hat i\)

Velocity, \(\overrightarrow{v}=-v\;\hat k\)

image

Angular momentum :

\(\overrightarrow{L}=m(\overrightarrow{r}×\overrightarrow{v})\)

\(=m[r \;\hat i×(-v\;\hat k)]\)

\(=mvr\;\hat j\)

 

Magnitude of angular Momentum :

\(|\overrightarrow{L}|=mvr\)

Direction of angular momentum :

\(\overrightarrow{L}\) is along \(+\hat j\)

Torque about point C :

image

Now, resolving T along the radius and perpendicular to plane of circle.

\(\because\) Sum of forces perpendicular to the plane of motion is zero,

\(\therefore\displaystyle\sum \overrightarrow{F}_{\perp} =0\)

Also, \(Tcos\,\theta=mg\)

image

Torque, \(\overrightarrow{\tau}\)  = Force × perpendicular distance

\(\overrightarrow{\tau}=r×mg-(Tcos\theta)r+Tsin\theta×0=0\)

\(\therefore \overrightarrow{\tau}=\dfrac{d\overrightarrow{L}}{dt}=0\)

Find the angular momentum and torque of a particle moving with uniform speed \(\overrightarrow{v}\) in conical pendulum about C, when it is at position A, as shown in figure.

image
A

\(\overrightarrow{L}=mvr\; \hat j,\overrightarrow{\tau}=0\)

.

B

\(\overrightarrow{L}=mvr\;sin\,\theta\; \hat j,\overrightarrow{\tau}=0\)

C

\(\overrightarrow{L}=mvr\; \hat k,\overrightarrow{\tau}=0\)

D

\(\overrightarrow{L}=mvr\;sin\,\theta\; \hat k,\overrightarrow{\tau}=0\)

Option A is Correct

Practice Now