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.

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.

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.

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.

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.

A \(H_{max}\)

B Range, R

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

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

Angular momentum is the rotational equivalent of linear momentum.

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

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.

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

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.

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

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 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.

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

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

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

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

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.

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

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

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.

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

- 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.

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**

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The rate of change of angular momentum is Torque.

Formula of Torque :

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

\(=mrv\;\hat j\)

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

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

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.

\(\therefore \tau_c=0\)

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

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