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Angular Momentum Of Body And Conservation Laws

Learn law of conservation of angular momentum. Practice to calculate conservation of angular momentum when external torque is zero and relation between moment of inertia and angular velocity.

Angular Momentum of a Rigid Body Rotating about a Fixed Axis

• Consider a body, rotating with angular velocity $$\omega$$ about an axis, as shown in figure.
• The velocity of each particle on body is different.

Angular Momentum of $$i^{th}$$ Particle

Angular momentum of $$i^{th}$$ particle is given by:

$$L_i=m_iv_ir_i$$

where, $$r_i$$ is the perpendicular distance of $$i^{th}$$ particle from the axis and $$v_i$$ is the velocity of $$i^{th}$$ particle.

Angular momentum of the body

If axis of rotation is same for all particles:

$$L_{body}=\Sigma L_i$$

$$=\Sigma m_iv_ir_i$$

$$=\Sigma m_i(r_i\omega)r_i$$

$$L_{body}=(\Sigma m_ir_i^2)\omega$$

$$L_{axis}=I_{axis}\,\omega$$

A ring of mass $$m$$ and radius $$r$$, rotates about an axis passing through its circumference, perpendicular to the plane, as shown in figure. If the angular velocity is $$\omega$$, find angular momentum ($$L_{axis}$$).

A $$2mr^2\omega$$

B $$3mr^2\omega$$

C $$\dfrac{1}{2}mr^2\omega$$

D $$4mr^2\omega$$

×

Calculation of $$I_{axis}$$

Moment of inertia of ring about a perpendicular axis passing through center of mass,

$$I_0=mr^2$$

Applying Parallel-Axis Theorem,

$$I=I_0+mr^2$$

$$=mr^2+mr^2$$

$$=2mr^2$$

Calculation of $$L_{axis}$$(angular momentum)

$$L_{axis}=I_{axis}\;\omega$$

$$=2mr^2\omega$$

A ring of mass $$m$$ and radius $$r$$, rotates about an axis passing through its circumference, perpendicular to the plane, as shown in figure. If the angular velocity is $$\omega$$, find angular momentum ($$L_{axis}$$).

A

$$2mr^2\omega$$

.

B

$$3mr^2\omega$$

C

$$\dfrac{1}{2}mr^2\omega$$

D

$$4mr^2\omega$$

Option A is Correct

Kinetic Energy of a System when Moment of Inertia is Changed

• If net torque acting on a body is zero, the angular momentum remains conserved.

$$\vec{\tau}_{net}=0$$

$$\dfrac{dL}{dt}=0$$

• If the moment of inertia of a body is changed without changing the axis of rotation i.e., mass distribution about the axis is changed,

then, $$I_1\omega_1=I_2\omega_2$$

Kinetic Energy of a Rotating Body

• Kinetic energy of a rotating body is given by

$$K=\dfrac{1}{2}I\omega^2$$

$$\Rightarrow K=\dfrac{1}{2}(I\omega)\omega$$  or  $$\dfrac{1}{2} \dfrac{(I^2\omega^2)}{I}$$

$$\Rightarrow K=\dfrac{1}{2}L\omega$$  or  $$\dfrac{1}{2} \dfrac{L^2}{I}$$

• When angular momentum ($$L$$) is conserved, then moment of inertia of a body increases as angular velocity decreases and hence its kinetic energy decreases.

If moment of inertia of a rotating rod gets doubled due to its elongation by some internal process, then choose the correct option.

A Its angular velocity gets doubled and kinetic energy remains same.

B Its angular velocity remains same and kinetic energy gets doubled.

C Its angular velocity and kinetic energy gets halved.

D Its angular velocity and kinetic energy gets doubled.

×

Given : $$I_2=2I_1$$

If external torque on a system is zero, then angular momentum remains conserved.

$$I_1\omega_1=I_2\omega_2$$

$$I_1\omega_1=2I_1\omega_2$$    [$$\because I_2=2I_1$$]

$$\omega_2=\dfrac{\omega_1}{2}$$

Kinetic energy, $$K=\dfrac{1}{2}I\omega^2$$

$$K_{initial}=\dfrac{1}{2}I_1\omega_1^2$$

$$K_{final}=\dfrac{1}{2}I_2\omega_2^2$$

$$=\dfrac{1}{2}(2I_1)\left(\dfrac{\omega_1}{2}\right)^2$$

$$=\dfrac{1}{2}\left(\dfrac{1}{2}I_1\omega_1^2\right)$$

$$=\dfrac{1}{2}K_{initial}$$

If moment of inertia of a rotating rod gets doubled due to its elongation by some internal process, then choose the correct option.

A

Its angular velocity gets doubled and kinetic energy remains same.

.

B

Its angular velocity remains same and kinetic energy gets doubled.

C

Its angular velocity and kinetic energy gets halved.

D

Its angular velocity and kinetic energy gets doubled.

Option C is Correct

When a Body is Thrown Tangentially from a Stationary Disk

• Angular momentum of individual disks change due to friction which is an internal force for the system of two disks.

$$\vec{\tau}_{sys}=0$$

A kid having mass $$M$$ stands at the edge of a platform of radius $$R$$. This platform can be freely rotated about its axis. The moment of inertia of the platform is $$I$$. The kid holds a ball of mass $$m$$ when the system is at rest. If the kid throws the ball tangentially with velocity $$v$$, what is the angular velocity of the platform after the event?

A $$\dfrac{mvR}{I}$$

B $$\dfrac{mvR}{MR^2+I}$$

C $$\dfrac{2mv}{MR}$$

D $$\dfrac{mv}{MR}$$

×

Torque, $$\tau_{ext}=0$$

$$L_{sys}$$ remains conserved.

$$L_i=L_f$$

$$0=mvR-(I+MR^2)\omega$$

on solving,

$$\omega=\dfrac{mvR}{MR^2+I}$$

A kid having mass $$M$$ stands at the edge of a platform of radius $$R$$. This platform can be freely rotated about its axis. The moment of inertia of the platform is $$I$$. The kid holds a ball of mass $$m$$ when the system is at rest. If the kid throws the ball tangentially with velocity $$v$$, what is the angular velocity of the platform after the event?

A

$$\dfrac{mvR}{I}$$

.

B

$$\dfrac{mvR}{MR^2+I}$$

C

$$\dfrac{2mv}{MR}$$

D

$$\dfrac{mv}{MR}$$

Option B is Correct

Identification of Motion of a Body

Translational motion

• If the velocity of all the particles of a body in 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 with 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.

Which of the following bodies are performing translational motion?

A

B

C

D

×

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

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

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.

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

Which of the following bodies are performing translational motion?

A
B
C
D

Option C is Correct

Relation between Moment of Inertia and Angular Velocity

• When torque along any axis is zero, then angular momentum about that axis remains constant.
• If  $$\tau_{axis}=0$$

$$\Rightarrow \dfrac{d\;L_{axis}}{dt}=0$$

$$\Rightarrow L_{axis}$$ = constant

• Thus, $$L_{axis}$$ is conserved.
• $$I_\omega$$ is constant.
• If mass distribution of a rotating body with respect to axis of rotation changes during rotation, then its moment of inertia changes and therefore, its angular velocity changes.
• If moment of inertia increases, angular velocity decreases and if moment of inertia decreases, angular velocity increases.

The moment of inertia of a rotating body is doubled about the same axis of rotation by some mechanism. No external torque is applied. Its angular velocity will :

A Double

B Remain same

C Reduce to half

D Become 4 times

×

$$\tau_{axis}=0$$

$$\therefore$$ $$L_{axis}$$ is conserved

$$\Rightarrow I_1\omega_1=I_2\omega_2$$

Given : $$I_2=2I_1$$

$$\therefore \;I_1\omega_1 = 2I_1\omega_2$$

$$\omega_2=\dfrac{\omega_1}{2}$$

The moment of inertia of a rotating body is doubled about the same axis of rotation by some mechanism. No external torque is applied. Its angular velocity will :

A

Double

.

B

Remain same

C

Reduce to half

D

Become 4 times

Option C is Correct

Conservation of Angular Momentum when External Torque is Zero

• Consider two points on a body.

here,

$$\vec{F_{12}}$$ is a force applied on particle 1 due to particle 2

$$\vec{F_{21}}$$ is a force applied on particle 2 due to particle 1

• The forces $$\vec{F_{12}}$$ and $$\vec{F_{21}}$$ will act along the line joining particles 1 and 2.
• By Newton's IIIrd law -

if $$\vec{F_{12}}=\vec{F}$$

then $$\vec{F_{21}}=-\vec{F}$$

Torque on the Body due to these Two Internal Forces

Torque, $$\tau_0=\vec{r_1}×\vec{F_{12}}+\vec{r_2}×\vec{F_{21}}$$

$$=\vec{r_1}×\vec{F}+\vec{r_2}×(-\vec{F})$$

$$=\vec{r_1}×\vec{F}-\vec{r_2}×\vec{F}$$

$$=(\vec{r_1}-\vec{r_2})×\vec{F}$$

$$\because$$ The vector $$(\vec{r_1}-\vec{r_2})$$ is along $$\vec{F}$$

$$\therefore\tau_0=0$$

Torque due to all Pairs of Internal Forces

Torque $$\vec{\tau}$$ due to all pairs of internal forces about any point is zero.

$$\therefore\vec{\tau}_{int}=0$$

• If external torque is zero,

$$\vec{\tau}=\dfrac{d\vec{L}}{dt}$$

$$\Rightarrow\sum \vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}$$

If $$\sum \vec {\tau}_{ext}=0$$

then $$L_{sys}$$ is constant.

Thus, if net external torque on a system is zero, its angular momentum remains conserved.

A wheel of moment of inertia $$(I_1)\,0.1\,kg\;m^2$$ is rotating about a shaft at an angular velocity $$(\omega_1)$$ $$160\;rev/min$$. A second stationary wheel of moment of inertia $$(I_2)\,0.06\,kg\;m^2$$  is coupled to the same shaft so that both the wheels finally rotate with a common angular speed ($$\omega\,'$$). Find the common angular speed.

A $$200\;rev/min$$

B $$100\;rev/min$$

C $$300\;rev/min$$

D $$150\;rev/min$$

×

Given :

Moment of inertia of wheel 1, $$I_1=0.1\,kg\;m^2$$

Angular velocity of wheel 1,   $$\omega_1=160\;rev/min$$

Moment of inertia of wheel 2, $$I_2=0.06\,kg\;m^2$$

If external torque on a system is zero, then angular momentum remains conserved.

$$I_1\omega_1=I'\omega\,'$$

$$\because I'=I_1+I_2$$

$$I_1\omega_1=(I_1+I_2)\omega\,'$$

$$0.1×160=(0.1+0.06)\omega\,'$$

$$\omega\,'=\dfrac{16}{0.16}$$

$$\omega\,'=100\;rev/min$$

A wheel of moment of inertia $$(I_1)\,0.1\,kg\;m^2$$ is rotating about a shaft at an angular velocity $$(\omega_1)$$ $$160\;rev/min$$. A second stationary wheel of moment of inertia $$(I_2)\,0.06\,kg\;m^2$$  is coupled to the same shaft so that both the wheels finally rotate with a common angular speed ($$\omega\,'$$). Find the common angular speed.

A

$$200\;rev/min$$

.

B

$$100\;rev/min$$

C

$$300\;rev/min$$

D

$$150\;rev/min$$

Option B is Correct

Angular Velocity of Rod when a Bullet Moving Horizontally, Embeds into it, Tangentially

External torque $$(\vec{\tau}_{ext})$$

$$\Sigma\vec{\tau}_{ext}=\dfrac{d\vec{L}_{sys}}{dt}$$

• If  $$\vec{\tau}_{ext}=0$$

then $$\vec{L}_{sys}$$ is a constant.

• If net external torque on a system is zero, then angular momentum of the system remains conserved.

Angular Momentum during Collision

• During collision or impulsive collision events, the impulsive force and its torque become internal to the system and do not affect the momentum and angular momentum of the system.
• Thus, angular momentum remains conserved.

A rod $$AB$$ of mass $$M$$ and length $$L$$ is hinged at point $$A$$ and hangs vertically. A bullet of mass $$m$$, moving with velocity $$v$$ gets embedded in the rod at point $$P$$, as shown in figure. Find the angular velocity of the rod after the event.

A $$\dfrac{mv}{(m+M)L}$$

B $$\dfrac{2mv}{ML}$$

C $$\dfrac{2mv}{(M+m)L}$$

D $$\dfrac{6mv}{(3M+4m)L}$$

×

Given :

Mass of bullet = $$m$$

Length of rod = $$L$$

Mass of rod = $$M$$

Velocity of bullet = $$v$$

$$\vec{\tau}_{ext}=0$$

$$\Rightarrow L_{sys}$$ remains conserved.

$$L_i=L_f$$

$$mv{r_{\perp}}=I\omega$$

$$mv\dfrac{2L}{3}=\left[\dfrac{ML^2}{3}+m\left(\dfrac{2L}{3}\right)^2\right]\omega$$

on solving,

$$\omega=\dfrac{6mv}{3ML+4mL}$$

A rod $$AB$$ of mass $$M$$ and length $$L$$ is hinged at point $$A$$ and hangs vertically. A bullet of mass $$m$$, moving with velocity $$v$$ gets embedded in the rod at point $$P$$, as shown in figure. Find the angular velocity of the rod after the event.

A

$$\dfrac{mv}{(m+M)L}$$

.

B

$$\dfrac{2mv}{ML}$$

C

$$\dfrac{2mv}{(M+m)L}$$

D

$$\dfrac{6mv}{(3M+4m)L}$$

Option D is Correct

Conservation of Angular Momentum

As $$\overrightarrow{\tau}_{ext}=0$$

$$L_{sys}$$ remains conserved.

• When the event is impulsive

When the event is impulsive, then the forces become internal and do not produce any torque. Thus, angular momentum of a body does not change and remains conserved.

• Mechanical energy is conserved

Mechanical energy of a system is conserved if there are no external forces and the work done by internal non-conservative forces, is zero.

• Mechanical energy is not conserved

Mechanical energy of a system is not conserved during impulsive situations like collision, jumping and landing.

Conservation of Mechanical Energy can be Applied

The conservation of mechanical energy can be applied after the collision/event has taken place and dissipative and non-conservative forces are absent.

A rod $$AB$$ of mass $$(M)\,4\,kg$$ and length $$(L)\,3\,m$$ is hinged at point A and hangs vertically. A bullet of mass $$(m)\,1\,kg$$ moving with velocity $$(v)\,8\,m/s$$ gets embedded in the rod at point $$P$$, as shown in figure. Find the maximum angle rotated by the rod after the event. ( $$g=10\,m/s^2$$)

A $$\dfrac{\pi}{2}$$

B $$\dfrac{\pi}{6}$$

C $$\dfrac{\pi}{3}$$

D $$\dfrac{\pi}{4}$$

×

Given :

Mass of bullet, $$m=1\,kg$$

Velocity of bullet, $$v=8\,m/s$$

Mass of rod, $$M=4\,kg$$

Length of rod, $$L=3\,m$$

$$\vec{\tau}_{ext}=0$$

$$\Rightarrow L_{sys}$$ remains conserved.

$$L_i=L_f$$

$$mvr_\perp=I\omega$$

$$mv\dfrac{2L}{3}=\left[\dfrac{ML^2}{3}+m\left(\dfrac{2L}{3}\right)^2\right]\omega$$

on solving,

$$\omega=\dfrac{6mv}{3ML+4mL}$$

on putting values,

$$\omega=\dfrac{6×1×8}{3(4)(3)+4(1)(3)}$$

$$\omega=1\;rad/sec$$

After the event has taken place,

conservation of mechanical energy

$$K_{initial}+P_{initial}=K_{final}+P_{final}$$

Initial kinetic energy

$$K_{initial}=\dfrac{1}{2}I\omega^2$$

$$=\dfrac{1}{2}\left\{\dfrac{ML^2}{3}+m\left(\dfrac{2L}{3}\right)^2\right\}\omega^2$$

$$=\dfrac{1}{2}\left\{\dfrac{4×3^2}{3}+1×2^2\right\}1^2$$

$$=8\,J$$

Taking point $$A$$ as the reference for gravitational potential energy

{(Potential energy)A=0}

Initial potential energy,

$$P_{initial}=-mg\;X_{cm}$$

$$=-mg\left[\dfrac{m×\dfrac{2L}{3}+M×\dfrac{L}{2}}{m+M}\right]$$

$$=-1×10\left[\dfrac{1×2+4×1.5}{5}\right]$$

$$=-16\,J$$

Final kinetic energy,

$$K_{final}=0$$

The angle is maximum, when the rod comes momentarily to rest.

Final potential energy,

$$P_{final}=-mg\;X_{cm}\;cos\;\theta_0$$

$$=-16\;cos\;\theta_0\,J$$

By Mechanical Energy Conservation

$$K_{initial}+P_{initial}=K_{final}+P_{final}$$

$$-16+8=-16cos\;\theta_0+0$$

$$cos\;\theta_0=\dfrac{1}{2}$$

$$\theta_0=\dfrac{\pi}{3}$$

A rod $$AB$$ of mass $$(M)\,4\,kg$$ and length $$(L)\,3\,m$$ is hinged at point A and hangs vertically. A bullet of mass $$(m)\,1\,kg$$ moving with velocity $$(v)\,8\,m/s$$ gets embedded in the rod at point $$P$$, as shown in figure. Find the maximum angle rotated by the rod after the event. ( $$g=10\,m/s^2$$)

A

$$\dfrac{\pi}{2}$$

.

B

$$\dfrac{\pi}{6}$$

C

$$\dfrac{\pi}{3}$$

D

$$\dfrac{\pi}{4}$$

Option C is Correct