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Angular Momentum And Conservation Collision

Learn angular momentum in combined motion, collision between rod & particle. Practice direction of torque, total torque, conservation of angular momentum, and rotational motion.

Angular Momentum

Combined Motion

  • Combined motion can be treated as the sum of two motions i.e., motion of center of mass (translational motion) and pure rotation about center of mass.
  • The motion of a moving wheel is an example of combined motion. 

Angular Momentum 

  • It is the measure of angular motion contained in the body.
  • It is calculated differently for point object and body (made up of infinite points).
  • It is denoted by L.

Angular momentum of point object

  • Consider a point object O of mass m, moving with velocity v, as shown in figure.

  • Its angular momentum about point O can be given as-

L = mv (perpendicular distance from A to the line of velocity)

\(\Rightarrow L = m\,v\,r_\bot\)

  • Its direction can be judged by taking a handycam at A and shoot the point object.
  • The direction in which camera is rotated, is the direction of angular momentum. 
  • Here, camera is rotated in clockwise direction.
  • Thus, the direction of angular momentum is clockwise.

Angular momentum of a body

  • To calculate the angular momentum of a body, two types of motion are to be considered, whether the body is hinged about a point or not.

\(Case - I\)

If a rigid body is hinged at a point then the angular momentum is given by 

\(L = I_\text{hinge} \,\omega\)

where \(I_{hinge}\) is moment of inertia of the body about hinge 

\(\omega\) is angular velocity of body

\(Case -II\)

If the rigid body is not hinged about any point i.e., it is in combined motion, then its motion is divided into two motions:

1. Linear or translational motion of center of mass.

2. Rotational motion when body is hinged about center of mass.

  • So, the angular momentum of the body when it is in combined motion, is the sum of angular momentum of the body hinged at center of mass and the angular momentum of the center of mass as point object.

Total angular momentum of body about A

\(= I_c\,\omega + m \,v_\text{cm} \,r_\bot\)

Illustration Questions

A body of mass \(m= 0.2\,kg\), radius \( r=1\,m\) and moment of inertia  \(I = 2\,kg\,m^2\), is moving with linear velocity \(v= 2\,m/s\) and angular velocity \(\omega = 1\,rad/s\). Calculate the angular momentum of the body about A.  

A \(3 \,kg\,m^2 \,s^{-1} \)

B \(2.4 \,kg\,m^2 \,s^{-1} \)

C \(5 \,kg\,m^2 \,s^{-1} \)

D \(6 \,kg\,m^2 \,s^{-1} \)

×

Since, the body is in combined motion

thus, to calculate its angular momentum about A, the motion of body is considered as the sum of rotational and translational motion.

image

Angular momentum of a body about A  = Angular momentum of body hinged at center of mass+Angular momentum of center of mass as point object

\(L_\text{about A} = I_c\,\omega + m\,v_{cm} \,r_\bot\)

Given: \(m= 0.2\,kg,\,r= 1\,m, \,I = 2\,kg\,m^2, \,v_{cm} = 2\,m/s, \,\omega = 1 \,rad/s\) 

\(L _\text{about A} = 2×1+0.2×2×1\)

\(=2+0.4\)

\(=2.4\,kg\,m^2\,s^{-1}\)

image

A body of mass \(m= 0.2\,kg\), radius \( r=1\,m\) and moment of inertia  \(I = 2\,kg\,m^2\), is moving with linear velocity \(v= 2\,m/s\) and angular velocity \(\omega = 1\,rad/s\). Calculate the angular momentum of the body about A.  

image
A

\(3 \,kg\,m^2 \,s^{-1} \)

.

B

\(2.4 \,kg\,m^2 \,s^{-1} \)

C

\(5 \,kg\,m^2 \,s^{-1} \)

D

\(6 \,kg\,m^2 \,s^{-1} \)

Option B is Correct

Angular Momentum of a Body in Combined Motion

Angular momentum of a body

  • To calculate the angular momentum of a body, two types of motion are to be considered, whether the body is hinged about a point or not.

\(Case - I\)

If a rigid body is hinged at a point then the angular momentum is given by 

\(L = I_\text{hinge} \,\omega\)

where \(I_{hinge}\) is moment of inertia of the body about hinge 

\(\omega\) is angular velocity of body

\(Case -II\)

If the rigid body is not hinged about any point i.e., it is in combined motion, then its motion is divided into two motions:

1. Linear or translational motion of center of mass.

2. Rotational motion when body is hinged about center of mass.

  • So, the angular momentum of the body when it is in combined motion, is the sum of angular momentum of the body hinged at center of mass and the angular momentum of the center of mass as point object.

Total angular momentum of body about A

\(= I_c\,\omega + m \,v_\text{cm} \,r_\bot\)

Illustration Questions

A disk of mass m = 1 kg and radius R= 1 m is rolling on a surface with velocity \(v= 3\,m/s\) and angular velocity \(\omega= 2\,rad/s\). Calculate the angular momentum of the disk about A.

A \(4 \,kg \,m^2 \,s^{-1}\)

B \(5 \,kg \,m^2 \,s^{-1}\)

C \(3 \,kg \,m^2 \,s^{-1}\)

D \(2 \,kg \,m^2 \,s^{-1}\)

×

Since, the body is in combined motion

thus, to calculate its angular momentum about A, the motion of body is considered as the sum of rotational and translational motion.

image

Angular momentum of body about A = Angular momentum of body hinged at center of mass+Angular momentum of center of mass as point object

\(L_\text{about A} = I_c\,\omega + m\,v_{cm} \,r_\bot\)

Given: \(m = 1\,kg, \, R= 1\,m , \,v= 3\,m/s , \,\omega = 2\,rad/s\)

Moment of inertia of disk about center of mass

\(I_c = \dfrac{m\,R^2}{2} = \dfrac{(1)(1)^2}{2} = \dfrac{1}{2} kg\,m^2\)

\(L_\text{about A} = \dfrac{1}{2} × 2 + 1 × 1× 3\)

\(= 4 \,kg\,m^2 \,s^{-1}\)

image

A disk of mass m = 1 kg and radius R= 1 m is rolling on a surface with velocity \(v= 3\,m/s\) and angular velocity \(\omega= 2\,rad/s\). Calculate the angular momentum of the disk about A.

image
A

\(4 \,kg \,m^2 \,s^{-1}\)

.

B

\(5 \,kg \,m^2 \,s^{-1}\)

C

\(3 \,kg \,m^2 \,s^{-1}\)

D

\(2 \,kg \,m^2 \,s^{-1}\)

Option A is Correct

Angular Momentum of Rod in Combined Motion

  • To calculate angular momentum of rod, two types of motion are to be considered, whether the body is hinged about a point or not.

\(Case - I:\) Angular momentum of rod hinged at center of mass

  • If a rod is hinged at a point then the angular momentum is given by 

\(L = I_{cm} \,\omega\)

where \(I_{cm}\) is moment of inertia of rod about center of mass

\(\omega\) is angular velocity of rod

\(Case- II:\) Angular momentum of rod not hinged about center of mass

  • If the rod is not hinged about any point i.e., it is in combined motion, then its motion is divided into two motions:

1. Linear motion of center of mass. 

2. Rotational motion when it is hinged about center of mass.

  • So, the angular momentum of the rod, when it is in combined motion, is the sum of angular momentum of the rod hinged at center of mass and the angular momentum of the center of mass as point object.

Total angular momentum of rod about A

\(L = I_c \,\omega + m\,v_{cm} \,r_\bot\)

Illustration Questions

A rod of mass m = 1 kg and length L = 2 m is rolling with velocity \(v= 2 \,m/s\) and angular velocity \(\omega= 1 \,rad/s\). Calculate the angular momentum of the rod about A.

A \(\dfrac{7}{3}\, kg\,m^2 \,s^{-1}\)

B \(5\,kg\,m^2 \,s^{-1}\)

C \(6\, kg\,m^2 \,s^{-1}\)

D \(\dfrac{2}{3}\, kg\,m^2 \,s^{-1}\)

×

Since, the rod is in combined motion

thus, to calculate its angular momentum about A, the motion of rod is considered as the sum of rotational motion and translational motion.

image

Angular momentum of rod about A = Angular momentum of rod hinged at center of mass + Angular momentum of center of mass as point object

\(L_\text{about A} = I_c \,\omega + m\,v_{cm} \,r_\bot\)

Given: \(m= 1\,kg\)\(L= 2\,m\)\(v= 2\,m/s\)\(\omega = 1\,rad/s\)

Moment of inertia of rod about center of mass

 \(I_c = \dfrac{m\,L^2}{12} = \dfrac{1× (2)^2}{12} = \dfrac{1}{3}\, kg \,m^2\)

\(L_\text{about A} = \dfrac{1}{3} × 1 + 1× 2× 1\)

\(= \dfrac{1}{3} +2\)

\(= \dfrac{7}{3} \,kg\,m^2\,s^{-1}\)

image

A rod of mass m = 1 kg and length L = 2 m is rolling with velocity \(v= 2 \,m/s\) and angular velocity \(\omega= 1 \,rad/s\). Calculate the angular momentum of the rod about A.

image
A

\(\dfrac{7}{3}\, kg\,m^2 \,s^{-1}\)

.

B

\(5\,kg\,m^2 \,s^{-1}\)

C

\(6\, kg\,m^2 \,s^{-1}\)

D

\(\dfrac{2}{3}\, kg\,m^2 \,s^{-1}\)

Option A is Correct

Torque about a Point

  • Consider a body on which forces F1, F2 and Fare acting, as shown in figure.

  • Suppose torque about A is to be calculated.

Step 1 

Draw free body diagram (FBD) which shows exact position of each force acting on the body.

Step 2

 Extend each line of force and take perpendicular distance from the point.

Step 3 

Determine the direction of torque, by taking a handycam at A.

The direction in which handycam rotates while shooting, is the direction of torque. 

  • Take either clockwise or anticlockwise as positive direction and other as negative direction. Then calculate total torque on body.
  • If total torque about a point is zero, then the angular momentum of the body about that point is conserved.

\(\because \,\vec \tau = \dfrac{d \,\vec L}{d\,t}\)

if  \(\vec \tau =0\)

then, \(\dfrac{d\,\vec L}{d\,t} =0\)

\(\Rightarrow\vec L = \text{constant}\)

  • To apply conservation of angular momentum, it is really difficult to find at least one point in space about which torque is zero. 

Illustration Questions

A disk of mass m and radius R is given an angular velocity \(\omega\) and gently placed on a horizontal rough surface. About which point, torque is zero?

A C

B A

C Both 

D None

×

Forces acting on body

Three forces are acting on the body :

(i) mg

(ii) Normal force

(iii) Gravitational force

image

Free body diagram of the body

image

Direction of forces about C :

The direction of forces N and mg are opposite to each other and they are equal in magnitude.

Thus, they cancel each other.

image

Direction of forces about A :

image

Perpendicular distance of line of frictional force from A is R and from C is 0.

Thus, torque about \(A = R\,f\)  

Torque about \(C= 0.f\)

\(=0\)

image

Hence, torque about C is zero.

A disk of mass m and radius R is given an angular velocity \(\omega\) and gently placed on a horizontal rough surface. About which point, torque is zero?

image
A

C

.

B

A

C

Both 

D

None

Option A is Correct

Final Velocity of Body when Rolling Starts 

  • To calculate final velocity of body, law of conservation of angular momentum can be used.
  • The law of conservation of angular momentum is not applied on every point but on a point about which torque of system is zero.
  • Thus, firstly find a point where torque of all forces is zero.
  • For example:- The torque about A due to all the forces is zero.

  • Consider a disk which is given an angular velocity \(\omega\) and gently placed on a horizontal rough surface. The mass and radius of disk are m and R, respectively.
  • The value of \(v\,'\) is to be found out as rolling starts.

Step 1:

 Draw the diagrams before and after the event.

Step 2:

Find a point about which torque is zero.

Here, A is the point about which torque is zero.

Step 3:

 Write angular momentum about that point, before and after the event.

Before the event: \(L_B = I_c \,\omega\) 

After the event:  \(L_A = I_c \,\omega\,' + m\,v\,' \,R\)

Here, \(v\,' = \omega\, ' \,R\)  [Rolling constraint]

Step 4:

Equate these angular momentum.

\(L_A =L_B\)

Illustration Questions

Choose the incorrect option.

A \(I_c \,\omega _1 + m(0)\,R = I_c \,\omega_2 + m\,v_2 \,R\)

B Torque about C is zero

C \(v_2 = \omega _2 \,R\)

D \(\omega _1 \,R = 0\)

×

Forces acting on body 

Three forces are acting on body :

(i) mg

(ii) Normal force

(iii) Gravitational force

image

Free body diagram of body

image

Direction of forces about C :

The direction of forces N and mg are opposite to each other and they are equal in magnitude  thus, they cancel each other.

image

Perpendicular distance of line of frictional force from C is zero.

Thus, torque about C is zero.

Hence, option (B) is correct.

image

Angular momentum 

Before rolling: \(L_B = I_c \,\omega _1\)

After rolling:  \(L_A= I_c \,\omega _2 + m\,v_2 \,R\)

Since torque about C is zero thus,

Angular momentum before rolling = Angular momentum after rolling

\(L_B = L_A\)

\(I_c \,\omega_1 =I_c \,\omega_2 + m\,v_2 \,R\)

Hence, option (A) is correct.

When rolling starts, its rolling constraint will be

\(v_2 = \omega_2\,R\)

Hence, option (C) is correct.

Since \(\omega_1\) is non -zero and R is non- zero,

thus,  \(\omega _1 \,R \neq 0\)

Hence, option (D) is incorrect. 

Choose the incorrect option.

image
A

\(I_c \,\omega _1 + m(0)\,R = I_c \,\omega_2 + m\,v_2 \,R\)

.

B

Torque about C is zero

C

\(v_2 = \omega _2 \,R\)

D

\(\omega _1 \,R = 0\)

Option D is Correct

Laws which are Conserved when a Particle Strikes a Rod Perpendicularly and Elastically

  • In elastic collision between rod and particle, the kinetic energy of system before event is equal to the kinetic energy of system after event.
  • During elastic collision, following three laws are applicable:

1. Law of conservation of linear momentum

2. Law of conservation of angular momentum

3. Law of conservation of kinetic energy

1. Law of Conservation of Linear Momentum

This law is applicable only when there is no external force acting on the system.

For example:

  • If a body is hinged, external force acts on it and thus, this law is not applicable.
  • If a body is not hinged, no external force acts on it thus, this law can be applied. 

2. Law of Conservation of Angular Momentum

  • If torque about a point is zero for a system, then law of conservation of angular momentum can be applied.
  • If a body is hinged, then this law can be applied about hinge, as all the external forces act on the hinge.

3. Law of Conservation of Kinetic Energy

Law of conservation of kinetic energy is only applied when there is elastic collision because in elastic collision the kinetic energy before collision is equal to the kinetic energy after collision. 

Illustration Questions

A rod is hinged as shown in figure. If a ball collides elastically with the rod, then choose the incorrect option.

A Angular momentum of system is conserved about hinge

B Linear momentum of system is conserved

C Kinetic energy before collision = Kinetic energy after collision

D Torque about hinge is zero during collision

×

Since the collision between rod and ball is elastic, thus

1. Angular momentum of system is conserved about hinge as all external forces act on hinge.

2. Kinetic energy before collision is equal to kinetic energy after collision.

3. Torque about hinge is zero during collision.

Hence, option (B) is incorrect as law of conservation of linear momentum is not applied on hinged bodies.

A rod is hinged as shown in figure. If a ball collides elastically with the rod, then choose the incorrect option.

image
A

Angular momentum of system is conserved about hinge

.

B

Linear momentum of system is conserved

C

Kinetic energy before collision = Kinetic energy after collision

D

Torque about hinge is zero during collision

Option B is Correct

Elastic Collision between Rod and Particle

  • In elastic collision between rod and particle, the kinetic energy of system before event is equal to the kinetic energy of system after event.
  • During elastic collision, following three laws are applicable:

1. Law of conservation of linear momentum

2. Law of conservation of angular momentum

3. Law of conservation of kinetic energy

1. Law of conservation of linear momentum

This law is applicable only when there is no external force acting on the system.

For example:

  • If a body is hinged, external force acts on it and thus, this law is not applicable.
  • If a body is not hinged, no external force acts on it and thus, this law can applied. 

2. Law of conservation of angular momentum

  • If torque about a point is zero for a system, then law of conservation of angular momentum can be applied.
  • If a body is hinged, then this law can be applied about hinge, as all the external forces act on the hinge.

3. Law of conservation of kinetic energy

Law of conservation of kinetic energy is only applied when there is elastic collision because  in elastic collision, the kinetic energy before collision is equal to the kinetic energy after collision. 

Illustration Questions

A rod of length L and mass m1 is hinged, as shown in figure. If a bullet of mass m2 strikes the rod elastically with velocity \(u\), then choose the incorrect option.

A \(m_2 \,u \,\dfrac{2L}{3} = m_2 \,v \left(\dfrac{2L}{3}\right) + \dfrac{m_1\,L^2}{3} \omega\)

B \(m_1 \,u \,\left(\dfrac{2L}{3}\right) = m_1 \,v \left(\dfrac{2L}{3}\right) + \dfrac{m_2\,L^2}{3} \omega\)

C \(\dfrac{1}{2}m_2 \,u^2 =\,\dfrac{1}{2} (m_1+m_2) \left[\omega\left(\dfrac{2L}{3}\right)\right]^2 \)

D \(m_2 \,v = m_2 \,u + m_1 \left(\dfrac{\omega L}{2}\right) \)

×

Applying law of conservation of angular momentum

\(m_2 \,u \,\left(\dfrac{2L}{3}\right) = m_2 \,v \left(\dfrac{2L}{3}\right) + \left(\dfrac{m_1\,L^2}{3}\right) \omega\)

Applying law of conservation of linear momentum

\(m_2 \,v \ = m_2 \,u + m_1\left(\dfrac{\omega\,L}{2}\right) \)

Since the collision is elastic, thus 

\((KE)_\text{before}=(KE)_\text{after}\)

\(\dfrac{1}{2} m_2\,u^2 = \dfrac{1}{2} (m_1 +m_2) \left[\omega\left(\dfrac{2L}{3}\right)\right]^2\)

image

Hence, option (B) is incorrect.

image

A rod of length L and mass m1 is hinged, as shown in figure. If a bullet of mass m2 strikes the rod elastically with velocity \(u\), then choose the incorrect option.

image
A

\(m_2 \,u \,\dfrac{2L}{3} = m_2 \,v \left(\dfrac{2L}{3}\right) + \dfrac{m_1\,L^2}{3} \omega\)

.

B

\(m_1 \,u \,\left(\dfrac{2L}{3}\right) = m_1 \,v \left(\dfrac{2L}{3}\right) + \dfrac{m_2\,L^2}{3} \omega\)

C

\(\dfrac{1}{2}m_2 \,u^2 =\,\dfrac{1}{2} (m_1+m_2) \left[\omega\left(\dfrac{2L}{3}\right)\right]^2 \)

D

\(m_2 \,v = m_2 \,u + m_1 \left(\dfrac{\omega L}{2}\right) \)

Option B is Correct

Perfectly Inelastic Collision of a Bullet with a Stationary Rod

  • Consider a rod of length \(\ell\) hinged at one end. A bullet collides with the rod with velocity \(v_1\). The rod is stationary before collision.
  • After the collision, the angular momentum of rod becomes \(\omega\).
  • The final velocity of rod and bullet becomes same, as bullet sticks to the rod. 
  • The value of \(v_2\) becomes equal to \(\omega × \text{(distance of bullet from hinge)}\)

  • In this event,
  • Angular momentum remains conserved.
  • Linear momentum remains conserved.
  • Kinetic energy is not conserved as the collision is inelastic.
  • If the mass of rod is m1 and bullet is m2 then applying law of conservation  of linear momentum, 

\(m_2 \,v = m_2 \,u + m_1 \,\omega \ell\)

  • Applying law of conservation of angular momentum,

\(m_2 \,u(\ell) = m_2 \,v(\ell)\)

Illustration Questions

A rod of length L and mass m1 is hinged, as shown in figure. If a bullet of mass m2  strikes the rod inelastically with velocity u and sticks to the rod, then choose the incorrect option.

A \(m_2\,u \left(\dfrac{2L}{3}\right) = m_2 \,v \left(\dfrac{2L}{3}\right)+\dfrac{m_1\,L^2}{3} \omega\)

B \(v= \omega \,L\)

C \(v= \omega \left(\dfrac{2L}{3}\right)\)

D \(m_2 \,v = m_2 \,u + m_1 \left(\dfrac{\omega \,L}{2}\right)\)

×

Since the collision is inelastic, thus 

On applying law of conservation of linear momentum,

\(m_2\,v = m_2 \,u + m_1 \,\omega \left(\dfrac{L}{2}\right)\)

On applying law of conservation of angular momentum,

\(m_2 \,u\left(\dfrac{2L}{3}\right)=m_2 \,v\left(\dfrac{2L}{3}\right)+\dfrac{m_1\,L^2}{3} \omega\)

Since the bullet sticks to the rod, thus \(v= \omega \left(\dfrac{2L}{3}\right)\)

image

Hence, option (B) is incorrect.

image

A rod of length L and mass m1 is hinged, as shown in figure. If a bullet of mass m2  strikes the rod inelastically with velocity u and sticks to the rod, then choose the incorrect option.

image
A

\(m_2\,u \left(\dfrac{2L}{3}\right) = m_2 \,v \left(\dfrac{2L}{3}\right)+\dfrac{m_1\,L^2}{3} \omega\)

.

B

\(v= \omega \,L\)

C

\(v= \omega \left(\dfrac{2L}{3}\right)\)

D

\(m_2 \,v = m_2 \,u + m_1 \left(\dfrac{\omega \,L}{2}\right)\)

Option B is Correct

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