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

- 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.

- 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.

- 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

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

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

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

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

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

- 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.

Total angular momentum of body about A

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

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

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

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

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

- 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

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

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

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

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

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

- Consider a body on which forces F
_{1}, F_{2}and F_{3 }are acting, as shown in figure.

- Suppose torque about A is to be calculated.

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

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

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.

- 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.

\(L_A =L_B\)

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

- 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

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.

- 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.

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.

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

- 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

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.

- 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.

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.

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

- 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 m
_{1}and bullet is m_{2}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,

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

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