### Equation of the Angular momentum

Objects which moving straight have momentum that can be calculated using the equation:

p = m v

p = momentum (kg m/s), m = mass (kg), v = speed (m/s)

The quantity of the rotational motion, which is identical to mass (m) in the linear motion, is the moment of inertia (I). The quantity of the rotational motion, which is identical to the velocity (v) in the linear motion, is the angular velocity (ω). Thus, the rotating object has angular momentum that can be calculated using the equation:

L = I ω

L = angular momentum (kg m^{2}/s), I = moment of inertia (kg m^{2}), ω = angular velocity (rad/s)

6.2. Sample problems of the angular momentum

Sample problem 1.

A particle with a mass of 0.5 grams moves in a circle with a constant angular velocity of 2 rad/s. Determine the angular momentum of the particle if the radius of the particle’s path is 10 cm.

Solution:

The moment of inertia of the particle:

I = m r^{2 }= (0.5 x 10^{-3 }kg)(1 x 10^{-1} m)^{2 }= (0.5 x 10^{-3} kg)(1 x 10^{-2 }m^{2}) = 0.5 x 10^{-5} kg m^{2}

The angular speed of the particle:

ω = 2 rad/s

The angular momentum of the particle:

L = (0.5 x 10^{-5 }kg m^{2})(2 rad/s) = 1 x 10^{-5} kg m^{2}/s

6.3 The law of conservation of angular momentum

The law of conservation of angular momentum states that if the resultant moment of force on a rigid body when rotating is zero, then the angular momentum of the rigid body when rotating is always constant. The law of conservation of angular momentum can be derived mathematically by modifying the equation of Newton’s second law of angular momentum. Here is the equation of Newton‘s second law on the angular momentum:

If the resultant moment of force is zero, then the equation above changes to:

I_{t }= the final moment of inertia, I_{o} = the initial moment of inertia, ω_{t }= the final angular speed, ω_{o} = the initial angular speed, L_{t }= the final angular momentum, L_{o }= the initial angular momentum.

### 6.4. Sample problems of the law of conservation of angular momentum

Sample problem 1.

A homogeneous solid cylinder disk is initially rotating on its axis with a speed of 4 rad/s. The mass and radius of the disk are 1 kg and 0.5 m. If above the plate is placed a ring that has a mass and radius of 0.2 kg and 0.1 m and the center of the ring,

just above the center of the disk, then the disc and ring will rotate together with the angular velocity of…..

Solution:

Moment of inertia of solid cylinder : I = 1⁄2 m r^{2 }= 1⁄2 (1 kg)(0.5 m)^{2 }= (0.5)(0.25) = 0.125 kg m^{2}

Moment of inertia of ring : I = m r^{2 }= (0.2 kg)(0.1 m)^{2 }= (0.2)(0.01) = 0.002 kg m^{2}

Initial angular momentum (L_{1}) = Final angular momentum (L_{2})

I_{1} ω_{1} = I_{2} ω_{2}

(0.125 kg m^{2})(4 rad/s) = (0.125 kg m^{2 }+ 0.002 kg m^{2})(ω_{2})

(0.5) = (0.127)(ω_{2})

ω_{2} = 0.5 : 0.127

ω_{2 }= 4 rad/s