Hello dear we have know that in translational motion, linear momentum is provided by mass and velocity of particle in motion. Thus the linear momentum is considered as the product of mass and velocity.

In rotational motion, inertia is provided by the moment of inertia. Linear velocity of particle is replaced by angular velocity, and then the momentum in rotational motion is also replaced by the angular momentum.

Angular momentum of a rotating particle is a vector quantity defined as the product of linear momentum and perpendicular distance of particle from the axis of rotation.

**Let’s find the angular momentum of rotating body……………!**

__Angular momentum__**:**

__Angular momentum__

Consider a body consists n^{th} number of particles of masses m_{1}, m_{2,} ………………. m_{n }situated at distances r_{1}, r_{2} …………… r_{n} respectively from the axis of rotation passing through point ‘O’ is rotating with constant angular velocity ‘ω’ as shown below.

The particle of mass m_{1} having linear velocity v_{1} = r_{1 }_{ω}

The linear momentum is P_{1} = m_{1}v_{1}

∴ P_{1} = m_{1}r_{1}ω_{1}

The angular momentum i.e. moment of linear momentum is product of linear momentum.

∴ L_{1}= P_{1}r_{1}

∴ L_{1}= m_{1}r_{1}ω.r_{1}

The sum of angular momentum of particle is **called angular momentum of body.**

L = L_{1} + L_{2} + ………….. + L_{n}

i.e. Thus angular momentum of a body is equal to the product of M.I. of body & its angular velocity.

S.I. unit is **Kg.m ^{2}/s**

CGS unit is **g.cm ^{2}/s**

Dimensions of L = **[M ^{1}L^{2}T^{-1}]**

**Let’s try to solve following numerical……………!**

1.) A ring of mass 50 g and radius of 4 cm rotating about its centre at 240 rpm decreases to 60 rpm. Find the change in angular momentum of disc.

Solution:

Here, m= 50 g =0.05 kg, R= 4 cm = 0.04 cm

f_{1}= 240 rpm= 4 rps, f_{2}= 60 rpm= 1 rps,