Angular Momentum In Case of Rotation about a Fixed Axis:
We know that `"L" = "r"xx"p"`
Now, `"v"=omega"r"` ⇒`"L"=mr^2omega`
`"L" = "I"omega`
Connservation of angular momentum:
We know that, `"dL"/"dt"=d/dt("I"omega)="r"`
If `tau_"ext"=0 "then" "I"omega="constant"`
When a person is rotating with hands stretched, so the moment of inertia will be more because distribution of mass is far from the axis of rotation.
As the person brings his arms close to body, moment of inertia decreases because the mass is now distributed close to axis.
In this situation no external torque is applied, means angular momentum is conserved.
Iω = constant.
As I decrease, angular velocity ω
Example- A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to `2/5`times the initial value? Assume that the turntable rotates without friction.
Solution: When the child folds his hand back, there is no external torque involved, so the angular momentum is conserved in this case.
We can apply the equation
Hence we can find `omega`