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Angular Momentum in Case of Rotation About a Fixed Axis

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Angular Momentum In Case of Rotation about a Fixed Axis:

We know that `"L" = "r"xx"p"`

`"L"="rx"("mv")`

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"`

Example:

  • 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

`Iomega="constant"`

`Iomega=(2/5I)omega'`

Hence we can find `omega`

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