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Deduce the relation between angular momentum and angular velocity. |
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Answer» Solution :Let US consider the point mass m is at a distance r from the axis of rotation. Its linear momentum at any instant is tangential to the circular path. Then the angular momentum `vecL` is perpendicular to `vecr` and `vecp`. Hence it is directed along the axis of rotation. The angle `theta` between `vecr` and `vecp` in this case is `90^(@)`. The magnitude of the angular momentum L could be written as `LK=r mv sin 90^(@)=r mv` where v is the linear velocity. The relation between linear velocity v and angular velocity `OMEGA` in a circular MOTION is `v=r omega`. Hence `L=r m r omega` `L=(mr^(2))omega` THe directions OFL and `omega` are along the axis of rotation. the abvoe EXPRESSION can be written in the vector notation as `vec=L(=mr^(2))vec(omega)` `vec(L)=(Sigma m_(i)r_(i)^(2)vec(omega)` `vecL=Ivec(omega)` |
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