Obtain the relation between angular momentum of a particle and torque acting on it.

Explanation

Rotational kinetic energy of a rigid body
`K=(1)/(2)Iomega^(2)`
The rate at which work done on the body equal to the rate at which kinetic energy increases.
The rate of increases of kinetic energy is
`(dK)/(dt)=(d)/(dt)((1)/(2)Iomega^(2))`
`P=(1)/(2)I(d)/(dt)(omega^(2))` [`because` I = constant]
`therefore P=(1)/(2)Ixx2omega(domega)/(dt)`
`therefore P=Iomegaalpha" "[because (domega)/(dt)=alpha]`
but `P=tauomega`
`therefore tauomega=Iomegaalpha`
`therefore tau=Ialpha`
This equation is similar to Newton.s second law for linear motion `F=ma`
Hence, `alpha=(tau)/(I)`
The angular is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body.