Question

prove that the rate of change of angular momrntum of a particle is equal to da torque acting on it.

Answer 1

Consider a particle of mass m and linear momentum at a p bar position r bar relative to the origin O. The angular momentum L of the particle with respect to the origin O is defined to be

L bar= r (bar) × p (bar)

Magnitude of the angular momentum vector is

L = rp sinθ

Where, p is the magnitude of  p(bar) and θ is the angle between r(bar)and p(bar)

L bar= r (bar) × p (bar)

Differentiating with respect to time,
dL÷dt=d÷dt { r (bar) × p (bar)}

Then, the velocity of the particle is v= dr(bar)÷ dt and p= mv(bar)

Because of this, (dr÷dt)×p(bar)=mv=0

∴Equation (i) becomes

r×(dp÷dt)=r×F=tou bar

Hence, 

dL÷dt=torque

I'd suggest you to properly go through the study material. The same has been explained there.

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