Answer
It has been proved that $L'(t)=\tau(t)=0$ when $\tau(t)=0$ and $L$ is constant.
Work Step by Step
Angular linear momentum is given as: $L(t)=mr(t) \times v(t)$
Need to use differentiation cross product rule.
This suggests that $L'(t)=mr'(t) \times v(t)+mr(t) \times v'(t)$
Thus, $ L'(t)=mv(t) \times v(t)+mr(t) \times a(t)$
and $ L'(t)=0+r(t) \times (ma(t))$
Finally, $ L'(t)=r(t) \times F(t)$
Use the definition of torque. Torque,$\tau(t)=r(t) \times F(t)$
Therefore, $L'(t)=\tau(t)$
It has been proved that $L'(t)=\tau(t)=0$ when $\tau(t)=0$ and $L$ is constant.
