Answer
$L'(t)=\tau(t)$
When $\tau(t)=0$, then $L'(t)=\tau(t)=0$. This means that $L$ is constant.
Work Step by Step
Since, $L(t)=mr(t) \times v(t)$
Apply differentiation rule, we get
$L'(t)=mr'(t) \times v(t)+mr(t) \times v'(t)$
$\implies L'(t)=mv(t) \times v(t)+mr(t) \times a(t)$
$\implies L'(t)=0+r(t) \times (ma(t))$
$\implies L'(t)=r(t) \times F(t)$
As we know, Torque,$\tau(t)=r(t) \times F(t)$
Hence, $L'(t)=\tau(t)$
When $\tau(t)=0$, then $L'(t)=\tau(t)=0$. This means that $L$ is constant.
