A particle of mass m is released from point P at x = x _0 on the … - Vidyadip

Question

A particle of mass $m$ is released from point P at $x = x _0$ on the X -axis from origin O and falls vertically along the Y-axis, as shown in Fig.

i. Find the torque $t$ acting on the particle at a time $t$ when it is at point Q with respect to O .
ii. Find the angular momentum L of the particle about O at this time t .
iii. Show that $\tau=\frac{d L}{d t}$ in this example.

✓

Answer

i. The force of gravity, $F = mg$ produces the torque $\tau$. Let $\vec{r}$ be the position vector of Q . Then the magnitude of the torque is given by
$
\begin{aligned}
& \tau=r F \sin \theta \\
& =r \times m g \times \frac{x_0}{r}=m g x_0 \quad\left[\because \sin \theta=\frac{x_0}{r}\right]
\end{aligned}
$
The direction of the torque is directed into the plane of paper and perpendicular to it, as shown by $\otimes$.
ii. The magnitude of the angular momentum is $L=r p \sin \theta=r m v \sin \theta$
But the velocity v at point Q is given by $v = u + at =0+ gt = gt$
$
\therefore \quad L=r m g t . \frac{x_0}{r}=m g x_0 t
$
The direction of angular momentum is the same as that of torque.
iii. Now $L = mgx _0 t$
Differentiating both sides with respect to $f$, we get
$
\frac{d L}{d t}=\frac{d}{d t}\left(m g x_0 t\right)=m g x_0=\tau
$
Hence the relation $\tau=\frac{d L}{d}$ holds in this example.

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