1. Set up the motion of the insect

Let the rod have length $L$. Since the insect starts from $O$ and moves with constant speed $v$ relative to the rod, its distance from the pivot at time $t$ is

$r(t)=vt, \qquad 0\le t\le T$

where

$T=\frac{L}{v}.$

After $t=T$, it reaches the end and stops relative to the rod, so its distance becomes constant:

$r=L \quad (t>T).$

2. Angular speed remains constant

The whole system rotates with constant angular speed $\omega$. Hence the insect has angular motion with the same angular speed $\omega$ while it is moving outward.

In polar coordinates, for a particle with coordinates $(r,\theta)$ and $\dot\theta=\omega=$ constant,

$\vec a=(\ddot r-r\omega^2)\hat r+(r\ddot\theta+2\dot r\omega)\hat \theta.$

Here,

• $\dot r=v$ for $0<t<T$
• $\ddot r=0$
• $\dot\theta=\omega$
• $\ddot\theta=0$

So,

$\vec a=(-r\omega^2)\hat r+(2v\omega)\hat \theta.$

3. Torque about the pivot due to the insect

Torque about $O$ is

$\vec \tau = \vec r \times m\vec a.$

Since $\vec r=r\hat r$, only the transverse ($\hat\theta$) component of acceleration contributes to torque. Therefore,

$|\tau| = r\,(m a_\theta)=r\,(m\cdot 2v\omega).$

Thus,

$|\tau|=2m\omega v\, r.$

Using $r=vt$,

$|\tau|=2m\omega v^2 t, \qquad 0\le t\le T.$

So the torque increases linearly with time from zero.

4. What happens after the insect reaches the end?

For $t>T$, the insect stops relative to the rod, so

$\dot r=0.$

Then the tangential acceleration becomes

$a_\theta=2\dot r\omega=0.$

Hence,

$|\tau|=0, \qquad t>T.$

So at $t=T$, the torque suddenly drops from

$|\tau(T^-)|=2m\omega v^2T=2m\omega vL$

to zero.

5. Graph description

Therefore the torque-time graph is:

• starts from zero at $t=0$
• increases linearly with $t$ up to $t=T$
• suddenly becomes zero for $t>T$

This corresponds to Option B.

6. Comparison with stored answer

Derived answer: B
Stored correct answer: B

They agree.