Question

An ant travels along a long rod with a constant velocity $\bar{u}$ relative to the rod starting from the origin. The rod is kept initially along the positive x-axis. At t = 0, the rod also starts rotating with an angular velocity $\omega$ (anticlockwise) in x-y plane about origin. Then

• A. The position of the ant at any time t is $\bar{r} = ut[\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}]$.
• B. The speed of the ant at any time t is $u\sqrt{1+\omega^2t^2}.$
• C. The magnitude of the tangential acceleration of the ant at any time t is $\frac{\omega^2 t u}{\sqrt{1+\omega^2t^2}}$.
• D. The speed of the ant at any time t is $\sqrt{1+2\omega^2t^2}\;u.$

Answer 1

Answer: (A), (B), (C)

The position of the ant at any time t is $\bar{r} = ut[\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}]$.; The speed of the ant at any time t is $u\sqrt{1+\omega^2t^2}.$; The magnitude of the tangential acceleration of the ant at any time t is $\frac{\omega^2 t u}{\sqrt{1+\omega^2t^2}}$.

Answer 2

1. Position vector
Since the ant moves a distance $ut$ along the rod which itself has rotated by angle $\omega t$,

$$\bar r(t)=ut\bigl(\cos\omega t\,\hat i+\sin\omega t\,\hat j\bigr).$$

2. Speed
Differentiate $\bar r(t)$ to get velocity $\bar v$. One finds

$$|\bar v|^2 = u^2(1+\omega^2t^2) \quad\Longrightarrow\quad |\bar v|=u\sqrt{1+\omega^2t^2}.$$

3. Tangential acceleration
The tangential component is $d|\bar v|/dt$:

$$a_{\rm tan} =\frac{d}{dt}\bigl(u\sqrt{1+\omega^2t^2}\bigr) =\frac{\omega^2t\,u}{\sqrt{1+\omega^2t^2}}.$$