Question

A wheel is rolling without slipping on ground along a straight line. If the axis of the wheel has a linear speed $v$, the instantaneous velocity of point $P$ at an angle $\theta$ from vertical on the rim, relative to the ground will be

• A. $v\cos \left(\frac{\theta }{2}\right)$
• B. $2v\cos \left(\frac{\theta }{2}\right)$
• C. $v(1+\sin \theta )$
• D. $v(1+\cos \theta )$

Answer 1

The correct option is B $2v\cos \left(\frac{\theta }{2}\right)$

The velocity at point $P$ on the wheel will be a vector sum of translational velocity of $\text{COM}$ of wheel and tangential velocity of point $P$ due to rotational motion.


In case of pure rolling, $v=r\omega ....\left(i\right)$
$\Rightarrow$The angle between translational velocity $\left(v\right)$ and tangential velocity $\left(r\omega \right)$ of point $P$ is $\theta$ as shown in figure.
$\Rightarrow |{\overrightarrow{v}}_P|=\sqrt{v^2+(r\omega {)}^2+2(v\left)\right(r\omega )\cos \theta }....\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$:
$|{\overrightarrow{v}}_P|=\sqrt{r^2{\omega }^2+\left(r^2{\omega }^2\right)+2\left(r^2{\omega }^2\right)\cos \theta }$
$|{\overrightarrow{v}}_P|=r\omega \sqrt{2(1+\cos \theta )}$
$\because 1+\cos \theta =2{\cos }^2\frac{\theta }{2}$
$\Rightarrow |{\overrightarrow{v}}_P|=r\omega \sqrt{2\times 2{\cos }^2\frac{\theta }{2}}$
$\therefore |{\overrightarrow{v}}_P|=2r\omega \cos \frac{\theta }{2}=2v\cos \left(\frac{\theta }{2}\right)$