Question

A wheel of radius $r=1\text{m}$ rolls without slipping with an angular velocity of $\omega =\frac{\pi }{4}\text{rad/s}$ about its centre and velocity of its centre of mass is $v_{COM}=2\text{m/s}$. If the point $P(0,0)$ is in contact with the ground initially, find its position vector after time $t=1\text{s}$.

• A. $(1-\frac{1}{\sqrt{2}})\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$
• B. $(1-\frac{1}{2})\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$
• C. $\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$
• D. $(2-\frac{1}{\sqrt{2}})\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$

Answer 1

The correct option is D $(2-\frac{1}{\sqrt{2}})\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$


We know that displacement of point $P$ in $x-$ direction is given as $\Delta x=(v_{COM}\times t)\hat{i}+\left(r\sin \theta \right)(-\hat{i})$
also,
$\theta =\omega \times t=\frac{\pi }{4}\text{rad}$
$\Rightarrow \Delta x=(2\times 1-1\times \sin \frac{\pi }{4})\hat{i}\Rightarrow \Delta x=(2-\frac{1}{\sqrt{2}})\hat{i}$

We know that displacement of point $P$ in $y-$ direction is given as $\Delta y=r-r\cos \theta$
$\Rightarrow \Delta y=1-1\times \cos \frac{\pi }{4}\Rightarrow \Delta y=(1-\frac{1}{\sqrt{2}})\hat{j}$

$\therefore$ Position vector of point $P$ at time $t=1\text{s}$ is
$(2-\frac{1}{\sqrt{2}})\hat{i}+(1-\frac{1}{\sqrt{2}})\hat{j}$