Question

A wheel of radius $r=1\text{m}$ is rolling with an angular velocity $\omega =\frac{\pi }{3}\text{rad/s}$ about its center and velocity of its center of mass is $v_{com}=2\text{m/s}$. Point $P(0,0)$ is in contact with ground initially, find its height (in $\text{m}$) from the surface after time $t=1\text{s}$.

• A. $0.4$
• B. $0.3$
• C. $0.5$
• D. $0.6$

Answer 1

The correct option is C $0.5$


Height of the point will be the displacement in $y-$axis.
So, analysing only rotational motion about COM,
We know that displacement of point $P$ in $y-$ direction is given as $\Delta y=r-r\cos \theta$
also,
$\theta =\omega \times t=(\frac{\pi }{3}\times 1)\text{rad}$
$\Rightarrow \Delta y=1-1\times \cos \frac{\pi }{3}\Delta y=(1-\frac{1}{2})\hat{j}=0.5\hat{j}\text{m}$
$\therefore$ Height of point $P$ after $1\text{s}=0.5\text{m}$