Question

A wheel of radius $1m$ is rolling purely on a flat, horizontal surface. It's centre is moving with a constant horizontal acceleration = $3m/s^2$. At a moment when the centre of the wheel has a velocity $3m/s$ then find the acceleration of a point $1/3m$ vertically above the centre of the wheel.

Answer 1

$a_{cm}=r\alpha$

$V=r\omega$

$\therefore \omega =\frac{v}{r}$

$\therefore a=\sqrt{(a_t+a_{cm}{)}^2+a{}_r^2}$

Here $a_t=\frac{1}{3}\alpha$

$a_r=\frac{1}{3}{\omega }^2$

$a_{cm}=3m/s^2$

$v=3m/s$

Putting the value, $a=5m/s^2$