Question

Due to slipping, points $A$ and $B$on the rim of the disk have the velocities shown. Determine the velocities of the centre point $C$ and point $F$ at this instant.

Answer 1

$v+R\omega =1.5.....\left(i\right)$ and $R\omega -v=3.0....\left(ii\right)$
From Eq. (i) and (ii), we have
$R\omega =2.25m/s$ and
$v=-0.75m/s$
Thus, velocity of point $C$ is $0.75m/s$ (towards left).
$v_F=\sqrt{v^2+(R\omega {)}^2+2v(R\omega \left)\cos \right(90+30)}$
$=\sqrt{0.5625+5.0625+2\times 0.75\times 2.25\times (-\frac{1}{2})}$
$=1.98m/s$