Question

The centre of mass of a disc of radius $\frac{8}{\sqrt{5}}$ m is moving with a velocity of $4\text{m/s}$ on a horizontal plane. The angular velocity of the disc about it’s centre is $\sqrt{5}\text{rad/s}$. Find the radius of curvature of the point $‘P^{\prime }$ shown in the figure in meter.

Answer 1

Velocity of point P $=\sqrt{V_0^2+(R{\omega }_0{)}^2}$
$=\sqrt{4^2+8^2}$


$V_p=4\sqrt{5}\text{m/s}$
$a_{centripetal}=\frac{(8{)}^2\sqrt{5}}{8}=8\sqrt{5}{\text{m/s}}^2$
And $a\perp =a_{cp}\sin \theta =16{\text{m/s}}^2$
$R=\frac{(V_p{)}^2}{a_{\perp }}=5\text{m}$