Question

The centre of mass of a disc of radius $\frac{8}{\sqrt{5}}m$ is moving with a velocity of $4\,ms^{- 1}$ on a horizontal plane. The angular velocity of the disc about its centre is $\sqrt{5}\,rad\,s^{- 1}$ . Find the radius of curvature of the point $P$ shown in the figure (in meter).

Answer 1

Velocity of point $'P'=\sqrt{V_{0}^{2} + \left(R \left(\omega \right)_{0}\right)^{2}}$

$V_{P}=4\sqrt{5}\,ms^{- 1}$
now, centripetal acceleration,
$a_{\text {centripetal }}=\frac{v^{2}}{r}=\frac{(8)^2}{\frac{8}{\sqrt{5}}}=8 \sqrt{5} \,m s ^{-2}$
and $a_{\bot}=a_{cp}sin\theta =16\,ms^{- 2}$
so the radius of the curvature,
$R=\frac{\left(V_{p}\right)^{2}}{a_{\bot}}=5\,m$