Question

The centre of mass of a disc of radius $\frac{8}{\sqrt{5}}m$ is moving with a velocity of $4m{s}^{-1}$ 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

Answer: 5

Velocity of point ${}^{\prime }{P}^{\prime }=\sqrt{V_0^2+{\left(R{\left(\omega \right)}_0\right)}^2}$

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