Question

A disc of radius R is under a pure rolling condition on a horizontal surface with a constant velocity. Radius of curvature of point P that is $\frac{R}{2}$ distance above C is $\frac{nR}{2}$. Then, n is

Answer 1

$a_C=\frac{v^2}{R_C}\Rightarrow R_C=\frac{v^2}{a_C}ForpointP,v=\omega (R+\frac{R}{2})=\frac{3}{2}R\omega and{a}_C={\omega }^2\frac{R}{2}\therefore R_C=\frac{{\left(\frac{3}{2}R\omega \right)}^2}{{\omega }^2\frac{R}{2}}=\frac{9}{2}R$