Question

A disk is rolling [pure] on a flat surface. Match for radius of curvatures of various points A, B, C & D on disk. R is radius of disk & $OC=\frac{R}{2}$
$\begin{array}{cccc}\text{(P)}& \text{Point A}& \left(1\right)& \text{Below C}\\ \text{(Q)}& \text{Point B}& \left(2\right)& \text{Above A}\\ \text{(R)}& \text{Point C}& \left(3\right)& \text{4 R}\\ \left(S\right)& \text{Point D}.& \left(4\right)& \text{0}\\ & & \left(5\right)& \text{0.5 R}\end{array}$

• A. P $\to$ 3, Q $\to$ 1, R $\to$ 5, S $\to$ 4,
• B. P $\to$ 5, Q $\to$ 1, R $\to$ 2, S $\to$ 4,
• C. P $\to$ 2, Q $\to$ 1, R $\to$ 5, S $\to$ 4,
• D. P $\to$ 3, Q $\to$ 1, R $\to$ 5, S $\to$ 2,

Answer 1

The correct option is A P $\to$ 3, Q $\to$ 1, R $\to$ 5, S $\to$ 4,

Centripetal acceleration of any point on disc is $a_c=w^{2}r$ here r is distance of point from 0.
Also $a_c=\frac{v_p^2}{r_c}$ here $v_p$ is velocity of point from ground & $r_c$ is radius of curvature.
also V = RW
Point A $V_p$ = 2 V $\Rightarrow$ $W^{2}R=\frac{4V^2}{r_c}\Rightarrow r_c=4R$
Point B $V_p$ = $\frac{3}{2}$ V $\Rightarrow$ $W^2\frac{R}{2}=\frac{9}{4}\frac{V^2}{r_c}\Rightarrow r_c=\frac{9}{2}R$
Point C $V_p$ = $\frac{V}{2}$ $\Rightarrow$ $W^2\frac{R}{2}=\frac{V^2}{4r_c}\to r_c=0.5R$
For point D $V_p=0\Rightarrow r_c=0$