Question

A particle is rotating in a circle of radius $R=\frac{2}{\pi }m,$ with constant speed $1m{s}^{-1}.$ Match the following two columns for the time interval when it completes $\frac{1}{4}th$ of the circle.

Answer 1

(A) Since speed is constant therefore average speed is going to be same
A $\to$ 4
(B) Average velocity $=\frac{\text{total displacement}}{\text{total time}}$

Displacement is equal to the hypotenuse made by the two radii at ${90}^o$

Displacement $AB=\sqrt{(OA{)}^2+(OB{)}^2}=\sqrt{R^2+R^2}=\sqrt{2R^2}$

$\Rightarrow AB=\sqrt{2}R$

Time for $\frac{1}{4}th$ of circle $t_{1/4}=\frac{1}{4}\times TimeperiodT$

Time period = $\frac{circumference}{speed}=\frac{2\pi R}{v}$

$t_{1/4}=\frac{1}{4}\times \frac{2\pi R}{v}=\frac{\pi R}{2v}$

Average velocity $=\frac{\sqrt{2}R}{\left(\frac{\pi R}{2v}\right)}=\frac{2\sqrt{2}v}{\pi }$

$v=1m/s$

So average velocity $=\frac{2\sqrt{2}}{\pi }$

B $\to$ 2

(C) Average acceleration = Centripetal acceleration= $\frac{v^2}{R}=\frac{1}{R}=\frac{\pi }{2}$
C $\to$ 3
(D) Displacement is $\sqrt{2}R=\frac{2\sqrt{2}}{\pi }$
D $\to$ 2