Question

At t₁ = 2.00 s, the acceleration of a particle in counter clockwise circular motion is It moves at constant speed. At time t₂ = 5.00 s, the particle's acceleration is What is the radius of the path taken by the particle if t₂ - t₁ is less than one period?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$At{t}_1=2s,\overrightarrow{a}=\left(5\sqrt{3}\frac{m}{s^2}\right)\hat{i}+5\hat{j}$
$At{t}_2=5s,\overrightarrow{a}=\left(5\frac{m}{s^2}\right)\hat{i}-\left(5\sqrt{3}\frac{m}{s^2}\right)\hat{i}+\hat{j}$
Given the body is moving in anti-clockwise direction

Since,the speed is constant, net acceleration will always be directed toward the centre and since it is

This will be the only path possible using this, we can calculate ω as we can find the change in angle


$tan{\theta }_1=\frac{1}{5\sqrt{3}}=\frac{1}{3}$
$tan{\theta }_2=\frac{3\sqrt{3}}{5}=\sqrt{3}$
$\therefore {\theta }_1={30}^{\circ }$
$\Rightarrow {\theta }_2={60}^{\circ }$

$\therefore$ We can see that angle between the vectors is ${90}^{\circ }$ which means particle would have covered an angle of $360-90={270}^{\circ }{=}^{3\pi /2}radin3s$
$\therefore \omega =\frac{\Delta \theta }{\Delta t}=\frac{\frac{3\pi }{2}}{3}=\frac{\pi }{2}rad/s$
$a={\omega }^{2}r$
$r=\frac{a}{{\omega }^2}=\frac{\sqrt{(5\sqrt{3}{)}^2}+5^2}{{\left(\frac{\pi }{2}\right)}^2}=\frac{\sqrt{100}}{\frac{{\pi }^2}{4}}$
=$\frac{4\times 10}{{\pi }^2}=\frac{40}{{\pi }^2}$