A particle moves along a circle of radius $\left(\frac{10}{\pi}\right) \mathrm{m}$ with constant tan

Question

A particle moves along a circle of radius $\left(\frac{10}{\pi}\right) \mathrm{m}$ with constant tangential acceleration. If the velocity of the particle is $60 \mathrm{~m} / \mathrm{s}$ at the end of the third revolution after motion has begun, the tangential acceleration is

Professor PC Thomas Classes · Answer

(a) : Given : $r=\frac{10}{\pi} \mathrm{~m}, v=60 \mathrm{~m} / \mathrm{s}, 	heta=3 \mathrm{~rev}=6 \pi \mathrm{~rad}$. \ From equation of motion $v^{2}=u^{2}+2 a_t s \Rightarrow v^{2}=2 a_t (r 	heta)$ (since $u=0$) \ $a_t=\frac{v^{2}}{2 r 	heta}=\frac{(60)^2}{2 	imes (\frac{10}{\pi}) 	imes 6\pi} = \frac{3600}{120} = 30 \mathrm{~m} / \mathrm{s}^{2}$