A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleratio

Question

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 	imes 10^{-4} \mathrm{~J}$ by the end of the second revolution after the beginning of the motion?\

Professor PC Thomas Classes · Answer

(c): Here, $m=10 \mathrm{~g}=10^{-2} \mathrm{~kg}, R=6.4 \mathrm{~cm}=6.4 	imes 10^{-2} \mathrm{~m}$, $K_{f}=8 	imes 10^{-4} \mathrm{~J}, K_{i}=0, a_{t}=$ ?\ Using work energy theorem,\ Work done by all the forces = Change in KE\ $W_{	ext {tangential force }}+W_{	ext {centripetal force }}=K_{f}-K_{i}$\ $\begin{aligned} \Rightarrow a_{t} & =\frac{K_{f}}{4 \pi R m}=\frac{8 	imes 10^{-4}}{4 	imes \frac{22}{7} 	imes 6.4 	imes 10^{-2} 	imes 10^{-2}} \ & =0.099 \approx 0.1 \mathrm{~m} \mathrm{~s}^{-2}\end{aligned}$ $$ =0.099 \approx 0.1 \mathrm{~m} \mathrm{~s}^{-2} $$