Question

A particle of mass $10 \, g$ moves along a circle of the radius $\frac{1}{\pi } \, cm$ with a constant tangential acceleration. What is the magnitude of this acceleration (in $m \, s^{- 1}$ ) if the kinetic energy of the particle becomes equal to $8\times 10^{- 4} \, J$ by the end of the second revolution after the beginning of the motion? (Particle starts from rest)

Answer 1

Tangential acceleration $a_{t}=r\alpha =$ constant $=K$
$\alpha =\frac{K}{r}$
At the end of the second revolution, angular velocity is $\omega $ then
$\omega ^{2}-\omega _{0}^{2}=2\alpha \theta $
$\omega^2-0^2=2\left(\frac{K}{r}\right)(4 \pi)$
$\omega ^{2}=\frac{8 \pi K}{r}$
K.E. of the particle is $=K.E.=\frac{1}{2}mv^{2}$
$K.E.=\frac{1}{2}mr^{2}\omega ^{2}$
$K.E.=\frac{1}{2}m\left(r^{2}\right)\left(\frac{8 \pi K}{r}\right)$
$8\times 10^{- 4}=\frac{1}{2}\times 10^{- 4}\times 8\times \frac{1}{\pi }\times \pi \times K$
$\Rightarrow K=2ms^{- 2}$