Question

A particle of mass $10g$ 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

Answer: 2

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 $=K.E.=\frac{1}{2}m{v}^2$
$K.E.=\frac{1}{2}m{r}^2{\omega }^2$
$K.E.=\frac{1}{2}m\left(r^2\right)\left(\frac{8\pi K}{r}\right)$
$8\times 10^{-4}=\frac{1}{2}\times 10^{-4}\times 8\times \frac{1}{\pi }\times \pi \times K$
$\Rightarrow K=2m{s}^{-2}$