Question

A particle of mass $10g$ moves along a circle of radius $6.4cm$ with a constant tangential acceleration. What is the magnitude of this acceleration 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 ?

• A. $0.15m/s^2$
• B. $0.18m/s^2$
• C. $0.2m/s^2$
• D. $0.1m/s^2$

Answer 1

Answer: (D)

or, $\frac{1}{2}m{v}^2=8\times 10^{-4}$
or, $\frac{1}{2}\times 10\times 10^{-3}{v}^2=8\times 10^{-4}$
or, $v^2=16\times 10^{-2}\Rightarrow v=0.4m/s$
initial velocity of particle, $u=0m/s$
we have to find Tangential acceleration at the end of 2 nd revolution.
total distance covered, $s=2(2\pi r)=4\pi r$
so, $v^2=2as$
$a=\frac{v^2}{2s}=\frac{(0.4{)}^2}{2(4\pi r)}$
$=\frac{16\times 10^{-2}}{\left(8\times 3.14\times 6.4\times 10^{-2}\right)}$
$=0.0995m/s^2\approx 0.1m/s^2$