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 $8 \times 10^{-4} \, J$ by the end of the second revolution after the beginning of the motion ?

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

Answer 1

Answer: (D)

or, $\frac{1}{2} mv ^{2}=8 \times 10^{-4}$
or, $\frac{1}{2} \times 10 \times 10^{-3} v ^{2}=8 \times 10^{-4}$
or, $v ^{2}=16 \times 10^{-2} \Rightarrow v =0.4 m / s$
initial velocity of particle, $u =0 m / s$
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, $v ^{2}=2 as$
$a=\frac{v^{2}}{2 s}=\frac{(0.4)^{2}}{2(4 \pi r)}$
$=\frac{16 \times 10^{-2}}{\left(8 \times 3.14 \times 6.4 \times 10^{-2}\right)}$
$=0.0995 m / s ^{2} \approx 0.1 m / s ^{2}$