Question

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

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

Answer 1

The correct option is D $0.1{\text{m/s}}^2$

Given $m=10\text{g}={10}^{-2}\text{kg}$
$R=6.4\text{cm}=6.4\times {10}^{-2}\text{m},$
Final Kinetic energy after two revolutions $K_f=8\times {10}^{-4}\text{J}$
$\frac{1}{2}m{v}^2=8\times {10}^{-4}\text{J}$
${10}^{-2}{v}^2=16\times {10}^{-4}$
$v=0.4\text{m/s}$
Final velocity after two revolutions $v=0.4\text{m/s}$
Given that tangential acceleration is constant, let it be $a$
Using kinematics equation,
$v^2-u^2=2as$
here,
Initial velocity $u=0$
Final velocity $v=0.4\text{m/s}$
Distance covered for two revolutions is $s=4\pi R$

$\Rightarrow {0.4}^2=2a\left(4\pi R\right)$
$\Rightarrow 0.16=8\pi a\left(0.064\right)$
$\Rightarrow a=\frac{0.16}{0.512\pi }=0.099\approx 0.1{\text{m/s}}^2$