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

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

Answer 1

Answer: (C)

Tangential acceleration $a_{t}=r\alpha =$ constant $=K$
$\alpha =\frac{K}{r}$
At the end of second revoluation angular velocity is $\omega $ then
$\omega ^{2}-\omega _{0}^{2}=2\alpha \theta $
$\omega^2=2\left(\frac{\mathrm{K}}{r}\right)(4 \pi)$
$\omega ^{2}=\frac{8 \pi K}{r}$
K.E. of the particle is $=K.E.=\frac{1}{2}mv^{2}$
$K.E.=\frac{1}{2}mr^{2}\omega ^{2}$
$\text{K}.\text{E}.=\frac{1}{2}\textit{m}\left(\right. r^{2} \left.\right)\left(\right. \frac{8 \pi K}{r} \left.\right)=\frac{1}{2}mr\left(\right. 8 \pi K \left.\right)$
$8 \times 10^{- 4} = \frac{1}{2} \times 10 \times 10^{- 3} \times 6.4 \times 10^{- 2} \times 8 \times 3.14 \times \text{K}$
$K=\frac{2}{6.4 \times 3.14}=0.1ms^{- 2}$