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 $=K.E.=\frac{1}{2}m{v}^2$
$K.E.=\frac{1}{2}m{r}^2{\omega }^2$
$\text{K}.\text{E}.=\frac{1}{2}\text{m}(r^2)(\frac{8\pi K}{r})=\frac{1}{2}mr(8\pi K)$
$8\times 10^{-4}=\frac{1}{2}\times 10\times 10^{-3}\times 6.4\times 10^{-2}\times 8\times 3.14\times \text{K}$
$K=\frac{2}{6.4\times 3.14}=0.1m{s}^{-2}$