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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/$s^2$
Explanation
Use v from KE, then v² = 2as along the circular path.
Detailed Solution
$\frac{1}{2}mv^2 = 8\times10^{-4}$ J $\Rightarrow v^2 = \frac{2\times8\times10^{-4}}{10\times10^{-3}} = \frac{16}{100} \Rightarrow v = 0.4$ m/s
Distance in two revolutions $s = 2\times2\pi R = 4\pi R$
$v^2 = u^2 + 2as \Rightarrow \left(\frac{4}{10}\right)^2 = 0 + 2a\times4\pi\times6.4\times10^{-2}$
$a = \frac{16}{100\times8\pi\times6.4\times10^{-2}} = \frac{160}{8\times3.14\times64} \approx 0.1\ m/s^2$
Distance in two revolutions $s = 2\times2\pi R = 4\pi R$
$v^2 = u^2 + 2as \Rightarrow \left(\frac{4}{10}\right)^2 = 0 + 2a\times4\pi\times6.4\times10^{-2}$
$a = \frac{16}{100\times8\pi\times6.4\times10^{-2}} = \frac{160}{8\times3.14\times64} \approx 0.1\ m/s^2$