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Physics Question on System of Particles & Rotational Motion

A metal coin of mass 5 g5\,g and radius 1 cm1\, cm is fixed to a thin stick ABAB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about ABAB at 2525 rotations per second in 5 s5\, s, is close to :

A

4.0×10−64.0 \times 10^{-6} Nm

B

2.0×10−52.0 \times 10^{-5} Nm

C

1.6×10−51.6 \times 10^{-5} Nm

D

7.9×10−67.9 \times 10^{-6} Nm

Answer

2.0×10−52.0 \times 10^{-5} Nm

Explanation

Solution

α=ΔωΔt=25×2π5=10πrad/sec⁡2\alpha =\frac{\Delta\omega}{\Delta t} = \frac{25\times2\pi}{5} =10\pi \text{rad}/\sec^{2}
τ=(54MR2)α\tau = \left(\frac{5}{4} MR^{2}\right)\alpha
=54×5×10−3×(10−2)2×10π= \frac{5}{4} \times5\times 10^{-3} \times \left(10^{-2}\right)^{2} \times 10\pi
=1.9625×10−5Nm= 1.9625 \times 10^{-5} Nm
≃2.0×10−5Nm\simeq 2.0 \times 10^{-5} Nm