JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Rotational Mechanics

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A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5 s is close to

(A) $2.0 \times 1 0^{- 5} Nm$ ✓ Correct
(B) $4.0 \times 1 0^{- 6} Nm$
(C) $1.6 \times 1 0^{- 5} Nm$
(D) $7.9 \times 1 0^{- 6} Nm$

Answer: (A)

Solution

The coin is mounted so that the rotation axis

A B

lies along a tangent in its own plane, giving a moment of inertia

I = \frac{m r ^{2}}{4} + m r^{2} = \frac{5 m r ^{2}}{4} = \frac{5 \times ( 5 \times 1 0 ^{- 3} ) \times ( 1 0 ^{- 2} ) ^{2}}{4} = 6.25 \times 1 0^{- 7} kg m^{2}

. Spinning up to

25

rotations per second means

ω = 25 \times 2 π = 50 π rad s^{- 1}

reached in

5 s

, so the torque needed is

τ = \frac{I ω}{t} = \frac{6.25 \times 1 0 ^{- 7} \times 50 π}{5} \approx 2.0 \times 1 0^{- 5} Nm

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Rotational Mechanics

Solution

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