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Rotational Motion question

2019 · 10 Apr · Shift 2 · Q61
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Rotational Motion question

2019 · 10 Apr · Shift 2 · Q61

JEE MainPhysicsRotational MotionMCQ+4 / −1
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 5s, is close to : JEE Main 2019 (Online) 10th April Evening Slot Physics - Rotational Motion Question 164 English
  1. A
    7.9 × 10–6 Nm
  2. B
    4.0 × 10–6 Nm
  3. C
    2.0 × 10–5 Nm
  4. D
    1.6 × 10–5 Nm
View written solutionFree

Correct answer: B: 4.0 × 10–6 NM

  1. Given data
  • Mass of coin: m=5 g=5×10−3 kgm = 5\,\text{g} = 5 \times 10^{-3}\,\text{kg}m=5g=5×10−3kg
  • Radius of coin: R=1 cm=10−2 mR = 1\,\text{cm} = 10^{-2}\,\text{m}R=1cm=10−2m
  • Final rotational speed: f=25 rev/sf = 25\,\text{rev/s}f=25rev/s
  • Time taken: t=5 st = 5\,\text{s}t=5s

The coin is fixed to a thin stick along a diameter ABABAB, so the system rotates about the axis ABABAB lying in the plane of the coin.

  1. Find angular velocity
ω=2πf=2π×25=50π rad/s\omega = 2\pi f = 2\pi \times 25 = 50\pi\,\text{rad/s}ω=2πf=2π×25=50πrad/s

Since it starts from rest,

α=ω−0t=50π5=10π rad/s2\alpha = \frac{\omega - 0}{t} = \frac{50\pi}{5} = 10\pi\,\text{rad/s}^2α=tω−0​=550π​=10πrad/s2
  1. Moment of inertia of the coin about diameter ABABAB

For a disc about any diameter in its plane,

I=14mR2I = \frac{1}{4}mR^2I=41​mR2

So,

I=14(5×10−3)(10−2)2I = \frac{1}{4}(5 \times 10^{-3})(10^{-2})^2I=41​(5×10−3)(10−2)2 I=14(5×10−3×10−4)=14(5×10−7)=1.25×10−7 kg m2I = \frac{1}{4}(5 \times 10^{-3} \times 10^{-4}) = \frac{1}{4}(5 \times 10^{-7}) = 1.25 \times 10^{-7}\,\text{kg m}^2I=41​(5×10−3×10−4)=41​(5×10−7)=1.25×10−7kg m2
  1. Torque required

Using

τ=Iα\tau = I\alphaτ=Iα τ=(1.25×10−7)(10π)\tau = (1.25 \times 10^{-7})(10\pi)τ=(1.25×10−7)(10π) τ=12.5π×10−7\tau = 12.5\pi \times 10^{-7}τ=12.5π×10−7 τ≈39.27×10−7=3.93×10−6 N m\tau \approx 39.27 \times 10^{-7} = 3.93 \times 10^{-6}\,\text{N m}τ≈39.27×10−7=3.93×10−6N m

So the torque is approximately

4.0×10−6 N m\boxed{4.0 \times 10^{-6}\,\text{N m}}4.0×10−6N m​
  1. Evaluate options
  • A: 7.9×10−67.9 \times 10^{-6}7.9×10−6 Nm — not correct
  • B: 4.0×10−64.0 \times 10^{-6}4.0×10−6 Nm — correct
  • C: 2.0×10−52.0 \times 10^{-5}2.0×10−5 Nm — not correct
  • D: 1.6×10−51.6 \times 10^{-5}1.6×10−5 Nm — not correct

Therefore, the correct option is:

B\boxed{\text{B}}B​
  1. Comparison with stored answer

Stored correct answer is C, but the calculated answer is clearly B.

The likely reason is that the wrong moment of inertia may have been used. For rotation about a diameter of a disc, the correct formula is:

I=14mR2I = \frac{1}{4}mR^2I=41​mR2

not 12mR2\frac{1}{2}mR^221​mR2 or some larger value.

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