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JEE Main 2019 (10 Apr, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2019 (10 Apr, Shift 2)

Q54JEE Main 2019 · 10 APR · SHIFT 2Medium

Rotational Kinematics & Dynamics

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:

Figure

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Answer: (c)

  1. AB is a tangent in the plane of the coin: I=14mR2+mR2=54mR2I = \frac{1}{4}mR^2 + mR^2 = \frac{5}{4}mR^2.
  2. I=54(5×10−3)(10−2)2=6.25×10−7 kg m2I = \frac{5}{4}(5\times10^{-3})(10^{-2})^2 = 6.25\times10^{-7}\text{ kg m}^2
  3. ω=25×2π=50π rad/s\omega = 25\times2\pi = 50\pi\text{ rad/s}, α=50π5=10π rad/s2\alpha = \frac{50\pi}{5} = 10\pi\text{ rad/s}^2
  4. τ=Iα=6.25×10−7×31.4≈2.0×10−5 N m\tau = I\alpha = 6.25\times10^{-7}\times31.4 \approx 2.0\times10^{-5}\text{ N m}
  5. So the answer is (c) 2.0×10−52.0\times10^{-5} N m.
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