1. 2×10⁻⁶ N m 2. 2×10⁻³ N m 3. 12×10⁻⁴ N m 4. 2×10⁶ N m

Answer: 1. 2×10⁻⁶ N m Find the angular deceleration from the rotation kinematics, then use τ=Iα . ω₀=3 rpm =3×2π/60=0.1π rad/s. Angle turned: θ=2π rev =4π² rad. ω²=ω₀²-2αθ⇒ α=0.01π²/8π²=1.25×10⁻³ rad/s ² . I=1/2MR²=1/2×2×(0.04)²=1.6×10⁻³ kg m ² . τ=Iα=1.6×10⁻³×1.25×10⁻³=2×10⁻⁶ N m.

Topic: Moment of force, torque· Rotational Motion

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is :

1. \(2\times10^{-6}\) N m

2. \(2\times10^{-3}\) N m

3. \(12\times10^{-4}\) N m

4. \(2\times10^{6}\) N m

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Answer: 1. \(2\times10^{-6}\) N m

Find the angular deceleration from the rotation kinematics, then use \(\tau=I\alpha\).

\(\omega_0=3\) rpm \(=\dfrac{3\times2\pi}{60}=0.1\pi\) rad/s.
Angle turned: \(\theta=2\pi\) rev \(=4\pi^2\) rad.
\(\omega^2=\omega_0^2-2\alpha\theta\Rightarrow \alpha=\dfrac{0.01\pi^2}{8\pi^2}=1.25\times10^{-3}\) rad/s\(^2\).
\(I=\tfrac12MR^2=\tfrac12\times2\times(0.04)^2=1.6\times10^{-3}\) kg m\(^2\).
\(\tau=I\alpha=1.6\times10^{-3}\times1.25\times10^{-3}=2\times10^{-6}\) N m.

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A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is :

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