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 × 10−6 Nm
  2. 2 × 10−3 Nm
  3. 12 × 10−4 Nm
  4. 2 × 106 Nm

Answer (Detailed Solution Below)

Option 1 : 2 × 10−6 Nm
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Detailed Solution

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CONCEPT:

  • Work energy theorem - The work-energy theorem defines that the net work done by the forces on the given object is equal to the change in the kinetic energy of the object.
  • When the object is in rotation and translational with a fixed axis and fixed direction. The work-energy theorem is written as;

             \(W = \frac{1}{2}I(\omega _f^2 - \omega _i^2)\)

         Here, \(\omega_f ,\omega_i\) is the final and initial angular velocity and 'I' is the moment of inertia.

CALCULATION:

According to the work-energy theorem we have;

\(W = \frac{1}{2}I(\omega _f^2 - \omega _i^2)\) -----(1)

and angular displacement, θ = 2π revolution

θ = 2π × 

θ = 4πrad

Now, the moment of inertia, \(I = \frac{mr^2}{2}\)

⇒ \(I = \frac{2\times (0.04)^2}{2}\)

⇒ I  = 1.6 \(\times\) 10-3 kgm2

Now, the initial angular velocity is written as 

\(\omega_i = 3\) rpm

\({\omega_i} = 3 × \frac{{2\pi }}{{60}}rad/s\)

In this, we also say that the rotational kinetic energy is equal to the product of the torque and angular displacement which is written as;

\(W=\frac{I\omega^2}{2} = \tau\times \theta\)

and the final angular velocity is equal to zero. Therefore putting all the values in equation (1) we have;

\(τ \theta = \frac{1}{2} × \frac{1}{2}m{r^2}(0^2 - \omega _i^2)\)

\( ⇒ \,τ = \frac{{\frac{1}{2} × \frac{1}{2} × 2(4 × {{10}^{ - 2}}){{\left( { - 3 × \frac{{2\pi }}{{60}}} \right)}^2}}}{{4{\pi ^2}}}\)

⇒ τ = 2 × 10-6 Nm

Hence, option 1) is the correct answer.

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