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 2 \pi revolution is : 

 

  • Option 1)

    2 \times 10 ^{-6} Nm

  • Option 2)

    2 \times 10 ^{-3} Nm

  • Option 3)

    12 \times 10 ^{-4} Nm

  • Option 4)

    2 \times 10 ^{6} Nm

 

Answers (1)

\omega _2 ^2 = \omega _1 ^2 + 2 \alpha \theta \\\\ 0 = ( 2 \pi*\frac{3}{60} ) ^2 - 2 \times 2 \pi \times 2 \pi \alpha \\\\ \alpha = \frac{1}{800} \\\\ T = I\alpha = \frac{mr ^2}{2} d \\\\ = 2 \times 10 ^{-6}


Option 1)

2 \times 10 ^{-6} Nm

Option 2)

2 \times 10 ^{-3} Nm

Option 3)

12 \times 10 ^{-4} Nm

Option 4)

2 \times 10 ^{6} Nm

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