# 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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