Question

A solid cylinder of mass 3.0 kg and radius 1.0 m is rotating about its axis with a speed of 40 rad s−1. Calculate the torque which must be applied to bring it to rest in 10s. What would be the power required?

Answer 1

Answer on Question #65592-Mechanics - Relativity

A solid cylinder of mass $m = 3.0 \, \mathrm{kg}$ and radius $r = 1.0 \, \mathrm{m}$ is rotating about its axis with a speed of $\omega = 40 \, \mathrm{rad~s^{-1}}$. Calculate the torque which must be applied to bring it to rest in $t = 10 \, \mathrm{s}$. What would be the power required?

Solution

The second Newton's law for rotation

$$J \frac {\Delta \omega}{\Delta t} = M.$$

Here $J = \frac{m r^2}{2} = \frac{3 \times 1^2}{2} = 1.5 \, \mathrm{kg} \cdot \mathrm{m}^2$ is the moment of inertia, $\Delta \omega = \omega - \omega_0 = 0 - 40 = -40 \, \mathrm{rad/s}$.

So the torque

$$M = J \frac {\Delta \omega}{\Delta t} = 1.5 \frac {- 40}{10} = -6 \, \mathrm{N} \cdot \mathrm{m}.$$

The work done is equal the change in kinetic energy

$$W = \Delta E = \frac {J \omega^ {2}}{2} - \frac {J \omega_ {0} ^ {2}}{2} = 0 - \frac {1.5 \times 40 ^ {2}}{2} = -1200 \, \mathrm{J}.$$

The power required

$$P = \frac {W}{t} = \frac {1200}{10} = 120 \, \mathrm{W}.$$

Answer $M = 6 \, \mathrm{N} \cdot \mathrm{m}$, $P = 120 \, \mathrm{W}$.

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