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 #65595 – Physics – Mechanics | Relativity

Question:

A solid cylinder of mass $3.0\,\mathrm{kg}$ and radius $1.0\,\mathrm{m}$ 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?

Solution:

The torque is:

$$\vec{\tau} = \vec{r} \times \vec{F} = \frac{d\vec{L}}{dt} = I \frac{d\vec{\omega}}{dt} = I \vec{\beta};$$

We can find the moment of inertia for cylinder:

$$I = \frac{m r^2}{2} = \frac{3 \cdot 1^2}{2} = 1.5\,\mathrm{kg} \cdot \mathrm{m}^2;$$

Consider that cylinder slows uniformly:

$$\beta = \frac{\Delta \omega}{\Delta t} = \frac{40}{10} = 4\,\frac{\mathrm{rad}}{\mathrm{s}^2};$$

Now we can find the torque and power:

$$\tau = \frac{dL}{dt} = I \frac{d\omega}{dt} = I\beta = 1.5 \cdot 4 = 6\,\mathrm{N} \cdot \mathrm{m};$$ $$P = \tau \omega = 6 \cdot 40 = 240\,\mathrm{W}.$$

Answer:

$$\tau = 6\,\mathrm{N} \cdot \mathrm{m};$$ $$P = 240\,\mathrm{W}.$$

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