Question

A solid cylinder of mass 3 kg and radius 1 m is rotating about its axis with a speed of 40 rad/s . Calculate the torque which must be applied to bring it to rest in 10 second. What would be the power required?

Answer 1

Answer on Question 66740, Physics, Mechanics, Relativity

Question:

A solid cylinder of mass $3\,kg$ and radius $1.0\,m$ is rotating about its axis with a speed of $40\,rad/s$. Calculate the torque which must be applied to bring it to rest in 10 seconds. What would be the power required?

Solution:

a) We can find the torque from the formula:

$$\tau = I \alpha,$$

here, $I$ is the moment of inertia of the solid cylinder, $\alpha$ is the angular acceleration of the solid cylinder.

The moment of inertia of the solid cylinder can be found from the formula:

$$I = \frac{1}{2} m r^2,$$

here, $m$ is the mass of the solid cylinder, $r$ is the radius of the solid cylinder.

Then, we get:

$$I = \frac{1}{2} m r^2 = \frac{1}{2} \cdot 3\,kg \cdot (1.0\,m)^2 = 1.5\,kg \cdot m^2.$$

We can find the angular acceleration of the solid cylinder from the kinematic equation:

$$\omega = \omega_i + \alpha t,$$

here, $\omega_i = 40\,rad/s$ is the initial angular speed of the cylinder, $\omega = 0\,rad/s$ is the final angular speed of the cylinder (when the cylinder is bring to rest), $\alpha$ is the angular acceleration of the cylinder and $t$ is the time.

Then, from this formula we can find the angular acceleration of the solid cylinder:

$$\alpha = \frac{\omega - \omega_i}{t} = \frac{0\,rad/s - 40\,rad/s}{10\,s} = -4\,rad/s^2.$$

The sign minus indicates that the cylinder decelerates.

Substituting $I$ and $\alpha$ into the first formula we can calculate the torque which must be applied to bring it to rest in 10 second:

$$\tau = I \alpha = 1.5 \, kg \cdot m^2 \cdot \left(-4 \, \frac{rad}{s^2}\right) = -6 \, N \cdot m.$$

The sign minus indicates that the torque acting in the opposite direction to the rotation of the cylinder. So, the magnitude of the torque will be $\tau = 6N \cdot m$.

b) We can find the power required from the formula:

$$P = \tau \omega,$$

here, $P$ is the power, $\tau$ is the torque applied to the cylinder, $\omega$ is the angular speed of the cylinder.

Then, we get:

$$P = \tau \omega = 6N \cdot m \cdot 40 \, \frac{rad}{s} = 240 \, W.$$

Answer:

a) $\tau = 6N \cdot m$.

b) $P = 240 \, W$.

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