Question

A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?​

Answer 1

Answer:

a) The angular acceleration is 1.875 radians per square second.

b) The time taken by the disk to reach the final angular speed is 8 seconds.

Explanation:

a) Let suppose that the disk accelerates uniformly, given that initial and final angular speed ($\omega_{o}$, $\omega_{f}$), in radians per second, and change in angular position ($\Delta \theta$), in radians, are known. The angular acceleration ($\alpha$), in radians per square second, are found by using this expression:

$\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}$ (1)

If we know that $\omega_{o} = 0\,\frac{rad}{s}$, $\omega_{f} = 15\,\frac{rad}{s}$ and $\Delta \theta = 60\,rad$, then the angular acceleration of the disk is:

$\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}$

$\alpha = 1.875\,\frac{rad}{s^{2}}$

The angular acceleration is 1.875 radians per square second.

b) The time taken by the disk to reach the final angular velocity is determined by the following kinematic formula:

$t = \frac{\omega_{f}-\omega_{o}}{\alpha}$ (2)

Where $t$ is the time, in seconds.

If we know that $\omega_{o} = 0\,\frac{rad}{s}$, $\omega_{f} = 15\,\frac{rad}{s}$ and $\alpha = 1.875\,\frac{rad}{s^{2}}$, then the time taken by the disk is:

$t = \frac{\omega_{f}-\omega_{o}}{\alpha}$

$t = 8\,s$

The time taken by the disk to reach the final angular speed is 8 seconds.