Question

2. A pulley of radius 7.0 cm is connected to a motor that rotates at a rate of 6000 rad s-1 and then decelerate uniformly at a rate of 1000 rad s-1 within 5 s.

(a) Calculate its angular acceleration.

(b) What is the number of rotations within the time range?

(c) How long is the string that winds it within the time range?

(d) Determine the tangential acceleration of the string.

(e) Calculate the time taken for the pulley to stop from 1000 rad s-1.

Answer 1

For the pulley of radius 7.0 cm connected to a motor that rotates at a rate of 6000 rad/s and then decelerates at 1000 rad/s within 5 seconds, we have:

a) Its angular acceleration is -1000 rad/s².

b) The number of rotations within the time range is 2785.2 revolutions.

c) The length of the string that winds it within the time range is 1225.5 meters.

d) The tangential acceleration of the string is -70 m/s².

e) The time taken for the pulley to stop from 1000 rad/s is 1 second.

a) The angular acceleration is given by:

$\omega_{f} = \omega_{i} + \alpha t$    (1)

Where:

$\omega_{f}$: is the final angular velocity = 1000 rad/s

$\omega_{i}$: is the initial angular velocity = 6000 rad/s

$\alpha$: is the angular acceleration =?

t: is the time = 5 s

Solving equation (1) for α we have:

$\alpha = \frac{\omega_{f} - \omega_{i} }{t} = \frac{1000 rad/s - 6000 rad/s}{5 s} = -1000 rad/s^{2}$

Hence, the angular acceleration is -1000 rad/s². Its negative value is because the motor is decelerating.

b) The number of rotations can be calculated with the following equation:

$\theta_{f} = \theta_{i} + \omega_{i}t + \frac{1}{2}\alpha t^{2}$

Where:

$\theta_{i}$: is the initial displacement or initial number of rotations = 0

$\theta_{f}$: is the final displacement or the final number of rotations =?

$\theta_{f} = 6000 rad/s*5 s - \frac{1}{2}1000 rad/s^{2}*(5 s)^{2} = 17500 rad*\frac{1 rev}{2\pi rad} = 2785.2 rev$

Then, the number of rotations within the time range is 2785.2 revolutions.

c) The length of a rope that is winded within the time range, can be calculated knowing the perimeter of the pulley (circumference):

$C = 2\pi r = 2\pi*0.070 m = 0.44 m$

Now, since the pulley rotates 2785.2 revolutions in the interval time of 5 seconds, the length of the string is:

$l = \frac{0.44m \: length}{1 \: rev}*2785.2 \: rev = 1225.5 m \: length$

Hence, the length of the string is 1225.5 meters.

d) The tangential acceleration of the string is given by:

$a = \alpha r$

Where:

a: is the tangential acceleration =?

$a = \alpha r = -1000 rad/s^{2}*0.070 m = -70 m/s^{2}$                    

Therefore, the tangential acceleration of the spring is -70 m/s².

e) The time taken for the pulley to stop from 1000 rad/s can be calculated with equation (1):

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

Where:

$\omega_{f}$: is the final angular velocity = 0 (when the pulley stops)

$\omega_{i}$: is the initial angular velocity = 1000 rad/s

$t = \frac{\omega_{f} - \omega_{i}}{\alpha} = \frac{0 - 1000 rad/s}{-1000 rad/s^{2}} = 1 s$

Then, the time taken for the pulley to stop is 1 second.

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I hope it helps you!