Question

A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How many radians did the disk turn while stopping ? C. how many revolutions? ​

Answer 1

A. The angular acceleration of the disk is -1.047 radians per square second.

B. The disk turns 4.715 radians while stopping.

C. The disk did 0.750 revolutions while stopping.

Step-by-step explanation:

A. In this case, the disk is deceleration at a constant rate. Hence, the angular acceleration experimented by the object (
`\alpha`), in radians per square second, can be found by means of this kinematic expression:

`\alpha = (\omega-\omega_(o))/(t)` (1)

Where:

`\omega_(o)` - Initial angular speed, in radians per second.

`\omega` - Final angular speed, in radians per second.

`t` - Time, in seconds.

If we know that
`\omega_(o) \approx 3.142\,(rad)/(s)`,
`\omega = 0\,(rad)/(s)` and
`t = 3\,s`, then the angular acceleration of the disk is:

`\alpha = (\omega-\omega_(o))/(t)`

`\alpha = -1.047\,(rad)/(s^(2))`

The angular acceleration of the disk is -1.047 radians per square second.

B. The change in position of the disk (
`\Delta \theta`), in radians, is determined by the following kinematic formula:

`\Delta \theta = (\omega^(2)-\omega_(o)^(2))/(2\cdot \alpha)` (2)

If we know that
`\omega_(o) \approx 3.142\,(rad)/(s)`,
`\omega = 0\,(rad)/(s)` and
`\alpha = -1.047\,(rad)/(s^(2))`, then the change in position is:

`\Delta \theta = (\omega^(2)-\omega_(o)^(2))/(2\cdot \alpha)`

`\Delta \theta = 4.715\,rad`

The disk turns 4.715 radians while stopping.

C. A revolution equals 2π radians, then, then number of revolutions done by the disk while stopping is found by simple rule of three:

`\Delta \theta = 4.715\,rad * (1\,rev)/(2\pi\, rad)`

`\Delta \theta = 0.750\,rev`

The disk did 0.750 revolutions while stopping.