Question

an 8.0-cm -diameter hard disk spinning at 7200 rpm can stop in 11 revolutions. what is the magnitude of the angular acceleration?

Answer 1

The magnitude of the angular acceleration is approximately
`\(8236.2 \text{ rad/s}^2\).`

To find the magnitude of the angular acceleration, we can use the following formula:

`\[ \text{Angular acceleration (\(\alpha\))} = \frac{\text{Change in angular velocity (\(\Delta \omega\))}}{\text{Time (\(t\))}} \]`

Where:

- Change in angular velocity
`(\(\Delta \omega\))` is the final angular velocity
`(\(\omega_f\))` minus the initial angular velocity
`(\(\omega_i\)).`

- Time
`(\(t\))` is the time it takes for the disk to stop.

First, let's find the initial angular velocity
`(\(\omega_i\))` and final angular velocity
`(\(\omega_f\)):`

1. The disk spins at 7200 revolutions per minute (rpm), which we need to convert to radians per second (rad/s):

`\[ \omega_i = \frac{7200 \text{ rpm}}{60} * \frac{2\pi \text{ rad}}{1 \text{ min}} * \frac{1 \text{ min}}{60 \text{ s}} = 753.98 \text{ rad/s} \]`

2. To find the final angular velocity
`(\(\omega_f\)),`we need to know how many radians the disk covers in 11 revolutions. Since one revolution is equivalent to
`\(2\pi\)` radians, 11 revolutions will be
`\(11 * 2\pi\)` radians.

`\[ \omega_f = \frac{\text{Change in angular position}}{\text{Change in time}} = \frac{11 * 2\pi \text{ rad}}{t} \]`

Now, we can calculate
`\(\Delta \omega\)` and then the angular acceleration
`(\(\alpha\)):`

`\[ \Delta \omega = \omega_f - \omega_i = \frac{11 * 2\pi \text{ rad}}{t} - 753.98 \text{ rad/s} \]`

We are given that the disk stops, so
`\(\omega_f = 0\)` when it comes to rest. Therefore:

`\[ 0 = \frac{11 * 2\pi \text{ rad}}{t} - 753.98 \text{ rad/s} \]`

Now, solve for \(t\):

`\[ t = \frac{11 * 2\pi \text{ rad}}{753.98 \text{ rad/s}} \]`

Calculate
`\(t\):`

`\[ t \approx 0.0916 \text{ s} \]`

Now that we have the time it takes for the disk to stop, we can calculate the angular acceleration
`(\(\alpha\)):`

`\[ \alpha = (\Delta \omega)/(t) = \frac{-753.98 \text{ rad/s}}{0.0916 \text{ s}} \]`

Calculate
`\(\alpha\):`

`\[ \alpha \approx -8236.2 \text{ rad/s}^2 \]`

The magnitude of the angular acceleration is approximately
`\(8236.2 \text{ rad/s}^2\).` Note that the negative sign indicates deceleration or slowing down.