Question

A flywheel in the form of a uniformly thick disk of radius 1.18 m has a mass of 94.6 kg and spins counterclockwise at 443 rpm. If the positive direction of rotation is counterclockwise. calculate the constant torque r required to stop it in 1.00 min.

Answer 1

The constant torque required to stop the counterclockwise-spinning flywheel with a radius of 1.18 m, mass 94.6 kg, and initial angular velocity 443 rpm in 1.00 minute is approximately -51.37 N·m.

To calculate the constant torque required to stop the flywheel in 1.00 minute, we can use the rotational analog of Newton's second law:

`\[ \tau = I \alpha \]`

where:

-
`\(\tau\)` is the torque,

- \(I\) is the moment of inertia of the flywheel,

-
`\(\alpha\)` is the angular acceleration.

The moment of inertia \(I\) for a uniform disk is given by:

`\[ I = (1)/(2) m r^2 \]`

where:

-
`\(m\)`is the mass of the flywheel,

-
`\(r\)` is the radius of the flywheel.

The angular acceleration
`\(\alpha\)` can be calculated from the angular velocity
`\(\omega\) and the time \(t\)`using the equation:

`\[ \alpha = (\Delta \omega)/(\Delta t) \]`

The change in angular velocity
`\(\Delta \omega\)` is given by the final angular velocity
`\(\omega_f\)` (when the flywheel stops) minus the initial angular velocity
`\(\omega_i\):`

`\[ \Delta \omega = \omega_f - \omega_i \]`

The final angular velocity
`\(\omega_f\)` is 0, as the flywheel comes to a stop.

Let's plug in the given values and calculate the torque:

Given data:

- Radius
`(\(r\))` = 1.18 m

- Mass
`(\(m\))` = 94.6 kg

- Initial angular velocity
`(\(\omega_i\))` = 443 rpm

- Time
`(\(t\))`= 1.00 min

First, convert the initial angular velocity to radians per second:

`\[ \omega_i = \frac{443 \, \text{rpm} * 2 \pi \, \text{rad}}{60 \, \text{s}} \]`

Then, calculate the angular acceleration:

`\[ \alpha = (-\omega_i)/(t) \]`

Finally, use the torque equation to find the constant torque:

`\[ \tau = I \alpha \]`

`\[ \tau = (1)/(2) m r^2 \alpha \]`

Now, let's proceed with the calculations:

1. Convert
`\(\omega_i\)`to radians per second:

`\[ \omega_i = (443 * 2 \pi)/(60) \, \text{rad/s} \]`

2. Calculate \(\alpha\):

`\[ \alpha = (-\omega_i)/(t) \]`

3. Calculate \(I\) using the moment of inertia formula:

`\[ I = (1)/(2) m r^2 \]`

4. Plug in the values into the torque formula:

`\[ \tau = I \alpha \]`

Now, let's perform the calculations.

1. Convert
`\(\omega_i\)`to radians per second:

`\[ \omega_i = (443 * 2 \pi)/(60) \, \text{rad/s} \]`

`\[ \omega_i \approx 46.32 \, \text{rad/s} \]`

2. Calculate
`\(\alpha\):`

`\[ \alpha = (-\omega_i)/(t) \]`

`\[ \alpha = (-46.32)/(60) \, \text{rad/s}^2 \]`

`\[ \alpha \approx -0.772 \, \text{rad/s}^2 \]`

3. Calculate \(I\) using the moment of inertia formula:

`\[ I = (1)/(2) m r^2 \]`

`\[ I = (1)/(2) * 94.6 * (1.18)^2 \, \text{kg} \cdot \text{m}^2 \]`

`\[ I \approx 66.52 \, \text{kg} \cdot \text{m}^2 \]`

4. Plug in the values into the torque formula:

`\[ \tau = I \alpha \]`

`\[ \tau = 66.52 * (-0.772) \]`

`\[ \tau \approx -51.37 \, \text{N} \cdot \text{m} \]`

The constant torque required to stop the flywheel in 1.00 minute is approximately
`\( -51.37 \, \text{N} \cdot \text{m} \).`The negative sign indicates that the torque is acting in the opposite direction of the initial rotation, which is counterclockwise.