A flywheel in the form of a uniformly thick disk of radius 1.68 m has a mass of 54.1 kg and spins counterclockwise at 209 rpm. Calculate the constant torque required to stop it in 1.00 min.
The Correct Answer And Explanation is:
Given:
- Radius of flywheel, R=1.68 mR = 1.68 \, \text{m}
- Mass of flywheel, m=54.1 kgm = 54.1 \, \text{kg}
- Initial angular velocity, ω0=209 rpm\omega_0 = 209 \, \text{rpm}
- Time to stop, t=1.00 mint = 1.00 \, \text{min}
We need to calculate the constant torque required to stop the flywheel in the given time.
Steps:
1. Convert the initial angular velocity to rad/s:
First, convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). The conversion factor is 2π rad2\pi \, \text{rad} per revolution and 60 s60 \, \text{s} per minute: ω0=209 rpm×2π rad1 rev×160 s/min=21.90 rad/s\omega_0 = 209 \, \text{rpm} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1}{60 \, \text{s/min}} = 21.90 \, \text{rad/s}
2. Moment of inertia of the flywheel:
The flywheel is a solid disk, and the moment of inertia II of a disk is given by: I=12mR2I = \frac{1}{2} m R^2
Substitute the known values: I=12(54.1 kg)(1.68 m)2=48.7 kg⋅m2I = \frac{1}{2} (54.1 \, \text{kg}) (1.68 \, \text{m})^2 = 48.7 \, \text{kg} \cdot \text{m}^2
3. Angular deceleration:
The final angular velocity ωf=0\omega_f = 0 because the flywheel is stopped, and the initial angular velocity ω0=21.90 rad/s\omega_0 = 21.90 \, \text{rad/s}. The time to stop the flywheel is given as 1 minute or 60 seconds. The angular deceleration α\alpha can be calculated using the equation for constant acceleration: ωf=ω0+αt\omega_f = \omega_0 + \alpha t
Substitute known values and solve for α\alpha: 0=21.90 rad/s+α(60 s)0 = 21.90 \, \text{rad/s} + \alpha (60 \, \text{s}) α=−21.9060=−0.365 rad/s2\alpha = – \frac{21.90}{60} = -0.365 \, \text{rad/s}^2
4. Torque required:
The torque τ\tau required to produce this angular deceleration is related to the moment of inertia and angular acceleration by: τ=Iα\tau = I \alpha
Substitute the known values: τ=(48.7 kg⋅m2)(−0.365 rad/s2)=−17.75 N⋅m\tau = (48.7 \, \text{kg} \cdot \text{m}^2) (-0.365 \, \text{rad/s}^2) = -17.75 \, \text{N} \cdot \text{m}
The negative sign indicates that the torque is acting in the opposite direction to the motion (i.e., it’s a stopping torque).
Final Answer:
The constant torque required to stop the flywheel in 1.00 minute is approximately: τ=17.75 N⋅m\tau = 17.75 \, \text{N} \cdot \text{m}
Explanation:
To calculate the torque required to stop the flywheel, we first converted the initial rotational speed from rpm to rad/s. We then calculated the moment of inertia of the flywheel, which depends on its mass and radius. Using the known time to stop and the angular acceleration formula, we determined the deceleration needed. Finally, applying the relationship between torque, moment of inertia, and angular acceleration, we found the required torque. This torque, acting in the opposite direction of the flywheel’s rotation, will bring the flywheel to rest in the specified time.