Question

Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
a. A motor spins up the flywheel with a constant torque of 50 N m. How long does it take the flywheel to reach top speed?
b. How much energy is stored in the flywheel?

c. The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.0 s. What is the average power delivered to the machine?
d. How much torque does the flywheel exert on the machine?

Short answer

a) It will take 176.7 sec for flywheel to reach top speed

b) Energy stored in the flywheel is 5.5 x 10⁵ J

c) Average power delivered to the machine is 1.39 x 10⁵ W

d) Torque exerted by the flywheel on the machine 1562 NM

Step-by-step solution

Part(a) Step1: Given information

mass of fly wheel = 250 kg

diameter = 1.5 m

max angular velocity = 1200 rpm

Part(a) Step2: Explanation

First convert angular velocity in rad/sec

$$\begin{gathered} \omega =1200\mathrm{rpm} \\ =1200rpm\times \frac{2\pi rad}{60sec} \\ =40\pi \mathrm{rad}/\mathrm{s} \end{gathered}$$

Now use this in equation of motion

$\omega ={\omega }_o+\frac{\tau }{I}t$

Moment of inertia is given as

$I=\frac{m{r}^2}{2}=\frac{250\times 0.{75}^2}{2}=70.31\mathrm{kg}{m}^2$

Substitute the values

$40\pi rad/s=0+\frac{50Nm}{70.31kg.m^2}t$

Solve for t, we get t=176.7 sec/

Part(b) Step1: given information

mass of fly wheel = 250 kg

diameter = 1.5 m

max angular velocity = 1200 rpm

Part(b) Step2: Explanation

Calculate energy using the equation (1), we get

$$\begin{gathered} E=\frac{1}{2}I{\omega }^2 \\ =\frac{1}{2}(70.31kg.m^2)\times (40\pi rad/sec{)}^2 \\ =5.55\times {10}^{5}J\dots \dots \dots \dots \dots \dots \dots (3) \end{gathered}$$

Part(c) Step1 : Given Information

As energy transferred to machine is half in 2 sec.

mass of fly wheel = 250 kg

diameter = 1.5 m

max angular velocity = 1200 rpm

Part(c) Step1 : Explanation

Only half energy is transferred so from equation (3) we get

Energy transferred is 2.78 x10⁵ J

Power can be calculated by P = E/t

$E_m=\frac{2.78\times {10}^{5\mathrm{J}}}{176.7\mathrm{s}}=1.39\times {10}^5\mathrm{W}$

Part(d) Step1: Given information

As energy transferred to machine is half in 2 sec.

mass of fly wheel = 250 kg

diameter = 1.5 m

max angular velocity = 1200 rpm

Part(d) Step2: Explanation

The average angular velocity gained is calculated using kinetic energy gained

$$\begin{gathered} E=\frac{1}{2}I{\omega }^2 \\ \omega =\sqrt{\frac{2E_m}{I}}=\sqrt{\frac{2(2.78x{10}^{5}J)}{70.31kg.m^2}}88.86\mathrm{rad}/s \end{gathered}$$

Now use this to find the torque

$$\begin{gathered} P_{avg}=\tau \times \omega \\ \tau =\frac{P_{avg}}{\omega }=\frac{1.39\times {10}^{5}W}{88.86rad/s}=1562\mathrm{Nm} \end{gathered}$$