# Torque Help: Flywheel Energy & Power Calculations

• tydychic
In summary: There seems to be a mistake in the calculation from the original poster. The correct answer should be 1.5 kNm.In summary, a flywheel is a large, massive wheel used to store energy. It can be spun up slowly and then release its energy quickly to accomplish tasks requiring high power. In this scenario, an industrial flywheel with a 1.5 diameter and a mass of 250kg has a maximum angular velocity of 1200 rpm. A motor spins up the flywheel with a constant torque of 50Nm. It takes the flywheel 176.763 seconds to reach its top speed and it stores 555168 J of energy. When disconnected from the motor and connected to a machine, the flywheel
tydychic

## Homework Statement

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 diameter and a mass of 250kg. it's max angular velocity is 1200 rpm. a)a motor spins up the flywheel with a constant torque of 50Nm. 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 2.0s. What is the average power delivered to the machine? D) How much torque does the flywheel exert on the machine?

3. The Attempt at a Solution
I've figured out the answers to A, B, and C. A) t=176.763 seconds. B) KE(r)=555168 J. C) Power= 138792 W. D) I read that P=torque*angular velocity; however, the answer in the book says 1.30 kNM and I continue to get 1104.47. Can calculator error really explain the difference or am I going wrong somewhere?

hi tydychic!

if you want us to check your work, you need to show us your calculations

Calculations as requested:

Omega final = alpha *time. omega final = 1200rpm * 1min/60sec * 2pi rad/1 rev=125.664 rad/s.
t = omega/alpha= 176.743

B) KE(r)= .5I* omega squared - (1/2)I* omega initial squared (which equals zero, so tossing that). KE(r)=(1/4)(250kg)(.75m)^2*(125.664rad/sec) = 555168 J

C) Half the power...555168/2= 277584J over two seconds...277584 J/2sec= 138792 Watts

D) 'if the equation I read is correct...138792 W/ (125.664 rad/sec) = 1104NM, but this isn't the correct answer, so I must be doing something wrong...any clues?

hi tydychic!

(have an alpha: α and a beta: β and a pi: π and try using the X2 tag just above the Reply box )
tydychic said:
Flywheels are large, massive wheels used to store energy.
tydychic said:
Calculations as requested:

ah, i think you're using the wrong β …

(btw, I've never seen it called β before … is that normal?)

a flywheel is essentially a wheel

(btw, can you see why they build them that way?)

see some pictures at http://en.wikipedia.org/wiki/Flywheel"

Last edited by a moderator:
If I was using the wrong Beta, then my answers for A, B, and C would be incorrect. However, they aren't, so I'm assuming Beta = 1/2 is correct...? Where does that leave D?

tydychic said:
If I was using the wrong Beta, then my answers for A, B, and C would be incorrect. However, they aren't, so I'm assuming Beta = 1/2 is correct...? Where does that leave D?

hmm … strange flywheel!

ok , let's check …
tydychic said:

i'm not getting that answer …

$$\alpha$$=2(50Nm)/(250kg*.75m$$^{2}$$)=.711rad/s$$^{2}$$...I guess parentheses might help. I'm sorry. Does this help?

Hi all, I am having the same exact problem... As a matter of fact, I think that we may be in the same physics class. Hehe.

Parts A,B, and C agree with OP and the back of the textbook. However, I cannot get part D to match. According to the book, the answer is 1.5kNm yet mine continues to come out as "1104.79 kw/rad(IIRC)". Is this some sort of rounding error?

You wouldn't happen to know a dude with a yo-yo, would you? Because if you did, then yes, same class. Definitely the same book.

That, I chalked up to book-error. I couldn't get any other answer and the book is known to be wrong occasionally.

tydychic said:
You wouldn't happen to know a dude with a yo-yo, would you? Because if you did, then yes, same class. Definitely the same book.

That, I chalked up to book-error. I couldn't get any other answer and the book is known to be wrong occasionally.

HAHA yes. I've checked my answer to the point of insanity and I am willing to bet that the book once again had a typo. Anyways, good luck with the last two. See you in class on Monday.

-Oscar

It would appear that in this case, for part D), the book is correct with its 1.30 kNm.

## What is torque and how is it related to flywheel energy?

Torque is a measure of the force that causes an object to rotate around an axis. In the case of flywheel energy, torque is the force that is applied to the flywheel, causing it to rotate and store energy. The greater the torque, the more energy the flywheel can store.

## How do you calculate the flywheel's moment of inertia?

The moment of inertia of a flywheel can be calculated by multiplying the mass of the flywheel by the square of its radius. This value is important in determining the flywheel's ability to store and release energy.

## What is the equation for calculating flywheel energy?

The equation for calculating flywheel energy is E = 1/2 * moment of inertia * angular velocity^2. This equation takes into account both the flywheel's moment of inertia and its rotational speed.

## How does power affect flywheel energy and torque?

Power is the rate at which energy is transferred, and it is directly related to both flywheel energy and torque. A higher power output means the flywheel can store and release energy at a faster rate, and it also requires a higher torque to achieve this level of power.

## Can flywheel energy and power calculations be used in real-world applications?

Yes, flywheel energy and power calculations are commonly used in various industries such as automotive, industrial, and renewable energy. Flywheels can be used to store energy and provide a power boost when needed, making them useful in applications where a consistent power output is required.

• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
6K
• Mechanical Engineering
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
44
Views
3K
• Introductory Physics Homework Help
Replies
23
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
6K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
328
• Mechanics
Replies
43
Views
4K