Torque Help: Flywheel Energy & Power Calculations

[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?

 

[tiny-tim]

hi tydychic!

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

 

[tydychic]

Calculations as requested:
A) T(motor)=I*alpha. I-Beta*Mass*Radius squared, Beta =1/2. Alpha=2* Torque(motor)/Mass*radius squared= .711 rad/seconds squared.

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?

 

[tiny-tim]

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 )

Flywheels are large, massive wheels used to store energy.

Calculations as requested:
A) T(motor)=I*alpha. I-Beta*Mass*Radius squared, Beta =1/2. Alpha=2* Torque(motor)/Mass*radius squared= .711 rad/seconds squared.

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"

 

[tydychic]

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?

 

[tiny-tim]

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 …

A) T(motor)=I*alpha. I-Beta*Mass*Radius squared, Beta =1/2. Alpha=2* Torque(motor)/Mass*radius squared= .711 rad/seconds squared.

i'm not getting that answer …

can you please show your calculations?

 

[tydychic]

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

 

[Lusos]

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?

 

[tydychic]

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.

 

[Lusos]

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

 

[gneill]

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