Yatayyat
Badges: 14
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
I have been doing this problem for a while and want to see if my working out for it is fine?

Here is the question itself:

A cylindrical flywheel of radius 0.93 m and mass 54 kg is used to store energy by spinning.

a) Calculate the rotation frequency (in Hz or s-1) required for it to store 5.6 MJ.
(The moment of inertia of a solid cylinder rotating about its centre is given by ½mr2 )

b) Calculate the time taken for a 412 N force acting tangentially at the flywheel’s radius to accelerate the flywheel to 1911 rpm from rest.

Name:  IMG_3232.jpg
Views: 8
Size:  489.1 KB

Name:  IMG_3234.jpg
Views: 9
Size:  496.1 KB

Much appreciated!
Attached files
0
reply
Yatayyat
Badges: 14
Rep:
?
#2
Report Thread starter 3 weeks ago
#2
bump
0
reply
mqb2766
Badges: 17
Rep:
?
#3
Report 3 weeks ago
#3
Does it say anything more about the flywheel? The inertia is only
I = mr^2
if all the mass is located at the edge of the wheel. If its uniform, the inertia should be 1/2 of that.

Edit - you have it in brackets that its mr^2/2, so your first page answers are missing this 1/2 factor.
Last edited by mqb2766; 3 weeks ago
0
reply
Yatayyat
Badges: 14
Rep:
?
#4
Report Thread starter 2 weeks ago
#4
(Original post by mqb2766)
Does it say anything more about the flywheel? The inertia is only
I = mr^2
if all the mass is located at the edge of the wheel. If its uniform, the inertia should be 1/2 of that.

Edit - you have it in brackets that its mr^2/2, so your first page answers are missing this 1/2 factor.
Yes I see, so in fact my first page answer should be because it's missing a 1/2 factor, the answer would be '77.938... Hz * 1/2 = 38.969... Hz', so my new value is 38.97 Hz (2 dp)?
0
reply
mqb2766
Badges: 17
Rep:
?
#5
Report 2 weeks ago
#5
(Original post by Yatayyat)
Yes I see, so in fact my first page answer should be because it's missing a 1/2 factor, the answer would be '77.938... Hz * 1/2 = 38.969... Hz', so my new value is 38.97 Hz (2 dp)?
Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?
0
reply
Yatayyat
Badges: 14
Rep:
?
#6
Report Thread starter 2 weeks ago
#6
(Original post by mqb2766)
Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?
Wait doesn't less inertia imply that the wheel rotates faster when spinning? Since it has less resistance to its change in motion when a torque (rotational equivalent of a linear resultant force) is applied
0
reply
Yatayyat
Badges: 14
Rep:
?
#7
Report Thread starter 2 weeks ago
#7
(Original post by mqb2766)
Have you worked this through or are you expecting me to?
You have now less inertia and a slower wheel .... ?
And sorry, I just thought that since I was missing a 1/2 factor in my moment of inertia, if I multiplied my final answer by half would fix the mistake but that isn’t the case here

I redone the problem again and I’m now getting frequency to be 110.2 Hz

Name:  CF57415B-E727-47A8-9413-EF8E96A35A4B.jpeg
Views: 7
Size:  115.4 KB
0
reply
mqb2766
Badges: 17
Rep:
?
#8
Report 2 weeks ago
#8
(Original post by Yatayyat)
And sorry, I just thought that since I was missing a 1/2 factor in my moment of inertia, if I multiplied my final answer by half would fix the mistake but that isn’t the case here

I redone the problem again and I’m now getting frequency to be 110.2 Hz

Name:  CF57415B-E727-47A8-9413-EF8E96A35A4B.jpeg
Views: 7
Size:  115.4 KB
Seems ok? Its sqrt(2) larger than before?
0
reply
Yatayyat
Badges: 14
Rep:
?
#9
Report Thread starter 2 weeks ago
#9
(Original post by mqb2766)
Seems ok? Its sqrt(2) larger than before?
79.938 * sqrt(2) gives 113.049 so close to mine of 110.2

Does my part b look ok?
0
reply
mqb2766
Badges: 17
Rep:
?
#10
Report 2 weeks ago
#10
(Original post by Yatayyat)
79.938 * sqrt(2) gives 113.049 so close to mine of 110.2

Does my part b look ok?
Sure, all i was saying is that if the inertia is halved w^2 must be doubled for the same energy. So new angular speed is sqrt(2) * old one. Any difference is a numerical from you on the either the old or new calculation.

Checking b now.

Edit: b) appears fine.
Last edited by mqb2766; 2 weeks ago
0
reply
Yatayyat
Badges: 14
Rep:
?
#11
Report Thread starter 2 weeks ago
#11
(Original post by mqb2766)
Sure, all i was saying is that if the inertia is halved w^2 must be doubled for the same energy. So new angular speed is sqrt(2) * old one. Any difference is a numerical from you on the either the old or new calculation.

Checking b now.

Edit: b) appears fine.
Thank you so much! In future I’d try to do it that way as it saves much more time then having to go back and replugging the numbers
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Bishop Grosseteste University
    Undergraduate Open Day Undergraduate
    Fri, 15 Nov '19
  • University of Hertfordshire
    All Subjects Undergraduate
    Sat, 16 Nov '19
  • University of Roehampton
    General Open Day Undergraduate
    Sat, 16 Nov '19

Are you registered to vote?

18-20 years old (yes) (263)
54.23%
18-20 years old (no) (63)
12.99%
20-25 years old (yes) (82)
16.91%
20-25 years old (no) (9)
1.86%
25-30 years old (yes) (25)
5.15%
25-30 years old (no) (0)
0%
30-40 years old (yes) (24)
4.95%
30-40 years old (no) (3)
0.62%
40+ years old (yes) (9)
1.86%
40+ years old (no) (7)
1.44%

Watched Threads

View All