Moment of inertia of a suspended cylinder

  • #1

Homework Statement



A uniform cylinder 20 cm long, suspended by a steel wire attached to its mid-point so that its long axis is horizontal, is found to oscillate with a period of 2 seconds when the wire is twisted and released. When a small disc, of mass 10 g, is attached to each end the period is found to be 2.3 seconds. Calculate the moment of inertia of the cylinder about the axis of oscillation.


Homework Equations



L = Iω
I = ∑(mr^2)



The Attempt at a Solution



angular velocity of cylinder, ω = 2π/2 = 3.14 rad/s

angular velocity of cylinder with discs, ω = 2π/2.3 = 2.73 rad/s

Conservation of angular momentum:

Iω = Iω(1) + 2 (mr^2)ω(1)

3.14I = 2.73I + 2 x (0.01 x 0.1^2) x 2.73

0.41I = 5.46 x 10^-4

I = 1.33 x 10^-3 kg m^2

The correct answer is 6.2 x 10^-4, so I am way off. Any help would be much appreciated. Thanks!
 

Answers and Replies

  • #2
284
41
Hey there. So, conserving momentum won't be of use here, since both situations are independent of one another. Instead, since this is a situation involving torsion pendulums, think of the equations relevant to those. The key one is
$$\omega^{2} = \frac{\kappa}{I}$$
Where ##\kappa## is the torsion constant of the wire. Since this isn't given, you can't use this equation to find the rotational inertia directly. Instead, if you have two equations, you can eliminate it to be able to find the inertia of the cylinder.
 
  • #3
Thank you for the quick reply.

I'm still a little confused here. Is the mass of the discs not significant for this problem? I don't see how I can simply use the above equation to find the moment of inertia, since I don't know either the moment of inertia or the torsion constant. Sorry, I'm obviously not picking up on something here.
 
  • #4
284
41
The idea is to create two equations to eliminate the torsion constant, since it's not given. So the first equation will use the period, torsion constant, and the moment of inertia of the cylinder, and the second one will use the other period, torsion constant, and the moment of inertia of the cylinder plus the inertia of the disks. Using both you can eliminate the torsion constant, and then solve for the moment of inertia of the cylinder since it will be the only unknown left.
 
  • #5
Ok, thank, I got it:

for the cylinder: c = ω^2 I

for the cylinder + discs: c = ω(1)^2 I + ω(1)^2 (2(mr^2))

So: ω^2 I = ω(1)^2 I + ω(1)^2 (2(mr^2))

I got to the correct answer from here.

Thanks a lot for your help!
 
  • #6
284
41
No problem! Glad you got it all worked out.
 

Related Threads on Moment of inertia of a suspended cylinder

  • Last Post
2
Replies
25
Views
10K
  • Last Post
Replies
3
Views
745
Replies
1
Views
6K
  • Last Post
Replies
6
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
1
Views
6K
Replies
3
Views
2K
Replies
2
Views
6K
Replies
7
Views
909
Top