Period of small oscillations for a pendulum

In summary: I think I get it. You're using the moment of inertia of the whole system, not just the pendulum? Yes, I'm using the moment of inertia of the system as a whole.
  • #1
mrojc
12
0

Homework Statement

A pendulum consists of a light rigid rod of length 250 mm, with two identical uniform solid spheres of radius of radius 50 mm attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centres and (b) along it.[/B]

Homework Equations

Equation I have come across is

db205bb67d7f7f685b18b19b451ac429.png
[/B]

The Attempt at a Solution

From the parallel axis theorum the moment of inertia of a pendulum and sphere is mr2+2/5mr2. As there is two spheres I added on another 2/5 mr2. The answers for both parts are (a) 1.011 s and (b) 1.031 s. I have gotten answers close to, but not quite these answers, ie 1.2 s or 1.3 s. I feel like this is an easy question and that I'm after doing something stupid, so any guidance would be greatly appreciated.[/B]
 
Physics news on Phys.org
  • #2
Hello mro, :welcome:

Just so we are really talking about the same pendulum:

upload_2015-11-21_21-24-45.png

you didn't tell us what m is and I don't understand how this is a pendulum. Where is the axis of rotation ?
How can it oscillate along the "line of centres"? and perpendicular to it ?
 
  • #3
Hi BvU,
We weren't given m, and it doesn't make much sense to me either. When I read the question initially, that was the picture I had in my head. However, from reading it again, I think they're attached on the same end? Just simply because it said "one on either side of its lower end," not ends. To be honest, the book is very poorly worded at times. As regards to m, could it not be canceled with the lower line m in the equation? Maybe it's total mass of the system? I apologise for the vagueness of the question by the way, it's not very clear.
 
  • #4
@BvU I think the two cases have the spheres at the same end of the rod, the difference being the plane of oscillation:

Fig1.png
 
  • Like
Likes BvU
  • #5
@gneill Hi, thanks for the diagram, that's the picture I had in my head. Am I on the right track with my equations or is there something I'm forgetting about?
 
  • #6
mrojc said:
@gneill Hi, thanks for the diagram, that's the picture I had in my head. Am I on the right track with my equations or is there something I'm forgetting about?
I think you're on the right track, but you should be careful about your moment of inertia expressions. Be sure to carefully account for the displacement distance in the parallel axis theorem for each case.

I just did a quick calculation, and I suspect that they may have forgotten to include both spheres when they calculated their "given" answers. Of course, it's also possible that I've mucked up, too! So let's see what you get (show your work), and in the meantime I'll go over my own work.

Edit: Nope, I mucked up :oops: I'm seeing their results now.
 
  • #7
Ok, just want to double check something: for R in the period equation, is that L/2? Or do I have to include the radius of the spheres in that R too? Thanks for all your help so far by the way, really appreciate it! :)
 
  • #8
mrojc said:
Ok, just want to double check something: for R in the period equation, is that L/2? Or do I have to include the radius of the spheres in that R too? Thanks for all your help so far by the way, really appreciate it! :)
The R in your formula is the distance from the center of mass of the physical pendulum to the pivot point. See:

Hyperphysics: Physical Pendulum
 
  • #9
I seem to be making a mess of things :H I'm inputting 0.125 m for the moment of inertia of the pendulum and 0.05 m for the spheres, and I am not getting the answers out. I'm definitely doing something really stupid and I'll be so annoyed when I figure out what it is!
 
  • #10
mrojc said:
I seem to be making a mess of things :H I'm inputting 0.125 m for the moment of inertia of the pendulum and 0.05 m for the spheres, and I am not getting the answers out. I'm definitely doing something really stupid and I'll be so annoyed when I figure out what it is!
I'm not sure how you're coming up with a value for the moment of inertia alone, since the sphere mass was not specified. The mass cancels out eventually in the period formula though.

Why don't you share the details of your work? Maybe someone can spot where it goes awry.
 
  • #11
https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xat1/v/t34.0-12/12282819_10204148176339731_1402560635_n.jpg?oh=e147dab589d46be63cf24a1a026741fc&oe=56535411
I apologise for this being sideways but this was my calculations part.
 
  • #12
This problem seems to hinge on obtaining the correct moment of inertia for the given setup, so let's focus on that.

I'm not understanding the moment of inertia you've used in your period formula. Can you type it out and explain the variables? If r is the radius of a sphere, then what is the variable for the offset a la parallel axis theorem?
 

What is the period of small oscillations for a pendulum?

The period of small oscillations for a pendulum is the time it takes for the pendulum to complete one full swing from side to side. It is determined by the length of the pendulum and the acceleration due to gravity.

What factors affect the period of small oscillations for a pendulum?

The period of small oscillations for a pendulum is affected by the length of the pendulum, the mass of the pendulum bob, and the acceleration due to gravity. Other factors such as air resistance and friction may also have a small impact.

Is the period of small oscillations for a pendulum the same everywhere on Earth?

No, the period of small oscillations for a pendulum may vary slightly depending on the location on Earth. This is due to variations in the acceleration due to gravity caused by differences in altitude and latitude.

Can the period of small oscillations for a pendulum be changed?

Yes, the period of small oscillations for a pendulum can be changed by altering the length of the pendulum or the mass of the pendulum bob. The period can also be affected by changing the acceleration due to gravity, which can be achieved by changing the location of the pendulum.

How is the period of small oscillations for a pendulum calculated?

The period of small oscillations for a pendulum can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. This formula assumes small oscillations and no air resistance or friction.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
796
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
28
Views
549
  • Introductory Physics Homework Help
Replies
2
Views
635
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
5K
Replies
1
Views
1K
Back
Top