Calculate the rotational inertia of the pendulum

[LandOfStandar]

Homework Statement

The pendulum consists of a uniform disk with radius r=10.0cm and mass 500g attached to a uniform rod with length L=500mm and mass 270g.
a. Calculate the rotational inertia of the pendulum about the pivot point.
(answer .205kg m^2)
b. What is the distance between the pivot point and the center of mass of the pendulum?
(answer 47.7 cm)
c. Calculate the period of oscillation.
(answer 1.5s)

Homework Equations

Ic for rod is 1/12 x M x L^2
Ic for disk is 1/2 x M x R^2
Io is Ic + mh^2, where h is from the point of pivot to the center of mas.
T=2 x pi x [Io / (mgh)]^(1/2)

The Attempt at a Solution

a. Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk
Io = Ic + mh^2, I don't know what to do with the two Ic, and cannot find the Io without h, which is to be found in question b. What am I missing?

b. Io = Ic + mh^2 = .205 kg m^2 = Ic + (.770kg)(h^2) to find h, but cannot find Ic with the two previous number I got. No Idea Need Help!

c. T= 2 x pi x [Io / (mgh)]^(1/2) =
2 x pi x [(.205kg m^2) / [(.770kg)(9.81m/s^2)(.477m)]]^(1/2)= 1.5s
can get with the answers for a and b

If someone had msn (horseland20@hotmail.com) I will be on if you can help if not post!
I would appreciate the help.

 

[LandOfStandar]

Any one able to help, I have spent over two hours getting no where. Suggestion may help!

 

[Doc Al]

The Attempt at a Solution

a. Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk

So far, you've calculated the moment of inertia of each piece about its own center of mass. Now use the parallel axis theorem to find the moment of inertia of each piece about the pivot point.

Io = Ic + mh^2, I don't know what to do with the two Ic, and cannot find the Io without h, which is to be found in question b. What am I missing?

You don't need the center of mass of the pendulum to answer this part of the question.

 

[LandOfStandar]

So far, you've calculated the moment of inertia of each piece about its own center of mass. Now use the parallel axis theorem to find the moment of inertia of each piece about the pivot point.

I don't mean to be dumb, but i thought I used the parallel axis theorem, because that is where I from the formulas.

 

[Doc Al]

Where did you use the parallel axis theorem? I just see your calculations for the two Ic's.

 

[LandOfStandar]

My textbook stats the parallel axis theorem for a disk is 1/2 x M x R^2
and a thin rod 1/12 x M x L^2, maybe I call is it Ic when it is not.

 

[Doc Al]

My textbook stats the parallel axis theorem for a disk is 1/2 x M x R^2
and a thin rod 1/12 x M x L^2, maybe I call is it Ic when it is not.

Please double-check your book--it can't be that bad!

Ic means the moment of inertia about the center of mass. Those are just the formulas for calculating the moment inertia of those two shapes about their centers.

The parallel axis theorem allows you to use Ic to find the momentum of inertia about some other (parallel) axis. That's what you need to use to find the moment of inertia about the pivot point. Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html"

 

[LandOfStandar]

You are correct, I am reading my book wrong. they are Ic formulas.

So what I am stuck on is what is the d(in the formula and the site) or h in the one I stated Io is Ic + mh^2.

I have no idea, except is h not what b is asking for?

 

[Doc Al]

For part a, you need the moment of inertia of each piece about the pivot point. So "d" in the parallel axis formula is the distance from the center of each piece to the pivot point.

In part b, you need the distance from the pivot to the center of mass of the pendulum. (So you would first figure out where the center of mass is.)

 

[LandOfStandar]

ok, thanks for clearing that up!

to make sure,

Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk

I= Ic of rod + Md^2 = (.005625 kg m^2) + (.270kg)(.250m)^2 = .0225kg m^2 both mass and I = Ic of disk + MD^2 = (.0025 kg m^2) + (.500kg)(.6)^2 = .1825kg m^2

.0225kg m^2 + .1825kg m^2 = .205 kg m^2 got it.Now help with question b

 

[Doc Al]

Ic = 1/12 x M x L^2 = (1/12) x (.270kg) x (.500m)^2 = .005625 kg m^2 the rod
Ic = 1/2 x M x R^2 = (1/2) x (.500kg) x (.100m)^2 = .0025 kg m^2 the disk

That looks good.

I= Ic of rod + Md^2 = (.005625 kg m^2) + (.270kg)(.250m)^2 = .0225kg m^2

That looks good.

or is M .770kg both mass

No.

and I = Ic of disk + MD^2 = (0025 kg m^2) + (.500kg)(.05)^2 = .00375kg m^2

What's the distance between the center of the disk and the pivot? (It depends on whether the rod ends at the edge or the center of the disk.)

 

[LandOfStandar]

Got it, thank! While I was waiting I reread and got it from what you said before, thank you so much. It feels good to have it!

 

[LandOfStandar]

I got part b and it feel so good to have got this problem!

Thanks So Much!