Introductory Physical Pendulum Problem

[Zipzap]

Homework Statement

A pendulum is constructed using a thin rod (m1 = 2.0 kg, L = 1.0m) and a uniform sphere (m2 = 1.0 kg, R = 0.50 m). The period in "s" for small oscillations is:
a) 1.5
b) 1.7
c) 2.0
d) 2.2
e) 2.5

Homework Equations

T = 2pi *sqrt(I/mgd)

The Attempt at a Solution

To start off, the rod is pivoting at an end attached to a ceiling with the sphere attached to its bottom.
I initially got (c) by assuming it was a simple pendulum ("thin rod", lol), but a more careful look reveals that the rod does not have negligible mass, and so this pendulum must be physical. My biggest dilemma with this problem is finding the center of mass for this system, and then finding "d" assuming that. Unfortunately, inertia is my weakest area in Physics, and so I am stuck as to how to use the parallel axis theorem correctly to get my answer.
Any hints?
If it helps, people in my class keep getting (d) or (e).

 

[ashishsinghal]

show an attempt on finding moment of inertia and center of mass.

 

[Zipzap]

All right, here goes nothing =P

I know that the rod has inertia of (1/3)ML^2, and the sphere has (2/5)MR^2...

I also know I = integral of r^2, r = L+R...

Am I on the right track here?

 

[supratim1]

take inertia about the point of suspension.

 

[Zipzap]

What do you mean by that? You mean where the rod is being suspended?

 

[ashishsinghal]

All right, here goes nothing =P

I know that the rod has inertia of (1/3)ML^2, and the sphere has (2/5)MR^2...

About which axis.
You have to take moment of inertia about line perpendicular to the plane in which pendulum swings and passing through point of suspension

 

[Zipzap]

So then I can just simply add inertias for both the rod and sphere to get my moment of inertia?

 

[samratk]

reply

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[supratim1]

yes right. proceed that way.

 

[Zipzap]

And I can add them both because the pivot point is unchanged and the sphere itself contributes to the overall inertia, right?

Anyhow, I get (1/3)ML^2 + (2/5)MR^2, giving me a numerical value of 0.7666666 when I sub in my values.

What do I do about "d" when I input it into the formula for periods? =S

 

[ashishsinghal]

Anyhow, I get (1/3)ML^2 + (2/5)MR^2, giving me a numerical value of 0.7666666 when I sub in my values.

No. The moment of inertia of the sphere is not (2/5)MR^2 about the required axis. It is (2/5)MR^2 about its. You will need to find its moment of inertia about the required axis using parallel axis theorem.

 

[Zipzap]

About its axis? But it's attached to the rod, how am I supposed to find it using parallel axis theorem??

 

[supratim1]

the distance of the centre of sphere from point of suspension is equal to length of rod.

 

[Zipzap]

Ok, now I'm confused:

I know that I have to add the inertias together. The sphere will have a different inertia that I have find using the parallel axis theorem (no clue how to do this), and without the center of gravity, I have no idea how to find "d"

Any other helpful advice?

 

[ashishsinghal]

ok, let us start from the basics. When you want to find angular acceleration, you find it about an axis of rotation. Now, how do you find it? You calculate torque about that axis and the moment of inertia about that axis. Now, you have a relation:
$$\tau$$ = I $$\alpha$$

In your question the axis to be taken is the one mentioned earlier. For using the above relation you have to calculate moment of inertia about about that axis. Now, you know that moment of inertia of the sphere is 2/5 Mr^2 about its diameter. also distance between diameter and that axis is the length of the string.

using parallel axis theorem moment of inertia of the sphere is 2/5 Mr^2 + ML^2 where L is the length of the string.