# Period of A Pendulum With Moveable Mass

1. Dec 8, 2013

### Wreak_Peace

1. The problem statement, all variables and given/known data

The figure below shows the pendulum of a clock in your grandmother's house. The uniform rod of length L = 2 m has a mass m = 0.8 kg. Attached to the rod is a uniform disk of mass M = 1.2 kg and radius 0.15 m. The clock is constructed to keep perfect time if the period of the pendulum is exactly 3.5 s.

a) What should the distance 'd' be so that the period of the pendulum is 2.5 seconds?

b)Suppose that the pendulum clock loses 5.00 min/d. To make sure your grandmother will not be late for her quilting parties, you decide to adjust the clock back to its proper period. How far and in what direction should you move the disk to ensure that the clock will keep perfect time?

2. Relevant equations

Parallel Axis Theorem: I = Icm + MR2
I of a rod around it's edge = (ML2)/3
I of a disk = (MR2)/2

Period of a physical pendulum = T = 2∏√(I/Lmg)

3. The attempt at a solution

The I of the entire system should be I= (ML2)/3 + (M(R+d)2)/2
We are solving for d, with T = 2.5 seconds, so we plug this in for our equation for period.

T = 2∏√(I/Lmg)

T = 2∏√(((ML2)/3 + (M(R+d)2)/2)/Lmg)

Where I am stuck, however, is what would I use for 'L' in the denominator of the equation for the period. When I assume it is 2 meters, the length of the entire stick, I get 2.8m for d, which doesn't make sense. mR is the mass of the rod, md is the mass of the disk.

I got that answer by rearranging for d.
I end up with T^2 *L*mg/(4∏2) = I

T^2 *L*mg/(4∏2) =
(mR *L2)/3 + (md (d+r)2/2)

Then, 2(T^2 *L*mg/(4∏2) - (mR *L2)/3)/md = (d+r)2

I plugged in numbers for everything that is given, and end up with 8.565 = (d+.15)2, and by solving that, I get ~2.8m, which would mean the disk is off the rod.

What am I doing wrong?

2. Dec 8, 2013

### voko

What does $L$ mean? You cannot just plug random numbers in for it. Find out.

3. Dec 8, 2013

### Wreak_Peace

I assumed L was 2 meters because that was the length of the rod. Is L the distance from the center of mass to the pivot point?

4. Dec 8, 2013

### voko

Yes it is. What is it here?

As a general note, do not guess anything. If some symbol in a formula is obscure, find out what it is exactly. These days, with the Internet, Google and Wiki, it is not difficult at all, so you don't have a valid excuse for not doing that.

5. Dec 8, 2013

### Wreak_Peace

So, because we don't know how far the disk will be, for calculating the center of mass,
Xm = .8*1 + d*1.2
and I got .3411m, which is also incorrect, though much more reasonable.

I think I was incorrect when stating the I of the disk, I stated the I of a disk around it's center + the M*d^2, due to the parallel axis theorem. The moment of inertia of this disk can be treated as the same as a point mass, so the I of the disk will only be m*d^2, not m*d^2 + m*r^2.

Using that, I got an answer of .359 m, which is also incorrect.

6. Dec 9, 2013

### voko

The formula for the center of mass should be $X_m = \dfrac {m_{rod} L_{rod} / 2 + m_{disk} d} { m_{rod} + m_{disk} }$. I do not think this is quite what you got.

I do not see your formula for the total moment of inertia, but I disagree with this: "the moment of inertia of this disk can be treated as the same as a point mass".