# Homework Help: Physical Pendulum Question (Mass on a Grandfather Clock)

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1. Jul 2, 2015

### BrettF

1. The problem statement, all variables and given/known data
You are at a furniture store and notice that a grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows donw; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of 0.51 kg, where should you place the mass to give the pendulum a period of two seconds exactly? Measure the distance in meters from the top of the pendulum.

2. Relevant equations
Equation for the period of a physical pendulum:

T = 2π*√(I/(Mgd)) I is the moment of inertia, m is the mass of the entire pendulum, d is distance to the center of mass
I = (1/3)ML^2 + mL^2 Inertia for the pendulum: I added the inertia of the rod and the inertia of the weight at the end, not sure if this is correct

3. The attempt at a solution
Since the period is 2, this is what I did:

2 = 2π*√(I/((M+m)gd)) I substituted M in equation 1 for (M+m), the mass of the rod and weight since it is the total mass
1/π^2 = I/((M+m)gd) Simplification
d = Iπ^2/((M+m)g)
d = ((1/3)ML^2+mL^2)π^2/((M+m)g)
d = ((1/3)*1.23kg*(1.25m)^2+0.51kg*(1.25m)^2)π^2/((1.23+0.51)kg*9.8m/s^2) = 0.832 m
Since d is the distance to the center of mass (the center of the rod), I added L/2 to get the distance to the end of the rod, so 0.832 m + (1.25 m / 2) = 1.46 m. The correct answer should be 1.19 m, what am I doing incorrectly?
Thank you in advance for the help.

2. Jul 2, 2015

### Orodruin

Staff Emeritus
The moment of inertia also depends on where the mass is located.

3. Jul 2, 2015

### BrettF

Ahh, that makes sense, the moment of intertia for the weight is not mL^2 since it's not at the end of the rod, it should be m(d+L/2)^2 right?
So the inertia should be I = (1/3)ML^2 + m(d+L/2)^2 [I think]

Except I solved for d using
d = Iπ^2/((M+m)g)
and substituting the I from above and got the quadratic

mπ^2d^2 + (mπ^2L - (M+m)g)d + π^2ML^2/3 + mπ^2L^2/4 = 0, and this gives me a non real answer when I use the quadratic formula. What else am I missing? (I think this is closer to the right track since I put d into the inertia equation, that makes more sense thank you for pointing that out)

4. Jul 2, 2015

### Orodruin

Staff Emeritus
Think again. Where is the mass located with respect to the center of mass? The problem will be easier to solve if you use the place where you put the mass as a variable instead.

5. Jul 2, 2015

### BrettF

What I'm thinking is that d is the distance from the center of mass, but the pivot point is the end of the rod. I guess I'm just confused, would you be able to rephrase the question please? Thank you for your help.

6. Jul 2, 2015

### Orodruin

Staff Emeritus
As long as you are consistent, it does not matter what you call things. But you used d as: The center of mass distance from the pivot, the distance of the mass from the center of the rod, and the distance of the center of mass from the center of the rod.

7. Jul 3, 2015

### BrettF

Ahh this makes sense and is very clear, thank you. In the calculation I just did I used d just as the distance from the mass to the pivot point and got the right answer. Thank you so much for your help!