- #1

BrettF

- 5

- 0

## Homework Statement

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.

## Homework 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

## 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.