# What percentage of the total mass of the pendulum is in the uniform thin rod?

1. Nov 28, 2009

### jayced

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

Christy has a grandfather clock with a pendulum that is 1.070 m long.
(a) If the pendulum is modeled as a simple pendulum, what would be the period?
Already solved

(b) Christy observes the actual period of the clock, and finds that it is 1.20% faster than that for a simple pendulum that is 1.070 m long. If Christy models the pendulum as two objects, a 1.070-m uniform thin rod and a point mass located 1.070 m from the axis of rotation, what percentage of the total mass of the pendulum is in the uniform thin rod?
Need help

2. Relevant equations
(a) T=2pi sqrt(L/g)
(b) Couldn't find an appropriate equation in my textbook.

3. The attempt at a solution

(a) T=2pi sqrt(1.070m/9.8m/s^2)=2.08 s

(b) I don't have a clue about the equation that correspond to this part of the problem,no similar examples or anything in my textbook. To start a equation would be great, because I have percentages and meters and they want me to calculate total mass of a pendulum. Any help is great thank you.

2. Nov 28, 2009

### ApexOfDE

My solution is: denoting mass of thin rod is m1 and point mass is m2.

==> torque eq: $$\tau = I \alpha$$

The contribution to torque is due to rod and point mass and similarly for I. Plug them into above eq and use small angle theta to find alpha. You also know that, alpha = -(omega)^2 * theta => you have angular velocity omega and T = 2*pi / omega => percentage

Hope this helps.

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