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

[jayced]

Homework Statement

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

Homework Equations

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

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.

 

[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.

 

[kerimek]

I would approach this problem by first understanding the concept of center of mass and how it relates to the distribution of mass in a pendulum. The center of mass is the point where the entire mass of an object can be considered to be concentrated. In a uniform thin rod, the center of mass is at the midpoint of the rod.

To find the percentage of the total mass of the pendulum that is in the uniform thin rod, we can use the equation for center of mass:

x_cm = (m_1x_1 + m_2x_2 + ... + m_nx_n) / (m_1 + m_2 + ... + m_n)

where x_cm is the center of mass, m is the mass of each component, and x is the distance of the component from the axis of rotation.

In this case, we have two objects: the uniform thin rod and the point mass. Let's assume that the total mass of the pendulum is M, and the mass of the uniform thin rod is m_1. The point mass would then have a mass of M - m_1.

Since the distance of the uniform thin rod from the axis of rotation is 1.070 m, we can rewrite the equation as:

1.070m = (m_1 * 1.070m + (M - m_1) * 0) / M

Solving for m_1, we get:

m_1 = 1.070m * M / (1.070m + 0)

m_1 = 1.070m * M / M

m_1 = 1.070m

Therefore, the percentage of the total mass of the pendulum that is in the uniform thin rod is:

(m_1 / M) * 100% = (1.070m / M) * 100%

So, if we know the total mass of the pendulum, we can calculate the percentage of that mass in the uniform thin rod. I hope this helps!