Two different answers for compound pendulum problem

In summary, the conversation discusses two different methods for calculating the time period of a pendulum. The first method uses the rotational analogue of Newton's Second Law, while the second method uses the standard T=2π√(l/g) relation. The discrepancy between the two methods is due to the difference in the moment of inertia used. While the first method uses the moment of inertia of the system, the second method uses the moment of inertia of the centre of mass, which is incorrect. This is because the moment of inertia of a system is not necessarily equal to the moment of inertia of its centre of mass.
  • #1
etotheipi
Homework Statement
A pendulum is made up of a light rigid beam of length 𝐿=0.50m and two point masses. The beam is attached to a fixed point at one end. One of the masses is of mass 𝑀=2.0kg and is attached to the beam at the opposite end to this fixed point. The other mass is of mass 𝑚=0.80kg and is attached to the beam a distance 𝑥=0.30m away from the fixed point.
Relevant Equations
$$I = \Sigma m r^{2}$$
I completed this problem in two different ways, and wonder why they give different answers.

Firstly, I calculate the moment of inertia of the system as [itex]I = 0.572 kg m^{2}[/itex], and the total torque acting on the system as [itex]12.152 N[/itex]. Thus I can apply the rotational analogue of NII to write $$-12.152\theta = 0.572\ddot{\theta}$$ which is the SHM condition, with time period of 1.36 seconds. This is the correct answer.

For the second method, I calculated the centre of mass of the rod/particles as being 0.443 m from the pivot, and worked this through in the normal way to obtain the standard [itex]T=2 \pi \sqrt{ \frac{l}{g} }[/itex] relation. which gives a value of 1.33 seconds.

For reference, this what I did explicitly:$$-m_{tot}g\theta = ma$$I used [itex]x = l\theta[/itex] where [itex]x[/itex] is the tangential displacement of the centre of mass and [itex]l[/itex] is the distance of the centre of mass of the pivot.

I know that the moment of inertia of a system is not necessarily equal to the moment of inertia of its centre of mass, so it would obviously wrong to use the moment of inertia of the centre of mass in the first method. I believe the mistake in the second method has something to do with this line of reasoning, but can't pinpoint it since my second method makes no reference to moment of inertia.

Why is it that using the centre of mass of the pendulum does not give the correct answer for time period?
 
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  • #2
etotheipi said:
calculated the centre of mass of the rod/particles
That is not relevant. The centre of mass is based on ##\frac{\Sigma m_ir_i}{\Sigma m_i}##. For radius of inertia you need ##\frac{\Sigma m_ir_i^2}{\Sigma m_i}##
 
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  • #3
haruspex said:
That is not relevant. The centre of mass is based on ##\frac{\Sigma m_ir_i}{\Sigma m_i}##. For radius of inertia you need ##\frac{\Sigma m_ir_i^2}{\Sigma m_i}##

This makes sense, though I still can't quite see why the compound pendulum isn't equivalent to a simple pendulum with a single mass at the centre of mass.
 
  • #4
etotheipi said:
This makes sense, though I still can't quite see why the compound pendulum isn't equivalent to a simple pendulum with a single mass at the centre of mass.
Because, as I showed, it does not have the same moment of inertia.
Consider the extreme cases: with one mass at the pivot you are left with a simple pendulum length L; with the two masses together you have a simple pendulum length L/2.
 

1. How do I determine the period of a compound pendulum?

The period of a compound pendulum can be determined by using the equation T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass of the pendulum, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass of the pendulum.

2. What is the difference between a simple pendulum and a compound pendulum?

A simple pendulum consists of a single mass attached to a string or rod, whereas a compound pendulum consists of multiple masses distributed along a rigid body. Additionally, a simple pendulum swings back and forth in a straight line, while a compound pendulum swings in a circular or elliptical motion.

3. How do I calculate the center of mass for a compound pendulum?

The center of mass for a compound pendulum can be calculated by dividing the sum of the products of each mass and its distance from the pivot point by the total mass of the pendulum. This can be represented by the equation xcm = (m1d1 + m2d2 + ... + mndn) / (m1 + m2 + ... + mn), where xcm is the position of the center of mass, mi is the mass of each component, and di is the distance of each component from the pivot point.

4. Why do I get two different answers when solving a compound pendulum problem?

There are two different ways to approach a compound pendulum problem: using the equation T = 2π√(I/mgd) or using the equation T = 2π√(I/mgd + k), where k is a correction factor for the distance between the pivot point and the center of mass. Depending on which equation is used, the resulting answer may vary slightly.

5. How can I account for air resistance in a compound pendulum problem?

In order to account for air resistance in a compound pendulum problem, you can use the equation T = 2π√(I/(mgd) + k), where k is a correction factor that takes into account the effects of air resistance. This factor can be determined experimentally by comparing the period of the pendulum with and without air resistance.

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