# Two different answers for compound pendulum problem

#### 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.
Homework 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 $I = 0.572 kg m^{2}$, and the total torque acting on the system as $12.152 N$. 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 $T=2 \pi \sqrt{ \frac{l}{g} }$ relation. which gives a value of 1.33 seconds.

For reference, this what I did explicitly:$$-m_{tot}g\theta = ma$$I used $x = l\theta$ where $x$ is the tangential displacement of the centre of mass and $l$ 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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#### haruspex

Homework Helper
Gold Member
2018 Award
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 $\frac{\Sigma m_ir_i^2}{\Sigma m_i}$

#### etotheipi

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.

#### haruspex

Homework Helper
Gold Member
2018 Award
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.

"Two different answers for compound pendulum problem"

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