# Physical Pendulum

#### joshuaa

A pendulum consists of a tiny bob of mass M and a uniform cord of mass m and length L. (a) Determine a formula for the period. (b) What would be the fractional error if one used the formula for a simple pendulum?

#### topsquark

Forum Staff
A pendulum consists of a tiny bob of mass M and a uniform cord of mass m and length L. (a) Determine a formula for the period. (b) What would be the fractional error if one used the formula for a simple pendulum?
Start with the torque equation, $$\displaystyle \tau = I \omega$$ and $$\displaystyle \vec{ \tau } = r \times \vec{F}$$

What is I? What is F and how does it change with the angle?

If that doesn't help much then please tell us what you have done so far.

-Dan

Edit: Come to think of it... Are you talking about a physical pendulum or an ideal pendulum?

#### joshuaa

Thanks topsquark for passing by. I am talking about a physical pendulum and I know that

T = 2π√[ I / mgh ]

I think that the pivot will be at the end of the uniform cord, therefore its moment of inertia, I, is (1/3)mL^2

is it correct to add the moment of inertia like this?

T = 2π√[ mL^2 / 3mgh ]

since h = L

T = 2π√[ L / 3g ]

If this is correct, how to add the mass M of the bob?

#### ebaines

I think you need to start with the fundamental equation, as topsquark suggested. Given

$$\displaystyle \sum \tau = I \alpha$$

You have torques due to (a) gravity acting on the center of the rod plus (b) gravity acting on the bob. And you have moments of inertia consisting of (a) the cord and (b) the bob. So:

$$\displaystyle mg \frac {L \sin \theta} 2 + Mg \sin \theta = (I_{cord} + I_{bob}) \alpha$$

As usual in these problems we can assume small angles of deflection, so $$\displaystyle \sin \theta \approx \theta$$. And since $$\displaystyle \alpha = \ddot \theta$$:

$$\displaystyle \ddot \theta - \frac {(\frac {mL} 2 + ML)}{I_{cord} + I_{bob}} \ \theta = 0$$

Can you take it from here?

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#### joshuaa

Thanks ebaines. Yes, I can.

I think that you forgot to add g in your last equation.

The period now should be

T = 2π√[ (Icord + Ibob) / gL(m/2 + M) ]

If (1/3)mL^2 the moment of inertia of the cord, what is the moment of inertia of the bob?

#### joshuaa

I simplified and got

T = 2π√[ 2L(m + 3M) / 3g(m + 2M) ]

What would be the fractional error if one used the formula for a simple pendulum?

the period in the simple pendulum is

T = 2π√[ L / g ]

Is the fractional error (1/3)?

#### ebaines

You are correct, I left 'g' out of my equation - sorry for any confusion that may have caused.

I simplified and got

T = 2π√[ 2L(m + 3M) / 3g(m + 2M) ]
Correct.

What would be the fractional error if one used the formula for a simple pendulum?

the period in the simple pendulum is

T = 2π√[ L / g ]

Is the fractional error (1/3)?
First let's define what "factional error" means. I think what it means is the amount of error (i.e. the difference between the estimate using $$\displaystyle T=2 \pi \sqrt { \frac L g }$$ and your "correct" answer) divided by the correct value, so in this case you have:

fractional error = $$\displaystyle \frac {2 \pi \sqrt { \frac L g } - 2 \pi \sqrt{ \frac L g \frac {(2m+6M)}{(3m + 6M)}}} { 2 \pi \sqrt{ \frac L g \frac {(2m+6M)}{(3m + 6M)}}}$$

Obviously this can be simplified to a much less messy final answer.

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#### joshuaa

When I simplify, I get this

fractional error = √[ 3(m + 2M) / 2(m + 3M) ] - 1

#### spachelord

I think you need to start with the fundamental equation, as topsquark suggested. Given

$$\displaystyle \sum \tau = I \alpha$$

You have torques due to (a) gravity acting on the center of the rod plus (b) gravity acting on the bob. And you have moments of inertia consisting of (a) the cord and (b) the bob. So:

$$\displaystyle mg \frac {L \sin \theta} 2 + Mg \sin \theta = (I_{cord} + I_{bob}) \alpha$$

As usual in these problems we can assume small angles of deflection, so $$\displaystyle \sin \theta \approx \theta$$. And since $$\displaystyle \alpha = \ddot \theta$$:

$$\displaystyle \ddot \theta - \frac {(\frac {mL} 2 + ML)}{I_{cord} + I_{bob}} \ \theta = 0$$

Can you take it from here?
How do you get Period from this equation?

#### topsquark

Forum Staff
How do you get Period from this equation?
The equation for the frequency of the physical pendulum is similar to the one for the simple pendulum:
$$\displaystyle \ddot{\theta} + \omega ^2 \theta = 0$$

in the small angle approximation. So find $$\displaystyle \omega$$ (just the square root of the $$\displaystyle \theta$$ coefficient) and convert that to a period.

-Dan