Why Does Mass Affect the Period in This Pendulum Problem?

[bcjochim07]

Pendulum-why is this wrong??

Homework Statement

A 14 cm long 230 g rod is pivoted at one end. A 17 g ball of clay is stuck on the other end. What is the period if the rod and ball swing as a pendulum?

Homework Equations

T= 2pi*sqrt(L/g)

The Attempt at a Solution

I used the above equation

T=2pi*sqrt(.14m/9.80)
T=.75 s which is wrong

I thought that the period of a pendulum didn't depend on mass, so why is this different?

 

[alphysicist]

bcjochim07,

The formula you posted in relevant equations applies to a simple pendulum, where a mass is contained all at the end of a massless string. This is a physical pendulum and has a different formula for the period.

 

[bcjochim07]

So would i find the center of mass by doing this:

let the position of the ball be x=0

cm= [(.02kg)(0m) + (.200kg)(.07m)]/.22kg

Then I would plug that into omega= sqrt[Mgl/I]
where l is the distance between the center of mass and the pivot

Then I would find the moment of inertia by first finding the moment of inertia for the rod and then for the ball and adding those together. Then I do 2pi/omega to get the period

 

[bcjochim07]

hmmm...I also tried to find the moment of inertia by using the total mass and the distance between the center of mass that I calculated and I didn't come up with the same moment of inertia as I did when I calculated them individually. Does this mean I calculated the center of mass wrong?

 

[alphysicist]

In your formula l is the distance between the center of mass and the pivot, so you probably want to make let the position of the pivot be x=0. (The way you did it is fine, but it gives the distance of the center of mass from the ball, so you would need to do a subtraction to find the l to plug in.)

c.o.m. = ( 0.017 * 0.14 + 0.230 * 0.07 ) / (0.0247)


I'm not sure I understood what you were doing in your last post (#4). Were you asking if you could find the moment of inertia by finding how far the center of mass is from the pivot and doing something like Mr^2? If so, that's not correct. It would predict, for example, that the I for a uniform solid sphere is 0 for an axis about it's center. (Because its center of mass is at its center.) But we know that's not true.

 

[bcjochim07]

oops had the wrong #s in there

Ok- so is this the correct way to do the problem:
for a pivoting rod: I=(1/3)ML^2
so I= (1/3)(.230kg)(.14m)^2
I= .0015 kgm^2 for rod

For the ball
I=mR^2
I=(.017kg)(.14m)^2
I=3.33E-4

add them together: = .0018 kgm^2

Then for the center of mass:
( 0.017kg * 0.14m + 0.230kg * 0.07m ) / (0.247kg) = .0748 m

omega= sqrt[Mgl/I]
so omega= sqrt [.247kg*9.80*.0748/.0018]
= 100.59 rad/s

2pi/100.59 = .062 sec?

 

[alphysicist]

Hi bcjochim07,

I noticed that's rather small for the period! I think you forgot to take the square root when you calculated omega.

 

[bcjochim07]

oh yep... so T=.6265s