Solve Physical Pendulum: Length, Mass, Period

[JolleJ]

[SOLVED] Physical pendulum

Homework Statement

A physical pendulum constists of a rod (length = 0,30 m og mass = 0,80 kg). The rod rotates around its center. On this rod there is a moveable mass (mass=0,80kg).
Now the period of the pendulum is T = 0,83 s.

The problem to solve is: How far below the center of the rod is the mass placed?

Homework Equations

$$T=2\pi*\sqrt{\frac{I}{m*g*a}}$$
From what I have read a is the distance between the center of the rod and the center of gravity.
Also: For a rod: I = 1/12 * m * l^2, where l is the length of the rod.
And for the mass shifted from the center I think it is I = m*x^2 - where r is the shifted distance. (From Steiners equation where I0 = 0).

The Attempt at a Solution

Moment of inertia I:
I = 1/12*0,8*0,3^2+0,8*x^2
Center of gravity a:
a = 0,80*x/1,6 = 0,5*x

Then this equation:
$$T=2\pi*\sqrt{\frac{I}{m*g*a}}$$
is solved with respect to for T = 0,83 s. However this gives L = Ø. That is, a non existing result.

I really hope that someone can help me, and tell me what I've done wrong.

Thanks in advance.

 

[Doc Al]

However this gives L = Ø. That is, a non existing result.

How did you conclude this?

 

[JolleJ]

Just by entering:
solve(0.83=2*%pi*sqrt((1/12*0.8*0.3^2+0.8*x^2)/(1.6*9.82*0.5*x)),x);
on a calculator. It says "false".

 

[Doc Al]

Crank it out by hand and check for yourself.

 

[JolleJ]

Okay, now I tried. I also get "false"...
First I did something wrong apparently and got 0,023 m / 0,31 m... Not sure what it was though.
However; still L = Ø. :s Something is wrong with the equation that I've made.

 

[Doc Al]

OK, you're right: Given the values that you posted, the resulting quadratic has no real solutions. I don't see anything wrong with the method used, so I suspect that there's a typo (or error) in one or more of the values.

 

[JolleJ]

Very strange, given that it's an examination task.
However, thank you very much (once again, again!). :D
I will write my teacher and ask him, if he's made a mistake. :)

 

[rl.bhat]

When the rod rotates around its center, the period is infinity. When a mass is attached to the rod the period is affected by only due to this mass, and its distance from the center. So T = 2*pi*sqrt(m*a^2/m*g*a) = 2*pi*sqrt(a/g)

 

[Doc Al]

When the rod rotates around its center, the period is infinity.

This is true.

When a mass is attached to the rod the period is affected by only due to this mass, and its distance from the center.

No, you can't just ignore the mass distribution of the rod. You must consider the rotational inertia of the entire system.

So T = 2*pi*sqrt(m*a^2/m*g*a) = 2*pi*sqrt(a/g)

No. That's the period of a simple pendulum; this must be treated as a physical pendulum.

 

[rl.bhat]

Moment of inertia I:
I = 1/12*0,8*0,3^2+0,8*x^2
This I is about the center of the rod. In the formula we want MI about CM.
So I = Icm + M(x/2)^2 or Icm = I - M(x/2)^2 where M is the total mass.
Just by entering:
solve(0.83=2*%pi*sqrt((1/12*0.8*0.3^2+0.8*x^2)/(1.6*9.82*0.5*x)),x);
on a calculator. It says "false". Rewright it as
0.83 = 2*pi*sqrt[(1/12*0.8*0.3^2 + 0.4*x^2)/(1.6*9.82*0.5x)] and solve.

 

[Shooting Star]

We all tried to, but b^2-4ac of the resulting quadratic is negative. Also, the thing, at least to me, is neater if you use CGS. The eqn I got is x^2 -(34.24)x + 600 = 0.

 

[rl.bhat]

The eqn I got is x^2 -(34.24)x + 600 = 0.
It should be x^2 -(34.24)x + 150 = 0.

 

[Doc Al]

Moment of inertia I:
I = 1/12*0,8*0,3^2+0,8*x^2
This I is about the center of the rod. In the formula we want MI about CM.

No, we need the moment of inertia about the point of suspension, which is the middle of the rod.

So I = Icm + M(x/2)^2 or Icm = I - M(x/2)^2 where M is the total mass.

Not sure what that's supposed to be. If you mean that to be the MI about the center of mass of the system, that's not quite right. Since the CM is at a point x/2 from the middle of the rod, the MI about that point would be: 1/12ML^2 + M(x/2)^2 + M(x/2)^2 = 1/12ML^2 + 1/2Mx^2. (Of course, this MI is irrelevant, as the CM is not the point of suspension.)

 

[JolleJ]

Thank you very much, all of you, however I just got respons from my teacher, who said there was an error in the question. It should be T = 0,835 s. And now everything works.

Thanks again !

 

[Shooting Star]

I have not repeated the calculation, but I have a strong doubt that changing the time period by such a small amount would resolve the problem. I ask the OP to do the calculation once more, even if he's fed up by this msg from me.

 

[JolleJ]

It works now, so there's no problem. :) Thanks though. :D

 

[Doc Al]

Yep, it works. Just by a hair.

 

[Shooting Star]

Well, it was a close shave then.

 

[JolleJ]

Haha. :)