# Physical pendulum

1. Dec 18, 2007

### JolleJ

[SOLVED] Physical pendulum

1. The problem statement, all variables and given/known data
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?

2. Relevant 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).

3. 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.

2. Dec 18, 2007

### Staff: Mentor

How did you conclude this?

3. Dec 18, 2007

### 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".

4. Dec 18, 2007

### Staff: Mentor

Crank it out by hand and check for yourself.

5. Dec 18, 2007

### 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.

6. Dec 18, 2007

### Staff: Mentor

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.

7. Dec 18, 2007

### 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. :)

8. Dec 19, 2007

### 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)

9. Dec 19, 2007

### Staff: Mentor

This is true.
No, you can't just ignore the mass distribution of the rod. You must consider the rotational inertia of the entire system.
No. That's the period of a simple pendulum; this must be treated as a physical pendulum.

10. Dec 19, 2007

### 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.

11. Dec 20, 2007

### 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.

12. Dec 20, 2007

### rl.bhat

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

13. Dec 20, 2007

### Staff: Mentor

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

14. Dec 20, 2007

### 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 !

15. Dec 20, 2007

### 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.

16. Dec 20, 2007

### JolleJ

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

17. Dec 20, 2007

### Staff: Mentor

Yep, it works. Just by a hair.

18. Dec 20, 2007

### Shooting Star

Well, it was a close shave then.

19. Dec 20, 2007

Haha. :)