What other forces act on a rod-pivot system besides centripetal force?

[SN1987a]

I have a little conceptual question about a rod-pivot system.

We have this rod (mass M, length l) pivoting about a fixed point. It is released at 30 degrees above the horizontal, and we're asked for the magnitude and direction force on the pivot when the rod reaches the horizontal.

So i figured one of the forces on the pivot would be the centripetal force keeping the rod in a circular motion. This I can "easily" calculate, and it's always pointing in the -r direction (opposite to the rod).

But I can't figure out what and if there are other forces. In fact, I "believe" there are no others. But the question is asked in such a way as to hint there might be something else.

If you have any insights, please share tehm with me.

 

[SN1987a]

further complications

It actually turns out even calculating the centripetal force is not as trivial as I had expected.

I'm trying to calculate the acceleration knowing that I \alpha=F r (center of mass). I know I = \frac {1}{2} m l^2, r=\frac {l}{2} and that F=-m g sin(\theta), where the angle is the angle between the rod and the horizontal.

But this means that
\frac {d^2\theta}{dt^2}=\frac{-3 g sin (\theta)}{2l}
and I don't know how to solve this ODE.

Usually, in class, the teacher approximates sin(\theta)=\theta, but in my case the angle is quite big for this.

What can I do about this?

 

[Icebreaker]

I don't think you should set it up as a second order differential equation. What class is it?

 

[SN1987a]

differential equation

This is an intro to classical mechanics problem.

I agree that I shouldn't run into this sort of differential equation. However, this is what comes out of the motion equations once I set them up. If you see any mistake in the setup, let me know.

On the other hand, this wouldn't be the first time the teacher expects us to know the solution of a crazy diffrential equation, even though we've never seen them before.

I still haven't figured out what other forces are applied on the pivot. Can it be that some of the weight of the rod also acts on it?

 

[Tide]

If it's a classical mechanics class then you have learned about energy conservation. Use it! :)

 

[SN1987a]

tentative solution

ok, so from the conservation of energy i know that K(\theta)+U(\theta)=E_T.

Since
U(\theta)=\int_{0}^{L}\!\frac{M}{L}g l sin(\theta){dl}=\frac{1}{2}MgLsin(\theta)

and
K(\theta)=\frac{1}{2}I\omega^2=\frac{1}{6}ML^2\left(\frac{d\theta}{dt}\right)^2

and initially all energy was potential, therefore
E_T=U(30\degree)=\frac{1}{4}MgL

I have that
\left(\frac{d\theta}{dt}\right)^2=\frac{3g}{L}\left(\frac{1}{2}-sin(\theta)\right)

Now, to come back to the total force exerted on my pivot, it should have two components: the gravitational component Mgsin(\theta), which becomes 0 for \theta=0, and a force opposite to centripetal force (which acts on the rod). This force should be
F(\theta)=-F_c=-Ma_r=-M\left(-r\left(\frac{d\theta}{dt}^2\right)\right)
where the effective radius r=\frac{L}{2}

Substituting for r and the derivative of theta found above, I get
F(\theta)=\frac{3Mg}{2}\left(\frac{1}{2}-sin(\theta)\right)
F(0)=\frac{3Mg}{4}
Does this make any sense?

 

[lightgrav]

(most people refer to the "time derivative of theta"
as the angular velocity, omega . I'm using "w" for it.)

in a simple pendulum at the bottom, T = Mg + Ma = Mg + Mrw^2 .

I think that your ½Iw^2 = ½MgL sin(30) + ½MgL ,
since the PE at bottom should equal -½MgL (pivot is h=0)

 

[SN1987a]

I'm not sure what you mean. The angle I'm considering is the angle with the horizontal. My starting point is 30 degrees above the horizontal and my endpoint is the horizontal. Therefore, I believe
\frac{1}{2}I\omega^2=U(0)-U(30)=\frac{1}{2}MgLsin(0)-\frac{1}{2}MgLsin(30)=0-\frac{1}{4}MgL

which I realize now comes to the same as what I was doing before, it's just simpler.

 

[lightgrav]

Whoa, sorry ... I read that one wrong!
I was going from your 30 to your -90
I'd better go to bed ...:zzz:

 

[SN1987a]

Don't worry man. No problem. Actually forced me to review my calculations, which is never a bad thing to do