# Vertical stick falls, rotates about CM, derive v(y,theta)

1. Nov 29, 2009

### frozenguy

1. The problem statement, all variables and given/known data
A uniform stick of mass m and length L, initially
upright on a frictionless horizontal surface, starts falling. The circle at the center of the
stick marks the center of mass. Derive an expression for the speed of the center of mass
as a function of y and θ if the stick falls as shown (with the center of mass moving
straight downward).

2. Relevant equations
$$v=\frac{dy}{dt}$$; $$\omega=\frac{d\theta}{dt}$$

$$v_{cm}$$$$=r\omega$$; $$I=\frac{1}{12}$$$$mL^{2}$$

$$K_{rot}$$=$$\frac{1}{2}$$$$I\omega^2$$

$$K=\frac{1}{2}mv^2$$

3. The attempt at a solution
There are no non-conservative forces so $$E_{mech}$$ is conserved.

Therefore I figure: $$U_{i}+K_{i}=U_{f}+K_{f}$$

So: $$mg\frac{1}{2}L=\frac{1}{2}mv^{2}_{cm}+\frac{1}{2}I\omega^2$$

Then subed in $$v=\frac{dy}{dt}$$ and $$\omega=\frac{d\theta}{dt}$$ and

$$I=\frac{1}{12}$$mL$$^{2}$$, canceled out the (1/2) and m and attempted to integrate the equation.

mg and L are all constants right? So I got $$0=2m\frac{dy}{dt}y+\frac{1}{12}2mL^2\frac{d\theta}{dt}\theta$$ which I don't think is right..

2. Nov 29, 2009

### Staff: Mentor

The basic idea of using conservation of mechanical energy is correct. Hint: Find the relationship between ω and vcm.

3. Nov 29, 2009

### frozenguy

Like $$v=r\omega$$ or $$v=\frac{1}{2}L\omega$$

So $$\frac{dy}{dt}=\frac{1}{2}L\frac{d\theta}{dt}$$ ?

Or is there another relationship I'm supposed to be looking for?

4. Nov 29, 2009

### Staff: Mentor

That's not quite right. Vcm is not the tangential speed of something rotating in a circle. (The center of mass falls straight down.)

Another hint: Find an expression for the height of the center of mass in terms of θ.

5. Nov 29, 2009

### frozenguy

Ahhh yes thats right.

Another hint: Find an expression for the height of the center of mass in terms of θ.[/QUOTE]

so like $$y=\frac{1}{2}L-cos(\theta)$$

6. Nov 29, 2009

### Staff: Mentor

You're getting warmer, but that expression is not quite right.

7. Nov 29, 2009

### frozenguy

hm.. Yeah, if I say $$\theta$$ is 0, then $$y=\frac{1}{2}L-1$$ which I don't want..

So is it $$y=\frac{1}{2}L-sin(\theta)$$ ?

8. Nov 29, 2009

### Staff: Mentor

Nope, not yet. (You can't have a sinθ term, which has no units, added to a length term. That's a tip off that something is wrong.)

Do this. Draw a diagram of the stick when it makes an angle θ with the ground. What's the height of the center? (Find the right triangle.)

9. Nov 29, 2009

### frozenguy

This is what is supplied with the problem. Well, its my version of it but it's pretty accurate. Um, neglecting that part of the angled stick is somehow below the surface of the table :uhh:

oh heh heh.
Ok so its $$y=\frac{1}{2}L-\frac{1}{2}Lcos(\theta)$$..

10. Nov 29, 2009

### Staff: Mentor

That will work. (My diagram would just measure the position of the center from the ground, thus y = L/2 cosθ.) Now take the derivative of both sides.

11. Nov 29, 2009

### frozenguy

So
$$\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)$$ .... in terms of my diagram? It would be negative for yours because as time increases, y is decreasing so the change is negative where as mine is positive because the change in y is increasing?

12. Nov 29, 2009

### Staff: Mentor

You left off dθ/dt.

13. Nov 29, 2009

### frozenguy

oops.. ya so

$$\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)\frac{d\theta}{dt}$$

So do I solve for omega and sub that into my energy equation $$mg\frac{1}{2}L=\frac{1}{2}mv^{2}_{cm}+\frac{1}{2}I\omega^2$$ ?

14. Nov 30, 2009

### Staff: Mentor

Yes. Express ω in terms of vcm.