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

[frozenguy]

Homework Statement

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).

Homework Equations

$$v=\frac{dy}{dt}$$; $$\omega=\frac{d\theta}{dt}$$

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

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

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

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..

 

[Doc Al]

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

 

[frozenguy]

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

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?

 

[Doc Al]

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

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

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

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

 

[frozenguy]

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

Ahhh yes that's right.

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

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

 

[Doc Al]

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

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

 

[frozenguy]

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

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

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

 

[Doc Al]

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

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

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.)

 

[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

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

 

[Doc Al]

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

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.

 

[frozenguy]

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.

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?

 

[Doc Al]

So
$$\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)$$ ... in terms of my diagram?

You left off dθ/dt.

 

[frozenguy]

oops.. you 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 $$mg\frac{1}{2}L=\frac{1}{2}mv^{2}_{cm}+\frac{1}{2}I\omega^2$$ ?

 

[Doc Al]

Yes. Express ω in terms of v_cm.