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

• frozenguy
In summary: 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.Yes. Express ω in terms of vcm.ω=\frac{dv}{dt}
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..

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

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

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?

frozenguy said:
Like $$v=r\omega$$ or $$v=\frac{1}{2}L\omega$$
That's not quite right. Vcm 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.

Doc Al said:
That's not quite right. Vcm 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 $$y=\frac{1}{2}L-cos(\theta)$$

frozenguy said:
so like $$y=\frac{1}{2}L-cos(\theta)$$
You're getting warmer, but that expression is not quite right.

Doc Al said:
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 $$y=\frac{1}{2}L-sin(\theta)$$ ?

frozenguy said:
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.)

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 $$y=\frac{1}{2}L-\frac{1}{2}Lcos(\theta)$$..

frozenguy said:
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.

Doc Al said:
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?

frozenguy said:
So
$$\frac{dy}{dt}=\frac{1}{2}Lsin(\theta)$$ ... in terms of my diagram?
You left off dθ/dt.

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

Yes. Express ω in terms of vcm.

## 1. How do you define the center of mass in a vertical stick?

The center of mass is the point at which the mass of an object is evenly distributed in all directions. In a vertical stick, the center of mass can be found at the midpoint of the stick.

## 2. What is the equation for calculating the velocity of the stick as it falls and rotates about its center of mass?

The equation for calculating the velocity of the stick is v(y,theta) = √(2gy(1-cosθ)), where g is the acceleration due to gravity, y is the distance from the center of mass to the bottom of the stick, and θ is the angle of rotation.

## 3. Is the velocity of the stick dependent on the length of the stick?

Yes, the velocity of the stick is dependent on the length of the stick. The longer the stick, the greater the distance y from the center of mass to the bottom, resulting in a higher velocity.

## 4. How does the angle of rotation affect the velocity of the stick?

The angle of rotation directly affects the velocity of the stick. As the angle increases, the velocity also increases. This is because the stick is falling a greater distance and therefore has a higher velocity.

## 5. How does air resistance impact the velocity of the stick?

Air resistance can have a significant impact on the velocity of the stick. As the stick falls, air resistance will increase, slowing down the stick's rotation and resulting in a lower velocity. However, for smaller angles of rotation, air resistance is usually negligible and can be ignored in calculations.

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