Angular speed/acceleration

1. Jun 21, 2005

texasgrl05

not sure how i get angular speed by knowing the linear speed??

On an open-reel tape deck, the tape is being pulled past the playback head at a constant linear speed of 0.337 m/s.
a) If the radius of the film taken up so far (as shown in the upper part of the drawing) is 0.0561 m, find the angular speed of the take-up reel.
(b) After 2.80 103 s, the take-up reel is almost full, as the lower part of the drawing indicates. The radius of film taken up so far is now 0.114 m. Find the average angular acceleration of the reel and specify whether the acceleration indicates an increasing or decreasing angular velocity.

2. Jun 21, 2005

Staff: Mentor

The linear speed of a point on a rotating object is given by: $v = \omega r$, where r is the distance from the axis.

3. Jun 21, 2005

wisredz

Isn't the angular speed the rate at which the angle theta which is formed by the line connecting the particle to the axis changes? Then I write the formula for displacement as follows using the Cosine rule:

$$\sqrt(2r^2-2cos\theta * r^2)$$

where r is the radius of the circle it is moving on. Then I derive this to get the velocity whixh yields

$$f'(\theta)=\frac{r^2sin\theta}{f(\theta)}*\frac{d\theta}{dt}$$

where $\theta$ is a differantiable function of time. But then I don't know what to do. Any help about this?

4. Jun 21, 2005

geosonel

you're correct about the angular speed (dθ/dt) ... which btw has units radians/second.

however, your cosine rule calculates the length of the triangle side opposite central angle θ (and connecting the 2 radii) between particle starting point (r, θ=0) and new position (r, θ). this is NOT the same as the distance along the circular arc on which the particle is moving.

because your formula involves the length of the triangle side between (r, θ=0) & (r, θ) and NOT the distance along the circular arc between (r, θ=0) & (r, θ), your derivative is NOT the particle's velocity.

this is the formula for particle velocity:

particle velocity = r*(dθ/dt) ... where vel in m/sec, r in m, and (dθ/dt) in radians/sec

your derivative is the rate of change of the triangle side ... which is not the particle velocity.

5. Jun 21, 2005

Staff: Mentor

Yes.
This is the displacement as measured from some particular point on the circle, not the center. It varies from 0 to 2r. It does not describe circular motion. It's rate of change will not be the speed of the particle about the center, except at theta = 0.

If you evaluate your derivative at $\theta = 0$ you will find that it equals $\omega r$.

Instead of what you did, try this. Imagine the particle tracing a circle at constant angular speed. The particle's x and y components can be written as:
$x = r \cos \theta$
$y = r \sin \theta$
Now take the derivative of these to find the components of the particle's velocity:
$v_x = - \omega r \sin \theta$
$v_y = \omega r \cos \theta$
Now find the magnitude of the velocity:
$v^2 = v_x^2 + v_y^2$
Thus verifying that:
$v = \omega r$

I hope that helps.

6. Jun 22, 2005

wisredz

Wow, I saw my mistake thanks a lot. Btw, using a parametric equation is really good. But if I think about the lenght of the arc I find the angular speed more easily. I think this one's right.

The circumference of the circle would be $2\pi*r$ and the length of the arc is $\frac{2\pi*r*\theta}{2\pi}=r\theta$ I derive this and get

$$v=r*\frac{d\theta}{dt}$$

Thanks a lot guys, I really appreciate it.

7. Jun 22, 2005

Staff: Mentor

Even better. Excellent.