# Homework Help: Speed of a point on a wheel as a function of θ

1. Oct 29, 2014

### bobred

1. The problem statement, all variables and given/known data
A point P on a wheel of a bike radius R, P is a distance r from the axle.
The speed of the bike is v, what is the speed of the point P as a function of θ?

2. Relevant equations
$$x=R\theta - r \sin\theta$$
$$y=R - \cos\theta$$
$$\rho=\frac{r}{R}$$
Distance travelled by the point in one revolution
$$D =R{\displaystyle \int_{0}}^{2\pi}\sqrt{\left(1+\rho^{2}\right)-2\rho\cos\theta}\,\textrm{d}\theta$$

3. The attempt at a solution
Speed is distance per unit time, but not sure where to begin as no function contains time.
Can I make
$$\theta=\omega t$$? If so how can I solve the integral?
James

2. Oct 29, 2014

### phinds

Seems like an oddly worded question, to me. The speed of the rim is going to be the same all the way around, not dependent on some particular point, as specified by an angle. Of course, you haven't actually said what θ IS so maybe I'm misinterpreting the problem.

Oh ... wait ... I'm thinking of the speed relative to the AXLE. I guess the problem is to specify a point by an angle and then get the speed of that point relative to the ground, yes?

3. Oct 29, 2014

### bobred

θ is the angle of rotation of the wheel.
I think you are right, as v is the speed of the bike itself.
James

4. Oct 29, 2014

### phinds

Well, in any case, you're going to need a defined reference position for θ. Maybe you're supposed to take the axle as the XY origin and the X axis as zero degrees?

5. Oct 29, 2014

### vela

Staff Emeritus
The y-equation can't be right. Looks like you forgot $r$.

Instead of $R\theta$, try just writing $vt$.

6. Oct 29, 2014

### LCKurtz

Where did you get these equations? Did you derive them or copy them from a book? Either way, what is your definition of $\theta$? Also, as vela pointed out, the $y$ equation can't be correct.

What does that equation have to do with it?

The problem asks for the rate of change with respect to $\theta$. Once you have the position vector $\vec P(\theta)$ you want $v(\theta) = |\vec P~'(\theta)|$.

7. Oct 29, 2014

### bobred

Hi, I missed the r from the expression for y

$$y=R-r\cos\theta$$

The expressions came from the question sheet.
Thanks Vela,LCKurtz.

8. Oct 29, 2014

### LCKurtz

There are a couple of other things you might like to observe about this problem. Take the case when $R=r$. Look at the speed of the point on the edge of the wheel when it touches the ground vs. when it is at the top of the wheel(whatever values of $\theta$ those are in your setup. You will see the speed isn't constant like it would be for just a spinning wheel. Yet the acceleration is of constant magnitude and directed towards the center in both cases.

9. Oct 30, 2014

### bobred

What I get is

$$v(\theta) = |\vec P~'(\theta)|=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}=\sqrt{(R-r\cos)^2+r^2\sin^2\theta}$$

$$v(\theta) = R\sqrt{(1+\rho^2)-2\rho\cos\theta}$$

10. Oct 30, 2014

### vela

Staff Emeritus
Your expression for $v(\theta)$ has units of length. The speed of the particle is actually given by
$$v_P(\theta) = \left| \frac{d\vec{P}}{dt} \right| = \left| \frac{d\vec{P}}{d\theta}\frac{d\theta}{dt} \right|.$$ The factor of $d\theta/dt$ is where the information about the speed $v$ of the bicycle enters.

11. Oct 30, 2014

### bobred

Yes I see thanks, but how do I get $$d\theta/dt$$, or can't I with the given information?

12. Oct 30, 2014

### bobred

I think I have it
$$d\theta/dt = \omega$$

Which has units s-1.

13. Oct 30, 2014

### vela

Staff Emeritus
Right. Now you just need to relate $\omega$ to $v$.

14. Oct 30, 2014

### bobred

Ok I think this is it

$$v(\theta) = |\vec P~'(\theta)|=\sqrt{\omega\left(\frac{dx}{d\theta}\right)^2+\omega\left(\frac{dy}{d\theta}\right)^2}=\sqrt{\omega (R-r\cos)^2+\omega r^2\sin^2\theta}$$

$$v(\theta) = \omega R\sqrt{(1+\rho^2)-2\rho\cos\theta}$$

which has the units of speed.

Last edited: Oct 30, 2014
15. Oct 31, 2014

### bobred

Cannot seem to be able to edit the above post

$$v(\theta) = |\vec P~'(\theta)|=\sqrt{\left(\frac{dx}{d\theta}\omega\right)^2+\left(\frac{dy}{d\theta}\omega\right)^2}=\sqrt{\omega^2 (R-r\cos)^2+\omega^2 r^2\sin^2\theta}$$

James

16. Oct 31, 2014

### vela

Staff Emeritus
So if the speed of the bike is $v$…

17. Oct 31, 2014

### bobred

Then the speed of the point would be

$$v+v(\theta)$$

18. Oct 31, 2014

### vela

Staff Emeritus
That's not right. In any case, you seemed to have missed my point. If I gave you numbers for $R$, $r$, and $v$, and $\theta$, could you give me a numerical answer for $v(\theta)$? Until you get to that point, you're not done. In other words, your answer should only depend on those variables.

19. Oct 31, 2014

### bobred

Yes, for $r=0$ the speed is $\omega R$ and independant of $\theta$ and for $r=R$, ($\theta=0$ is the point in contact with the ground) the speed is 0 for $\theta=0$ and $v=2\omega R$ for $\theta=\pi$.

20. Oct 31, 2014

### vela

Staff Emeritus
Ok, so what's the speed of P if θ=π/4, R=0.5 m, r=0.2 m, and v=3 m/s?

21. Oct 31, 2014

### bobred

Ok,so I need to consider the speed of the bike v aswell.

22. Nov 3, 2014

### bobred

Hi
$$\omega=\frac{v}{r}$$

With the values you supplied I get $$5.8 ms^{-1}$$
James

23. Nov 3, 2014

### vela

Staff Emeritus
Close. The angular speed of the wheel $\omega$ is independent of $r$: every point goes around the axis at the same angular rate. A point 10 cm from the axis, for example, goes around once in the same time a point 20 cm from the axis goes around once.

24. Nov 3, 2014

### bobred

Yes, so it should be
$$\omega=\frac{v}{R}$$

25. Nov 3, 2014

### vela

Staff Emeritus
Right.