Speed of a point on a wheel as a function of θ

[bobred]

Homework Statement

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?

Homework Equations

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

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

 

[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?

 

[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

 

[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?

 

[vela]

Homework Statement

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?

Homework Equations

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

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

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

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

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

 

[LCKurtz]

Homework Statement

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?

Homework Equations

$$x=R\theta - r \sin\theta$$
$$y=R - \cos\theta$$
$$\rho=\frac{r}{R}$$

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.

Distance traveled by the point in one revolution
$$D =R{\displaystyle \int_{0}}^{2\pi}\sqrt{\left(1+\rho^{2}\right)-2\rho\cos\theta}\,\textrm{d}\theta$$

What does that equation have to do with it?

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

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

 

[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.

 

[LCKurtz]

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.

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.

 

[bobred]

What I get is

$$v(\theta) = |\vec P~'(\theta)|=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}=\sqrt{(R-r\cos)^2+r^2\sin^2\theta}$$$$v(\theta) = R\sqrt{(1+\rho^2)-2\rho\cos\theta}$$

 

[vela]

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.

 

[bobred]

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

 

[bobred]

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

Which has units s^-1.

 

[vela]

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

 

[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.

 

[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

 

[vela]

So if the speed of the bike is $v$…

 

[bobred]

Then the speed of the point would be

$$v+v(\theta)$$

 

[vela]

Then the speed of the point would be

$$v+v(\theta)$$

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.

 

[bobred]

Yes, for $r=0$ the speed is $\omega R$ and independent 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 $\theta=\pi$.

 

[vela]

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?

 

[bobred]

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

 

[bobred]

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

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

 

[vela]

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.

 

[bobred]

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

 

[vela]

Right.

 

[bobred]

Big sigh, thank you for your help and patience.
James