Automobile speedometer physics test problem

1. Dec 2, 2003

acoult

Hello anyone that can help. I am trying to gain an understanding of rotational motion, including (tangential speed, angular momentum, centripetal force and so on). The first question I was hoping that someone could help me is:

An automobile speedometer is configured yo read speed proportional to the rotational speed of its wheels. If larger wheels are used, will the speedometer reading be higher, lower or the same.

I understand the question but I was hoping someone could help explain this to me in a mathamatical equation.

Thank you for your help. I promise not to bug you guys to much!

2. Dec 2, 2003

acoult

more explination

ok so in regards to my last post I am going to say what I know and what I am unclear on.

I know that angular velocity (speed) is W=change in theta over the change in time. So with this I can conclude that if one wheel has a theta of 1 rad over 1 min then it I out a larger wheel on it it twice the size (2/1) the W would = 2 vs the 1 for the 1st answer. So does this mean that the speedometer would read lower??? I am just not getting it.

3. Dec 3, 2003

gnome

If the speedometer is reading angular speed, then for a given ANGULAR speed the speedometer reading will be the same no matter how big the wheels are.

BUT a bigger wheel has a bigger circumference, right? Therefore, it covers more ground with a single revolution of the wheel. A wheel of radius 2x covers TWICE as much distance in 1 revolution compared to a wheel of radius x.

So, if the speedometer is calibrated to work correctly with wheels of radius x, when you put on wheels of radius 2x, you will get a speeding ticket because the speedometer will read 1/2 of your actual speed.

4. Dec 3, 2003

acoult

I see so instead of doubling the theta I should of taken half of it, right? That makes sense. thank you.

ok next question, this one is a question I have been struggling with for a while now.

I am not sure if you have seen the experiment where a person holding on to a bicycle wheel from poles on it's axis has an external force spin the wheel while the person is stitting on a chair that can spin. If the person flips the wheel over the person starts to spin in the chair. my question is why is there no change in angular velocity when you flip the bicycle wheel.

My guess: w=change in theta over change in time. the theta does not change when it is flips but is fliped in the oppsite direction which makes the person on the chair spin which means the angular momentum is conserved. I am just not sure I understand why that is... Am I way off here?

5. Dec 3, 2003

gnome

Excellent question. I hope Doc Al has the answer -- I don't.

I would put the question this way:
Treating the system as the student + the wheel, suppose the initial angular momentum is
Lsys = Lwheel(initial) = some vector quantity K

We know that after the wheel is inverted, conservation of momentum requires that
Lsys = K

and therefore

Lsys = Lstudent(final) + Lwheel(final) = K

But why is it necessary that
Lwheel(final) = -K
and Lstudent(final) = 2K ?

Why can't we end up with
Lwheel(final) = -.5K
and Lstudent(final) = 1.5K , or any other combination of values whose vector sum equals K?

Or to put it another way, is |&omega;wheel| necessarily constant? If so, why?

6. Dec 3, 2003

gnome

After thinking about it some more, I guess that in addition to conservation of momentum, we are also dealing with conservation of energy. Nothing has been done to alter the kinetic energy (a scalar quantity) of the wheel, so it must continue to spin at the same speed even as its direction of spin changes.

The student, on the other hand, using muscles to flip the axis of the wheel, converts some chemical potential energy to kinetic energy, in an amount sufficient to cause the student to spin fast enough to obey conservation of momentum. This way not only does the total energy of the system remain constant, but the total energy of each of the components remains constant as well, except for whatever amount of energy that radiates off as heat. That would be a very small amount from the muscular exertion at the moment that the wheel is flipped, and then continuing as both the wheel and the student gradually slow down due to friction.

How does that sound?

7. Dec 3, 2003

Staff: Mentor

I'll have to ponder this one. But my immediate comments are this: Rest assured that the total angular momentum is unchanged. To flip the wheel, the person must exert a torque on it; likewise the wheel exerts a torque on the person. If the torque (vector) is kept perpendicular to the angular momentum (vector) of the bike, the bike's angular momentum will change direction but not magnitude.

8. Dec 3, 2003

Chrishaum

The following is quoted from page 63 ofFundamentals of Physics, 6th ed.[\I}, written by David Halliday, Robert Resnick, and Jearl Walker:

The Key Ideas here are these:

1. The angular speed is related to the final angular momentum $L_b$ of the composite body about the stool's rotation axis by $L=I \omega$ .

2. The initial angular speed $\omega_b$ of the wheel is related to the an_gular momentum $L\itextrm{wheel}$of the wheel's rotation about its center by the same equation.

3. The vector addition of $L_b$ and $L\itextrm{wheel}$ gives the total angular momentum $L\itextrm{total}$of the system of student, stool, and wheel.

4. As the wheel is inverted, no net external torque acts on that system to change $L\itextrm{total}$about any vertical axis. (Torques due to forces between the student and the wheel as the student inverts the wheel are internal to the system.) So, the system's total angular momentum is conserved about any vertical axis.

$L_b,f$ + $L\itextrm{wheel},f$ = $L_b,i$ + $L\itextrm{wheel},i$ (Equation F)

where$i$ and $f$ refer to the initial state (before inversion ofthe wheel) and the final state (after inversion). Because inversion of the wheel inverted the angular momentum vector of the wheel's rotation, we substitute $-L\itextrm{wheel},i$ for $L\itextrm{wheel},i$. Then, if we set $L_b,i$ = 0 (because the sludent, the stool, and the wheel's center were initially at rest),

Equation F yields

$L__b,f = 2L\itextrm{wheel},f$

Using $L=I \omega$, $\omega_b = \frac{2I\itextrm{wheel}}{I_b}\omega\itextrm{wheel}$.

9. Dec 4, 2003

Staff: Mentor

I should have said that the angular momentum in the vertical direction is conserved. The chair is free to rotate about a vertical axis, but is constrained otherwise.