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Old Aug 9th 2009, 08:57 PM   #1
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Banked track

From the phrase the banked angle is designed for vehciles moving with speed v should indicate that there should be no friction to provide the centripetal acceleration.

Now the speed is increased to 2v, with the same angle, the component of normal reaction should be insufficient to provide required centripetal acceleration. So friction should be present so A and B can be eliminated.

Remaining C and D, I can't figure out how and why the normal reaction on the right wheel would increase? Is there any equation that helps me to prove this? Thanks for any help.
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Old Aug 10th 2009, 01:30 AM   #2
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The outer wheel has to move in a circle of larger radius for the car to move in a circle, and consequently has to move faster. This would require a larger centripetal force which can occur only if the normal reaction increases. So the car would slightly tilt outwards i guess.
I know very little about automobiles but this is what they refer to i guess when they talk about "differential" etc.
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Old Aug 10th 2009, 12:42 PM   #3
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The outer wheel goes under a lesser centripetal force because
of the equation:

$\displaystyle F_{ac} = \frac{mv^2}{r}$

Letting $\displaystyle r_1$ be for the inner wheel and $\displaystyle r_2$ be the outter wheel. Since the distance between the two wheels is not negligible.

So,

$\displaystyle r_2 > r_1$

Making,

$\displaystyle F_{ac2} < F_{ac1}$

Since,

$\displaystyle F_{ac} \propto {\frac {1}{r}}$

Since the road is at a angle $\displaystyle \theta$, we can say:

$\displaystyle F_N = F_g + F_{ac1x} + F_{ac2x}$

$\displaystyle F_N = m (g+ \frac {v^2}{r_1} \cos \theta + \frac {v^2}{r_2} \cos \theta)$

So if anything, the inner wheel should experience more force. Thoughts anyone?
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Last edited by Deco; Aug 10th 2009 at 12:45 PM.
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Old Aug 10th 2009, 09:51 PM   #4
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Let us consider an analogy.
Let us think of a long axle, one end of which is the centre around which the car moves in a circle. Let the two wheels be at the other end corresponding to the (front) inner and outer wheels respectively, and let them be free to move. When this axle moves in uniform circular motion, what is constant is the angular velocity $\displaystyle \omega$. Since the radii for the inner and outer wheels for this circular motion with respect to the centre are different, their velocities are different. (i dont mean the radius of each wheel which is the same). Since
$\displaystyle \omega = \frac{v} {r}$ , the outer wheel which is further away has to move faster to maintain $\displaystyle \omega$ constant.

The outer wheel should spin faster since it has to cover the circumference of a larger circle in the same time (else) there would be skidding (no kidding).
Since $\displaystyle F = m\omega^2 r$, m is the same for each wheel, and $\displaystyle \omega$ = constant, the wheel farther away experiences greater centripetal force not lesser i think.
The rest as in my earlier post.

Last edited by physicsquest; Aug 10th 2009 at 09:54 PM.
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Old Aug 10th 2009, 09:55 PM   #5
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$\displaystyle F=m \omega^2 r = m (\frac {v}{r})^2 r = \frac {mv^2}{r}$

Which is what I used.

Again, I see ambiguity. Because as shown;

$\displaystyle F \propto r$

or

$\displaystyle F \propto r^{-1}$
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Old Aug 10th 2009, 09:59 PM   #6
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But you arrive at a lower centripetal force for the outer wheel because from the equation $\displaystyle F = m \frac{v^2}{r}$ the effect of having $\displaystyle \omega$ constant is not evident.
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Old Aug 10th 2009, 10:03 PM   #7
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There is no ambiguity once it becomes clear that $\displaystyle \omega$ is constant and v is not as one moves away from the centre of the circular motion.
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Old Aug 10th 2009, 10:03 PM   #8
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How does that prove that I'm wrong? They are equivalent.

$\displaystyle \omega$ cannot remain constant according to your definition of it.
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Old Aug 10th 2009, 10:31 PM   #9
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In uniform circular motion, $\displaystyle \omega$ , the angular velocity by definition is constant irrespective of the radius.
Think of a spinning disc. The number of revolutions/sec made by each point on the disc is the same. Thus if the disc makes 10 revolutions per sec, so does each point. Thus each point makes an angular displacement of $\displaystyle 2\pi . 10$ or $\displaystyle 20\pi$ per sec and thus has the same angular velocity.
However, the linear velocity of each point on the disc is different. As r increases away from the centre, v increases accordingly to maintain $\displaystyle \omega$ constant.
When you use


and say there is the implicit assumption that v is constant which in fact it is not as it has r dependence.

However when you use $\displaystyle F = m\omega^2 r$ , both m and $\displaystyle \omega$ are constant and hence which is perfectly OK

Last edited by physicsquest; Aug 10th 2009 at 10:39 PM.
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Old Aug 11th 2009, 12:26 PM   #10
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Banked track

With no other choice available, D is the best option.

Physicsquest, your logic looks valid. Your first explanation suffices, however, the tilt you refer to must be caused by some sort of body force caused by the centripetal accelaration of the vehicle. It is therefor my opinion that this question (while thought provoking) has not been supplied with sufficient detail. However following the line of reasoning you have, you would predict the need for the increased force even though it is beyond the details of the current analysis.
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