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 Aug 9th 2009, 08:57 PM #1 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 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. Attached Thumbnails
 Aug 10th 2009, 01:30 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Aug 10th 2009, 12:42 PM #3 Physics Team     Join Date: Jul 2009 Posts: 310 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? __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC Last edited by Deco; Aug 10th 2009 at 12:45 PM.
 Aug 10th 2009, 09:51 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Aug 10th 2009, 09:55 PM #5 Physics Team     Join Date: Jul 2009 Posts: 310 $\displaystyle F=m \omega^2 r = m (\frac {v}{r})^2 r = \frac {mv^2}{r}$ Which is what I used. Again, I see ambiguity. Because as shown; $\displaystyle F \propto r$ or $\displaystyle F \propto r^{-1}$ __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
 Aug 10th 2009, 09:59 PM #6 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Aug 10th 2009, 10:03 PM #7 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Aug 10th 2009, 10:03 PM #8 Physics Team     Join Date: Jul 2009 Posts: 310 How does that prove that I'm wrong? They are equivalent. $\displaystyle \omega$ cannot remain constant according to your definition of it. __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
 Aug 10th 2009, 10:31 PM #9 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Aug 11th 2009, 12:26 PM #10 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 131 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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