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Old Jan 16th 2011, 04:21 PM   #1
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Issue with design speed

For circular motion to exist there must be an acceleration vector with direction towards the centre of circular path. On a banked track, the normal reaction force and weight force must add to give the net force directed to the centre of the path in order for the acceleration vector to be correct. We've simplified the situation so that it's similar to circular motion on a flat surface. Why is there a tendency to drift higher or lower if motion is not occuring at design speed?
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Old Jan 17th 2011, 01:41 AM   #2
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The force F_resultant that you obtain from the normal force and the weight is a constant on the bank. So, only this force can provide the centripetal acceleration.

If a larger force than F_resultant is required for circular motion, the F_resultant will not be able to 'match' this required force and the car will go higher up the bank.

The opposite is also true.

What affects the required force now? It's speed!

Recall the formula:

F = (mv^2) / r

r is constant, m is constant.

If the car goes faster that the designed speed, the car will require a larger force to keep it in circular motion, but since it cannot provide that force, it will go higher up.

It's like something is pulling you away from a wall and you don't have enough force to keep with the wall, the larger force will pull you away.

The same thing happens with a too low speed.
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Old Jan 18th 2011, 03:48 AM   #3
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Where does the acceleration lost go?
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Old Jan 18th 2011, 03:53 AM   #4
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The acceleration which doesn't contribute to circular motion, goes into the motion of the car in the left/right direction (ie, up or down the bank).
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Old Jan 19th 2011, 11:32 PM   #5
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Originally Posted by Unknown008 View Post
<snip>
If the car goes faster that the designed speed, the car will require a larger force to keep it in circular motion, but since it cannot provide that force, it will go higher up.

<snip>
The same thing happens with a too low speed.
So what you've trying to say is that r has to be increased/decreased to allow for circular motion.

The net force is provided by the gravitational and normal forces; what is the force that makes the object go up and down the bank composed of?

TIA,
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Old Jan 19th 2011, 11:44 PM   #6
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Yes, we can say it like that too.

The force that pulls the car down is a part of gravity, while the upward force is due to the reaction force of the centripetal force, that is the centrifugal force (this is mainly used in engineering so I was told). I don't think that you need to study that in your circular motion. (I didn't)
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Old Jan 21st 2011, 04:10 AM   #7
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In Uniform circular motion in the section beginning "Consider now the role that friction plays," why does f_s point down the slope and not up the slope?
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Old Jan 21st 2011, 05:35 AM   #8
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In this situation, we are considering friction and the maximum speed with which the car can go round with circular motion.

Also, if we're worried about the maximum speed at which we can go around the banked turn, if there was no friction the car would tend to slide towards the outside of the curve, so the friction opposes this tendency and points down the slope.
You remember earlier that I told you about centrifugal force? It becomes larger with increase in velocity (because centripetal force increases). So, at maximum speed round the back, the centripetal motion is very large and the car has a greater tendency to go up than to go down.
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