Easy Physics Question?? - BR

[ilovemedi]

Why does it become unsafe to take a turn in your car past a threshold speed?
Answer: Because the centripetal force can exceed the force holding the car to the road

Ok, so I understand that the centripetal force increases since the velocity increases, because Fc= v^2/r.. What i dont understand is why increasing the centripetal force would cause a car to slide. What's bothering me about the answer is that I thought centripetal force is the inward force, so increasing it would actually stabilize the car, not cause it to slide out. Also, in this case, the centripetal force is friction. So isnt the centripetal force the same as the the force 'holding the car to the road'?

---

Members don't see this ad.

 

[Patuxent78]

This may help you, I can't imagine that it wouldn't. It talks about stock cars and the physics behind their turns on the speedway.


http://www.stockcarscience.com/scienceTopics/scsRacing_CentrifugalForce.php

 

[V5RED]

Why does it become unsafe to take a turn in your car past a threshold speed?
Answer: Because the centripetal force can exceed the force holding the car to the road

Ok, so I understand that the centripetal force increases since the velocity increases, because Fc= v^2/r.. What i dont understand is why increasing the centripetal force would cause a car to slide. What's bothering me about the answer is that I thought centripetal force is the inward force, so increasing it would actually stabilize the car, not cause it to slide out. Also, in this case, the centripetal force is friction. So isnt the centripetal force the same as the the force 'holding the car to the road'?

Stuff relevant to your question:

Centripetal force is a metric of the force needed to cause circular motion. Centripetal force is not actually a force.

The equation for centripetal force gives you an amount of force needed to cause circular motion. Something has to provide this force (friction, tension, normal force, etc.).

If the centripetal force(ie the metric of how much force is needed to cause circular motion) is larger than the force available to provide circular motion(ie the friction in this case), then there will not be circular motion. In the case of the example, the car skids away from the track's center.

Extra stuff that is also important, but not relevant to your question:

Conversely, if the centripetal force is less than the provided force, the object would move toward the center(think about if you try to do a loop-de-loop on a bike and move too slowly, you fall because gravity provides too much force relative to what you needed for the circular motion)

This is why sources of force that can vary to meet the required force exactly(tension and static friction) are ideal for causing circular motion.

 

[johnnybravo19]

Simple approach

Imagine the car drifting. Centripital Acceleration is perpendicular to Frictional Force, that's why you have to turn in the opposite direction and rip the e-brake...should make it a little more clearer

 

[V5RED]

Simple approach

Imagine the car drifting. Centripital Acceleration is perpendicular to Frictional Force, that's why you have to turn in the opposite direction and rip the e-brake...should make it a little more clearer

This is neither simple nor correct.

The frictional force on a car that is undergoing centripetal acceleration points in the same direction as the acceleration, not perpendicular. That is why the centripetal acceleration happens. The friction provides the force for the acceleration. If the friction were perpendicular it would make the car fly into the air.

 

[Sports Junky]

Stuff relevant to your question:

Centripetal force is a metric of the force needed to cause circular motion. Centripetal force is not actually a force.

The equation for centripetal force gives you an amount of force needed to cause circular motion. Something has to provide this force (friction, tension, normal force, etc.).

If the centripetal force(ie the metric of how much force is needed to cause circular motion) is larger than the force available to provide circular motion(ie the friction in this case), then there will not be circular motion. In the case of the example, the car skids away from the track's center.

Extra stuff that is also important, but not relevant to your question:

Conversely, if the centripetal force is less than the provided force, the object would move toward the center(think about if you try to do a loop-de-loop on a bike and move too slowly, you fall because gravity provides too much force relative to what you needed for the circular motion)

This is why sources of force that can vary to meet the required force exactly(tension and static friction) are ideal for causing circular motion.

I would just like to say, this is the most concise explanation of centripetal force I have ever read. I think you might have cleared about every question I had about it! Thank you

For the longest time I was simply using the equation, but now I actually get it. This is the reason why you can use mg = mv^2/r to calculate the minimum speed at the top of the roller coaster. If the force of gravity exceeds the centripetal force required, it falls inward, of course!!

 

[member232]

Just to clarify-- if the frictional force in this instance requires, say 5 N, and the actual force (thus the centripetal force) is 6 N, then the car will skid away from the track? If centripetal force has an acceleration pointed towards the center of the circle, wouldn't you expect a greater force to cause the car to go inward. And vice versa, a centripetal force of, say 4 N to be too little to keep the car going in a circle, and thus the car skids off?

 

[V5RED]

Just to clarify-- if the frictional force in this instance requires, say 5 N, and the actual force (thus the centripetal force) is 6 N, then the car will skid away from the track? If centripetal force has an acceleration pointed towards the center of the circle, wouldn't you expect a greater force to cause the car to go inward. And vice versa, a centripetal force of, say 4 N to be too little to keep the car going in a circle, and thus the car skids off?

No.

Read my post.

 

[Lunasly]

Stuff relevant to your question:

Centripetal force is a metric of the force needed to cause circular motion. Centripetal force is not actually a force.

The equation for centripetal force gives you an amount of force needed to cause circular motion. Something has to provide this force (friction, tension, normal force, etc.).

If the centripetal force(ie the metric of how much force is needed to cause circular motion) is larger than the force available to provide circular motion(ie the friction in this case), then there will not be circular motion. In the case of the example, the car skids away from the track's center.

Extra stuff that is also important, but not relevant to your question:

Conversely, if the centripetal force is less than the provided force, the object would move toward the center(think about if you try to do a loop-de-loop on a bike and move too slowly, you fall because gravity provides too much force relative to what you needed for the circular motion)

This is why sources of force that can vary to meet the required force exactly(tension and static friction) are ideal for causing circular motion.

This is the most beautiful thing I have ever read I my entire life. Damn my professors... damn them.

 

[mehc012]

Just to clarify-- if the frictional force in this instance requires, say 5 N, and the actual force (thus the centripetal force) is 6 N, then the car will skid away from the track? If centripetal force has an acceleration pointed towards the center of the circle, wouldn't you expect a greater force to cause the car to go inward. And vice versa, a centripetal force of, say 4 N to be too little to keep the car going in a circle, and thus the car skids off?

You have it backwards.
The frictional force is the 'actual force'
The centripetal force is the 'required force'.

If the actual force is less than the force required to sustain circular motion, you...won't sustain circular motion.