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physics circular motion question..

I'll post the attachment below..
so the answer is B.. somehow, though the answer really seems like it could be any of the options.. Because the centripetal force =(mv^2)/r.. and force keeps the car moving around the bend.. which means if fewer passengers mass lower mass lower.. or if travel more slowly velocity is lower.. if drive further from centre, then radius is greater.. all of these factors decrease the centripetal force which would cause the car to skid outwards.. so why is the answer B??
Reply 2
Original post by Abraham_Otaku
so the answer is B.. somehow, though the answer really seems like it could be any of the options.. Because the centripetal force =(mv^2)/r.. and force keeps the car moving around the bend.. which means if fewer passengers mass lower mass lower.. or if travel more slowly velocity is lower.. if drive further from centre, then radius is greater.. all of these factors decrease the centripetal force which would cause the car to skid outwards.. so why is the answer B??


Because the centripetal force has to be supplied by the friction between tyres and road which is some coefficient of friction times the weight of the car
Original post by AlanT12
Because the centripetal force has to be supplied by the friction between tyres and road which is some coefficient of friction times the weight of the car


I guess i understand that, but then, why are the other options wrong?? They didn't say "most likely" but just 'more likely'... or am i interpreting this incorrectly.
Reply 4
Original post by Abraham_Otaku
I guess i understand that, but then, why are the other options wrong?? They didn't say "most likely" but just 'more likely'... or am i interpreting this incorrectly.

You have a limit to how much friction force can be supplied - the max value is the coefficient of friction times weight (normal reaction) and this has to supply the required centripetal accelaration. So e.g. if you go faster you would need more centripetal acceleration to go around bend and eventually you won't have enough friction to supply it. Hence you would skid off. If you use new tyres you increase coefficient of friction so it makes it less likely you would skid off. The mass cancels, so the number of passengers doesn't matter assuming the coefficient of friction is really a constant versus tyre loading.
(edited 1 year ago)
Original post by AlanT12
You have a limit to how much friction force can be supplied - the max value is the coefficient of friction times weight (normal reaction) and this has to supply the required centripetal accelaration. So e.g. if you go faster you would need more centripetal acceleration to go around bend and eventually you won't have enough friction to supply it. Hence you would skid off. If you use new tyres you increase coefficient of friction so it makes it less likely you would skid off. The mass cancels, so the number of passengers doesn't matter assuming the coefficient of friction is really a constant versus tyre loading.

If I'm not mistaken, then that means the centripetal force would be less than the maximum frictional force between the ground and tyres for all options a c d.. but for option b the worn tyres decreases the maximum frictional force which the ground can provide.. and the centripetal force would exceed that maximum frictional force, and hence car would not be able to remain on track- ie skids outwards?.. is that right?
Reply 6
Original post by Abraham_Otaku
If I'm not mistaken, then that means the centripetal force would be less than the maximum frictional force between the ground and tyres for all options a c d.. but for option b the worn tyres decreases the maximum frictional force which the ground can provide.. and the centripetal force would exceed that maximum frictional force, and hence car would not be able to remain on track- ie skids outwards?.. is that right?

Yes. Except it is important to understand that centripetal force is not a separate force of nature. It's just a force that's needed for something to go around in a circle, and has to be supplied by something. Hence if the friction cannot supply it then the car cannot go in a circle. Same as if you had a ball going around on a string and cut the string the ball would fly off because nothing is providing the centripetal force needed any more.
(edited 1 year ago)
Original post by AlanT12
Yes. Except it is important to understand that centripetal force is not a separate force of nature. It's just a force that's needed for something to go around in a circle, and has to be supplied by something. Hence if the friction cannot supply it then the car cannot go in a circle. Same as if you had a ball going around on a string and cut the string the ball would fly off because nothing is providing the centripetal force needed any more.


thank you for the help :biggrin:
@Abraham_Otaku
@AlanT12
You two might be interested to think about something here, and my opinion that this is yet another poor question.
Is it an exam question or a text book question?

Worn tyres give less grip in wet conditions.

In dry conditions, bald tyres give better grip. (See the tyres on Formula 1 cars in dry conditions.)
This is because, the amount of grip depends on the surface area in contact with the road.
A bald tyre has more rubber in contact with the road than an identical one with tread .
In wet conditions, a water film exists between the tyre and the road.
The tread on the tyre helps to disperse the water away from the contact area and improves grip.
its a textbook question..

Original post by Stonebridge
@Abraham_Otaku
@AlanT12
You two might be interested to think about something here, and my opinion that this is yet another poor question.
Is it an exam question or a text book question?
Original post by Abraham_Otaku
its a textbook question..


I can at least be grateful that this question didn't get into an examination. :smile:

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