Centripetal Force Question

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What is the minimum radius that a cyclist can ride around without slipping at 10 km per hours if the coefficient of static friction between his tires and the road is .5?

I got 1.6 meters. Kaplan says it is 4.1 meters.

Any thoughts?

 

[Snowy]

I also got 1.6 m when I tried it right now, so I don't know either... must be forgetting something.

Also, check out the substudy forum, which is reserved for questions just like this.

 

[auroraboy]

I also got 1.6m,

I used the following formula:

u*N= m*(v^2/r)


velocity would be (10000/3600)m/s



am I missing something?

 

[nikeshp]

i got 1.6 as well, using the same formula as the above poster, but i did not include the 1/2, im not sure where he got that from

 

[nikeshp]

wait a second, the problem gives you coefficient of static friction. you are moving, so that would be the wrong one to use. it doesnt give you the coefficient of rolling friction or the coefficient of kinetic friction?

 

[bluemonkey]

This problem came up in an earlier thread.

Apparently you need to use 10 miles/hour instead of km/hr.

 

[Kaustikos]

wait a second, the problem gives you coefficient of static friction. you are moving, so that would be the wrong one to use. it doesnt give you the coefficient of rolling friction or the coefficient of kinetic friction?

No, in centripetal force it is the static coefficient of friction that is used. Kinetic isn't used in centripetal force to hold the "body" around the circle.

And I honestly do not know how Kaplan did this or why, but this question is my high yield problem solving guide and it has the correct velocity (10 mph) and not that. But that's how you get your answer, as bluemonkey stated. I would honestly stray away from that book at the moment. 👎

 

[auroraboy]

i got 1.6 as well, using the same formula as the above poster, but i did not include the 1/2, im not sure where he got that from

haha, my bad, no 1/2, not sure why I put that in

 

[physics junkie]

wait a second, the problem gives you coefficient of static friction. you are moving, so that would be the wrong one to use. it doesnt give you the coefficient of rolling friction or the coefficient of kinetic friction?

The contact patch of the tire is not slipping on the surface of the road. That is why you use the coeff. of static friction. This is different from a box that has to slip to move. You would use coeff. of kinetic friction if you were doing a burn out. If your tires were slipping against the surface of the road when you drive you would wear out a set of tires everyday.

 

[nikeshp]

The contact patch of the tire is not slipping on the surface of the road. That is why you use the coeff. of static friction. This is different from a box that has to slip to move. You would use coeff. of kinetic friction if you were doing a burn out. If your tires were slipping against the surface of the road when you drive you would wear out a set of tires everyday.

thanks physics junkie, topic creator, i am sorry if i might have misled you.

however, another question for myself, why would you not use coeff of rolling friction. does this exist?

 

[auroraboy]

thanks physics junkie, topic creator, i am sorry if i might have misled you.

however, another question for myself, why would you not use coeff of rolling friction. does this exist?

In the question we are dealing with the boundary condition of rolling/slipping.

So I guess you could have used coefficient of rolling friction if it was mentioned.

 

[physics junkie]

The coefficient of rolling friction is an engineering topic. You would never see it on the MCAT unless it was explained in a passage. Silly engineers, always taking our purely theoretical discoveries and applying them to the real world.

 

[auroraboy]

The coefficient of rolling friction is an engineering topic. You would never see it on the MCAT unless it was explained in a passage. Silly engineers, always taking our purely theoretical discoveries and applying them to the real world.

lol, thanks!

I am an engineer 🙂