Force

[hansen44]

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

The answer is 4.1 m

How do you do this problem I am pulling my hair out with these kind of physics problems. Any help would be appreciated. Thank you

 

[sicboy188]

dont you need the weight of the cyclist to answer this question? as well as the bank angle?

 

[Kaustikos]

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

The answer is 4.1 m

How do you do this problem I am pulling my hair out with these kind of physics problems. Any help would be appreciated. Thank you

This is in relation to centripetal force. They basically want you to relate how the static coefficient of friction and centripetal force work. It sounds kind of tricky, but if you just draw a diagram, the answer will come to you.

The coeffecient of friction is what is making the cyclist stay inside of the circle, and the equation is equal to uk * Fn. The only other force is the F = m x a and the acceleration is the centripetal force.
Thus
fn = F
ukFn = m x a

uk mg = m a (a = centripetal force = v^2/r)

The masses cross each other out and you have
ukg = a
now all you need to do is calculate for R. Remember, you also need to convert km/h into m/s.

0.5 x 10 = X/R

10 km/hr * 1000m/km * 1hr/60min * 1min/60 seconds = 2.8 m/s
- edit -

wait 4.1 m/s? Did you give us the correct values?

 

[hansen44]

Yes I gave you the right answer, thats why im so confused. I got the same answer you did. Anyone understand this?

 

[Kaustikos]

Yes I gave you the right answer, thats why im so confused. I got the same answer you did. Anyone understand this?

Is this from the kaplan book? Because it says 10 miles/ hr.

 

[hansen44]

Its from the Kaplan MCAT High Yield Problem Solving guide and it says kilometers/hour (10 km/h).

 

[tncekm]

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

The answer is 4.1 m

How do you do this problem I am pulling my hair out with these kind of physics problems. Any help would be appreciated. Thank you

10km/hour = 10⁴m/hr x hr/60min x min/60s = 2.78 m/s

Fc=mv²/r = μmg= Fs
v²/r = μg

r=v²/μg = (2.78m/s)²/(0.5 x 10m/s&#178😉
r = 7.72m²/s² / 5m/s² = 1.54m

Well, that answer doesn't match 4.1m. Don't know what to tell ya.😕

 

[iA-MD2013]

10km/hour = 10⁴m/hr x hr/60min x min/60s = 2.78 m/s

Fc=mv²/r = μmg= Fs
v²/r = μg

r=v²/μg = (2.78m/s)²/(0.5 x 10m/s²)
r = 7.72m²/s² / 5m/s² = 1.54m

Well, that answer doesn't match 4.1m. Don't know what to tell ya.😕

This seems exactly right to me...but, the only way to get the right answer is if it was 10 miles/hour. One mile=1.6 kilometers, and you end up with r=4.1. But, basically, it's a stupid question (because of the conversion). If you understood what tnckem did, you're good to go.

 

[Kaustikos]

10km/hour = 10⁴m/hr x hr/60min x min/60s = 2.78 m/s

Fc=mv²/r = μmg= Fs
v²/r = μg

r=v²/μg = (2.78m/s)²/(0.5 x 10m/s²)
r = 7.72m²/s² / 5m/s² = 1.54m

Well, that answer doesn't match 4.1m. Don't know what to tell ya.😕

Because it is 10 miles/hr. Either the op got a typod book or someone gave him the question incorrectly.

 

[hansen44]

Ok well thanks everybody for the help, I also think this Kaplan book is flawed to be honest, some of the answers just don't add up.

 

[Maxwell Edison]

I second 1.542 m.