Physics question" circular motion"

[dorjiako]

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Physics question "uniform circular motion"
I need help solving this problem:1, What is the minimum radius that a cyclist can ride around without slipping at 10 kilometers per hour if the coefficient of friction between his tires and the road is 0.5?

 

[MD Odyssey]

Physics question "uniform circular motion"
I need help solving this problem:1, What is the minimum radius that a cyclist can ride around without slipping at 10 kilometers per second if the coefficient of friction between his tires and the road is 0.5?

What have you tried so far? And, as a serious cyclist, I'm a little offended that your hypothetical cyclist is cornering at 10,000 m/s.

 

[indianjatt]

I think that the static friction is what is causing the centripetal force which allows the cyclist to go in a circle. So, F friction = Nmeu = mg(0.5)

mg(.5)=mV^2/R
5 = V^2/R
5 m/s^2 = (10^4m/s)^2/R
R = 10^8/5 m

I hope I did it right.

 

[dorjiako]

The cyclist is going at 10000km per hour not second please. I tried solving it with mv^2/r =0.5mg but is not getting the answer. I am wondering if their answer is wrong. Well, their answer is 4.1m. I am getting r= v^2/.5g. V^2=10000m/3600s=2.77m/s^2=7.67/.5g =1.5m. I don't know how else to solve this question to get their 4.1m answer.

 

[Majik]

Physics question "uniform circular motion"
I need help solving this problem:1, What is the minimum radius that a cyclist can ride around without slipping at 10 kilometers per hour if the coefficient of friction between his tires and the road is 0.5?

Here's a hint:

Consider the forces driving the bike in circular motion:
- Static Friction (Rolling Friction)

The centripetal force (net force), in this case static friction, is propelling the tire into circular motion causing it to continually rotate.

Equating centripetal force to static friction, you could solve for R. Keep in mind that units must be in scientific notation.

 

[Majik]

The cyclist is going at 10000km per hour not second please. I tried solving it with mv^2/r =0.5mg but is not getting the answer. I am wondering if their answer is wrong. Well, their answer is 4.1m. I am getting r= v^2/.5g. V^2=10000m/3600s=2.77m/s^2=7.67/.5g =1.5m. I don't know how else to solve this question to get their 4.1m answer.

Step 1: Convert 10,000km/hour into m/s.

10,000km/hour x (1000 m/km) x (1 hour / 3600s)
10,000 x 1,000m / 3600s = 2,777m/s

Step 2: Write the relevant equation and solve:

Net Force = Static Friction
Net Force = msN
Net Force = ms x mg
Net Force = (0.5)(10 x mass); (mass is unknown)

Since Net Force = Centripetal Force which = mv^2/R then:
mv^2/R = msN, therefore rearranging for R we get:
v^2 / msg = R
(2,777)^2 / 5 = R
~1.5x10^6 m or 1,500 km

EDIT: Yup, looks like they made a calculation error.

 

[indianjatt]

The cyclist is going at 10000m per hour not second please. I tried solving it with mv^2/r =0.5mg but is not getting the answer. I am wondering if their answer is wrong. Well, their answer is 4.1m. I am getting r= v^2/.5g. V^2=10000m/3600s=2.77m/s^2=7.67/.5g =1.5m. I don't know how else to solve this question to get their 4.1m answer.

Haha my bad. I just copied what MDOdessey quoted.