Question about centripital force/friction/etc.

[tectactoe]

"What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 24 km/h and the coefficient of static friction between tires and track is 0.30?"


v = 24 km/h
us = 0.30


I know

(us)(Fn) = m(v^2/r)


but I am stumped as to how to get r from this, so I am thinking maybe I have the wrong equation? I just am so unsure of what to do!

x_x

 

[BishopUser]

in this case what is the normal force equal to?

 

[tectactoe]

I think that's what I'm not sure about...

Wouldn't the normal force be equal to the downward pull of gravity on the bike/person in this case (mg)? In which case, mass is not known. =[

 

[BishopUser]

well in that case you got mass on both sides of the equation, and they are both the same mass right?

 

[tectactoe]

>_<

Wow. I didn't even realize that. I'm an idiot. Thank you kindly, sir.

 

[tectactoe]

Wait, somethings wrong... my answers aren't correct.


having

(us)(Fn) = m(v^2/r)

(us)(mg) = m(v^2/r)

canceling masses

(us)g = (v^2/r)

solving for r

r = (v^2) / (us)g

r = (24^2) / (.30)(9.8)

r = 576 / 2.94

r = 195

and that's not the right answer... =[

Does Fn = something other than just mg? I'm not sure, really!

 

[Mthees08]

Equilibrium of forces, F(net)=0 so F(friction)+F(centripetal)=0

 

[tectactoe]

I guess I'm just confused as to how that will help me find the radius of the track... I mean, the Forces in the y direction cancel, but how does that help with r? Yikes this problem seems so simple but it's been stumping me for hours.

 

[Tedjn]

The forces in the y (vertical) direction cancel because the ground provides the normal force that counteracts the gravitational force. The forces in the horizontal direction, on the line through the center of the track, cancel because friction from the ground balances the centripetal force. Friction is dependent on the normal force, but it acts horizontally.

As Mthees08 said, F_{\text{friction}} = F_{\text{centripetal}} = \text{?}. From the equation for centripetal force, think about what happens when radius decreases. What must also happen to the friction? How long can friction keep balancing centripetal force?

 

[ZPower]

Wait, somethings wrong... my answers aren't correct.


having

(us)(Fn) = m(v^2/r)

(us)(mg) = m(v^2/r)

canceling masses

(us)g = (v^2/r)

solving for r

r = (v^2) / (us)g

r = (24^2) / (.30)(9.8)

r = 576 / 2.94

r = 195

and that's not the right answer... =[

Does Fn = something other than just mg? I'm not sure, really!

yep that's correct, but dotn forget that its: negative m*v^2/r, and negative Ff sice they both point inward toward the center