Question

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 27 km/h and the coefficient of static friction between tires and track is 0.22?

Answer 1

(mu)m(9.8)=m(v squared)/r
m is on both sides of the equal sign so cross it out
mu=0.22 given
v is given convert it out of km/h into meters per second
solve for r

friction = centrifugal
friciton is mu(normal force)
normal force = mg you dont need to know m it will cancel out
centrifugal = m(v squared)/r

Answer 2

It helps if you draw a force diagram of the biker. Since the track is unbanked, we know force of gravity and force normal cancel each other out, and we can ignore those. The other two forces are centripetal force (Fc), which is directed to the outside of the circle, and force of friction (Ff), which is directed, in this case, inward. If she is staying on a circle, the Fc must equal Ff. So, using the equations for force friction (Ff=u(Fn)) and centripetal force (Fc=(mv^2)/r), we can derive an equation to find r, the radius of the circle. So, here's what our work would look like:

u = 0.22
v = 27km/h = 7.5 m/s
Fc = Ff
Fc = (mv^2)/r
Ff = u(Fn)
Fn = mg

Ff = u(mg)
(mv^2)/r = u(mg)

The mass, m, cancels out, leaving:
(v^2)/r = u(g)

Plugging in what we know, we get:
(7.5^2)/r = .22(9.8)
65.6 = 2.93r
r = 26.08 m

So, the circle must have a radius of at least 26.08meters.

Answer 3

(mu)m(9.8)=m(v squared)/r
m is on both sides of the equal sign so cross it out
mu=0.27 given
v is given convert it out of km/h into meters per second
solve for r

r=v^2/mu*9.8

r=7.5*7.5/0.27*9.8

r=21.2m

f

Answer 4