A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.70 m/s2.The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data.
I thought the answer was 0.5449 but its wrong. no matter what I do i keep getting the same answer!
6 answers
Why would you think the answer is .5446? I will be happy to critique your work
you need to round off to three sig digits
α=a/r
the time it takes to travel π/2 radians is
related by
π/2=.5*a*t^2/r
solve for t
t=sqrt(π*r/a)
ω=α*t
Plug in t and a/r
ω=sqrt(a*π/r)
The force is m*ω^2*r
F=m*a*π
friction is
m*g*µs
then
m*g*µs=m*a*π
so basically....
µs=a*π/g
=1.70pi/9.8= 0.54497 ====> sig figs 0.545
the time it takes to travel π/2 radians is
related by
π/2=.5*a*t^2/r
solve for t
t=sqrt(π*r/a)
ω=α*t
Plug in t and a/r
ω=sqrt(a*π/r)
The force is m*ω^2*r
F=m*a*π
friction is
m*g*µs
then
m*g*µs=m*a*π
so basically....
µs=a*π/g
=1.70pi/9.8= 0.54497 ====> sig figs 0.545
and that answer comes up as incorrect...
From which book this tasks?
The reason behind the incorrect answer is because there is a tangential acceleration so there must exist a tangential force and we must consider that friction is the resultant of the tangential and centripetal force. so Fr=Sqrt(Fc^2+Ft^2) this should be right.