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!
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 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.
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
and that answer comes up as incorrect...
From which book this tasks?
To determine the coefficient of static friction between the car and the track, we can use the following steps:
1. Identify the relevant equations: In this case, we can use the equation for centripetal acceleration (a_c) and the equation for the maximum static friction force (f_s max).
2. Determine the variables: We are given the tangential acceleration (a_t), the radius of the circular track (r), and we need to find the coefficient of static friction (μ).
3. Use the equation for centripetal acceleration: The centripetal acceleration of the car is given by the equation a_c = rω^2, where ω is the angular velocity. Since the car starts from rest, ω = a_t / r. Substituting this value into the equation, we get a_c = r(a_t / r)^2 = a_t^2 / r.
4. Use the equation for maximum static friction force: The maximum static friction force between the car and the track can be calculated using the equation f_s max = μN, where N is the normal force exerted by the track on the car. In this case, N = mg, where m is the mass of the car and g is the acceleration due to gravity.
5. Equate the centripetal acceleration to the maximum static friction force: Set the value of a_c equal to f_s max, and solve for the coefficient of friction. This gives us a_t^2 / r = μN. Rearranging the equation, we find μ = (a_t^2 / r) / N.
6. Substitute the given values and calculate: Plug in the given values for a_t, r, and solve for μ. In this case, the given values are a_t = 1.70 m/s^2 and r = 1/4 of the circumference of the circle. The circumference of the circle is C = 2πr, so r = C/4 = πr/2. Substituting these values into μ = (a_t^2 / r) / N, we can solve for μ.
By following these steps and performing the calculations, you should obtain the correct answer for the coefficient of static friction.