A 1500 k g car starts from rest and drives around a flat 52- m -diameter circula
ID: 3892945 • Letter: A
Question
A 1500kg car starts from rest and drives around a flat 52-m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1100N .Part A What is the magnitude of the car's acceleration at t=12s?
Part B What is the direction of the car's acceleration at t=12s? Give the direction as an angle from the r-axis.
Part C If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle? A 1500kg car starts from rest and drives around a flat 52-m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1100N .
Part A What is the magnitude of the car's acceleration at t=12s? Part A What is the magnitude of the car's acceleration at t=12s?
Part B What is the direction of the car's acceleration at t=12s? Give the direction as an angle from the r-axis.
Part C If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle? Part B What is the direction of the car's acceleration at t=12s? Give the direction as an angle from the r-axis.
Part C If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?
Explanation / Answer
m=1500kg
d=52m
r = 26m
F=1100N
t= 12s
a) let's start with the tangential acceleration
a(tan)=F/m.
a(tan)=0.7333m/s^2
next is the speed of the car so we can get the radial acceleration...
vf = vi + a(tan)*(delta t)
vf = 0 + 0.7333*12
final velocity( vf)= 8.7996m/s
radial acceleration, a(rad) = v^2 / r
so a(rad) = 2.978m/s^2
total acceleration = sqrt(a(tan)^2 + a(rad)^2)
A = 3.067 m/s^2
b) (theta) = tan ^(-1) [a(tan) / a(rad)] that would be the inverse tangent of the tangential acceleration over the radial acceleration.
theta= tan ^(-1) [0.7333 / 2.978]
angle= 13.83 degrees.
c)
the centripetal force is given as the following:
(m*v^2)/r = (mu)*m*g, where (mu) is the constant for rubber on concrete which actually equals 1.
so v = sqrt{ (mu)*r*g }
substituting v = sq.root {(1)*(26)*(9.8)}
v = 15.96 m/s
using equation vf = vi + a(tan)*(delta t) and vi = 0 we get a result of
(delta t) = vf / a(tan)
substituting, (delta t) = 15.96 / 0.7333
(delta t) = 21.76 seconds.
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