Question

A 1700kg car starts from rest and drives around a flat 68-m-diameter circular track. The forward force provided by the car's drive wheels is a constant1300N .

What is the magnitude of the car's acceleration at t=12s?

What is the direction of the car's acceleration at t=12s? Give the direction as an angle from the r-axis.

If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

Answer 1

The mass of the car, m = 1700 kg

      The diameter of the circular track, D = 68 m

      The radius of the circular track, r = D/2

                                                            = 68 m/2

                                                            = 34 m

      The forward force provided by the car's drive wheels, F = 1300 N

      From Newton's second law of motion, we deined the force is

                       F = ma

      Therefore, the tangential acceleration is

                    a_t = F/m

                       = (1300 N)/(1700 kg)

                       = 0.7647 m/s²

The speed can be calculated as follows:

              v = at = 0.7647*12 = 9.1764 m/s

Radial component of the acceleration is,

             a_r = v²/r = (9.1746)² / 34 = 2.4766 m/s²

Net acceleration of the car is,

     a = [a_t² + a_r²]^1/2 = [0.7647^2 + 2.4766^2]^1/2 = 2.592 m/s²

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      Apply the Newtons second law of motion to the motion of the car in circular path,

      The net force acting on the car is

                       ?F_c = f_s

                     mv²/r = f_s

      From the definition of the frictional force,

                         f_s= ?_sN

      Therefore, the above equation becomes

                      mv²/r = ?_sN

                                = ?_smg

      Here, the coefficient of ststic friction between tires and track ?_s = 1.0

      Therefore, the velocity of the car is

                            v = ?rg

                               = ?(34 m)(9.8 m/s²)

                               = 18.25 m/s

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      Using the kinematic relation, we have

                          v = v₀+at

                              = 0 + at

      Therefore, the time at which the car begin to slide out of the circle is

                           t = v/a

                             = (18.25 m/s)/( 0.7647 m/s²)

                             = 24.18 s