Question

Chapter 11, Problem 002 An automobile traveling 78.0 km/h has tires of 75.0 cm diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in 31. 0 complete turns of the tires, what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the sliding) braking? (Note: automobile moves without Units Units Units (a) Number (b) Number (c) Number

Answer 1

Given, velocity v = 78km/hr = 21.66 m/s Diameter = 75cm = 0.75 m Radius(R) = 0.375 m a) Angular velocity(omega) = u/r = 21.66 m/s/0.375 m = 57.76 rad/sec b) initial displacement theta_1 = 0 rad final displacement theta_F = 31 rev times (2pi/1 rw) = 194.68 rad/sec initial angular velocity, omega_1 = 57.76 rad/sec final angular velocity, omega_f = 0 rad/sec Now, we know - alpha = omega_f^2 - omega_1^2/2(theta_f - theta_1) = 0^2 - (57.76)^2/2 (194.68 - 0) alpha = -8.56 rad/sec^2 c) no. of turns = 31 rev Displacement (x) = 31 rev times (2 pi r/1rev) = 73.005 m Ans