Question

A car initially traveling at 29.8 m/s undergoes a constant negative acceleration of magnitude 1.60 m after its brakes are applied (a) How many revolutions does each tre male before the comes to a stop, assuming the car does not sad and the tires have made of 0.330 ? (6) What is the angular speed of the wheels when the car has traveled half the total distance? Need Help? The 4 -14 points SerCP8 7.P.012, My Notes Ask Your Teacher . It starts from restatt , and a line drawn from the center of the disk to a point on the rim of A 50.)-cm diameter diskrotates with a constantangular acceleration of 2.7 r the disk m anage of 57.3" with the p a st the (a) Find the angular speed of the wheel at 2.30 rad/s of 2.30 ) Find the near velocity and tangential Inear velocity tangential acceleration Find the position of in degrees, with respect to the positive a t t 2.30

Answer 1

3. Initial angular speed, w_{​​​​​​o}​​​​​ = v/r = 29.8/0.330 = 90.3 rad/s

Angular acceleration, $\alpha$ = a/r = -(1.60/0.330) = - 4.85 rad/s^{​​​​​​2}​​​​​s2

Final angular speed, w = 0

(a) Using, w^{​​​​​​2} - w_{​​​​​​o}^​​​2 = 2$\alpha$$\theta$, we have,

0² - 90.3² = -2*4.85*$\theta$ => $\theta$ = 840.63

So, number of revolution = (840.63/2*3.14) = 133.86 rev

(b) When car travels half the distance, angular displacement will be, $\theta$ ' = (840.63/2) rad = 420.32 rad

Again using the same equation as above, we have,

w^{​​​​​​2} - 90.3² = -(2*4.85*420.32) = -4077.104

=> w^{​​​​​​2} = 4076.986 => w = 63.85

Thus, angular speed = 63.85 rad/s