Question

An automobile traveling at 50 km/h has tires of 75.0 cm diameter.

A) What is the angular speed of the tires on the wheels?

B) If the car is brought to a stop uniformly in 15 complete turns of the tire (without skidding), what is the magnitude of the angular acceleration of the wheels?

C) How far does the car move during the braking?

Answer 1

Part A:

Angular speed is given by:

V = w*r

w = V/r

V = 50 km/hr = 50*5/18 = 13.89 m/sec

r = 75/2 = 37.5 cm = 0.375 m

So,

w = 13.89/0.375 = 37.04 rad/sec

Part B

wi = 37.04 m/sec

wf = 0

theta = 15 rev = 15*2*pi = 94.25 rad, So

wf^2 = wi^2 + 2*alpha*theta

alpha = angular acceleration = ?

alpha = (0^2 - 37.04^2)/(2*94.25) = -7.28 rad/sec^2

|alpha| = 7.28 rad/sec^2

Part C

distance traveled will be

linear displacement = radius*angular displacement

s = r*theta

s = 0.375*94.25 = 35.34 m

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