A motorcycle initially traveling at 120.0 km/hr slows down uniformly to a speed

Question

A motorcycle initially traveling at 120.0 km/hr slows down uniformly to a speed of 80.0 km/hr. During the time that the motorcycle slows down, its wheels turn 88.0 revolutions. The diameter of the wheels is 70.0 cm.

a) What is the angular acceleration of the wheels (in rad/s2) while the motorcycle is slowing down?

b) What is the magnitude of the radial acceleration of a point on the outer rim of the wheel when the speed of the motorcycle is 80.0 km/hr?

c) What is the magnitude of the tangential acceleration of a point on the outer rim of the wheel while the motorcycle is slowing down?

d) How much distance does the motorcycle travel while it is slowing down from 120.0 km/hr to 80.0 km/hr?

e) If the motorcycle continues to decelerate at this rate, how much additional distance will it travel before coming to a stop?

Explanation / Answer

Given that initial velocity is Vo = 120 km /hr = 120*(5/18) = 33.33 m/sec

initial angular velocity is wi = Vo/r = (33.33)/(0.7/2) = 95.3 rad/sec

Final angular velocity is W = V/r = (80*5/18)/(0.7/2) = 63.5 rad/sec

angular dispalcement is theta = 88*2*3.142 = 553 rad

then using

W^2 - wi^2 = 2*alpha*theta

63.5^2 - 95.3^2 = 2*alpha*553

alpha = -4.56 rad/s^2 is the angular accelaration

b) a_rad = V^2/r = (80*5/18)^2)/(0.7/2) = 1411 m/s^2

c) a_tan = r*alpha = (0.7/2)*4.56 = 1.596 m/s^2

d) for one revolution distance moved is 2*pi*r

for 88 revolutions ,distnce moved is S = 88*2*pi*r = 88*2*3.142*0.35 = 193.5 m

e) Using W^2 - Wo^2 = 2*alpha*theta

0^2 - (63.5^2) = -2*4.56*theta

theta = 442.13 rad = 442.13/(2*3.142) = 70.35 rev

so distance moved is 70.35*2*3.142*0.35 = 154 m

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