To apply the concepts of impulse and momentum to problems involving unknown forc

Question

To apply the concepts of impulse and momentum to problems involving unknown forces, velocities, and times.

A new rear-wheel drive automobile design is being tested and you have been asked to estimate its performance. The car without wheels (i.e., the body) has a mass of mc=1350 kg. The wheels (including tires) have diameter of 61.0 cm and each wheel weighs mw=15.0 kg. The radius of gyration of the wheels about their centers is k=20.7 cm. Model the torque output (which acts in the counterclockwise direction) to each rear wheel as T(t)=80.0 t3/2 Nm. Assume that the wheels roll without slipping and neglect the air resistance and any other frictional effects.

(Figure 1)

Part A - Time for the Car to Reach 100 km/h

Find the time it will take for the car to reach v=100 km/h from rest.

Express your answer to three significant figures in seconds.

Hints

Part B - Force Exerted by the Brakes

After the car reaches 100 km/h the driver immediately steps on the brakes and comes to a stop in 2.85 s . The brakes are at a distance 14.5 cm from the center of the wheel and the force they apply acts in the y direction. Assume for this problem that the brakes apply an equal force to each wheel, that the torque to the rear wheels is zero and that the wheels continue to rotate until the car stops. Calculate the magnitude of the force, Fb, exerted by the brakes on each wheel during this time.

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

mass of car, mc ( without wheels) = 1350 kg
diameter of wheel including tyre, D = 0.61 m
mass of each wheel, mw = 15 kg
radius of gyration of wheels, k = 0.207 m
Assumptions
Torque, T(t) = 80t^(3/2)
No slipping, just rolling

let acceleration of car be a, angular acceleration of tyres be alpha
then a(t) = alpha(t)*D/2 ... (1)
and torque for rolling = T' ( Clockwise )
so, torque balance on one wheel
T(t) - T' = I*alpha(t)
here I = mw*k^2 ... (2)

also, from force balance on the car, 4f = (mc + 4mw) * a(t)
and f = 2T'/D
8T'/D = (mc + 4mw)*a(t)
8[T(t) - I*alpha(t)]/D = (mc + 4mw)*a(t)
8[80t^(3/2) - 2*(mw*k^2)*a(t)/D]/D = (mc + 4mw)*a(t)
8[80t^(3/2) - 2*(15*0.207^2)*a(t)/0.61]/0.61 = (1350 + 4*15)*a(t)
1049.18t^(3/2) = 1437.63*a(t) = 1437.63*dv/dt
1049.18t^(3/2)dt = 1437.63*dv
integrate from v = 0 to v = v , t = 0 to t = t
2*1049.18t^(5/2)/5 = 1437.63*v

so, v = 100 km/h = 27.778 m/s
t = 6.185 s

v = 27.778 m/s
final velocity, u = 0 m/s
t = 2.85 s
so, v = u + at
a = (v - u)/t = 9.746 m/s/s ( deceleration )
force exerted by friction on 4 wheels combined = 9.746*(1350 + 4*15) = 13742.8 N
Friction on each wheel, f = 3435.7 N
torque due to friction = fD/2
Force exerted by brake = Fb
so, fD/2 - Fb*0.145 = 15*0.207^2*9.746*2/D
Fb = 2034.981 N

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