The famous motorbike artist Evel Knievel enjoys jumping. He departs from an incl

Question

The famous motorbike artist Evel Knievel enjoys jumping. He departs from an inclined plane witha velocity of 100 km/h and no rotation. He wants to try a summersault while he is airborne within 2 s. Based on the conservation of angular momentum, he will start to rotate if his wheels' rotation is slowed down. His wheels are hollow cylinders with a radius of 0.4 m and a mass of 10 kg. The inertia of the bike and him is 200 kgm2. Calculate, how much the wheels' rotation has to be modified in order to make a full rotation of the bike during 2 s. Let's assume the angular velocity can be changed within 0.1 s Which angular acceleration does that correspond to?

Explanation / Answer

given
u = 100 km/h

radius of wheel, r = 0.4 m
holoow cylinders
m = 10 kg
I1 = 200 kg m^2

now, for one complete rotation in 2 s
w = 0 +alpha*t
and
2*pi = 0.5*alpha*t^2

2*pi = 0.5*w*t
t = 2 s
2*pi = 0.5*w*2
w = 2*pi rad/s

now,
u = 100 km/hr = 27.7778 m/s
angular speed of wheels W = u/r = 69.4445 rad/s

final angular speed = W'
form conservation of angular momentum
2*mr^2*W = 2**mr^2*W' + I1*w
hence
W' = -323.25408 rad/s
hence the angular speed of the wheels has to be changed by 69.445 - - 323.2504 = 392.698 rad/s

angular acceleration of the tyres = 392.698/0.1 = 3926.98 rad/s/s

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