A motorcycle stunt rider has just made a miscalculation and has been launched in

Question

A motorcycle stunt rider has just made a miscalculation and has been launched into the air. At a specific point during his flight through the air, his center of mass is 2.6 meters high, and has a linear velocity of 12.3 m/s. He also has an angular velocity of 2.1 rad/s and a moment of inertia of 5.9 kgm2. His body mass is 87.7 kg. When he hits the ground and starts sliding, the friction force between his leather clothing and the pavement is 207.5N. How far (m) does he slide from the point of impact?

Explanation / Answer

Given that

h = 2.6 m I = 5.9 kgm^2
v = 12.3 m/s m = 87.7 kg
w = 2.1 rad/s f = 207.4 N

m*g*h + 0.5*m*v^2 + 0.5*I*w^2 = f*d

(87.7*9.8*2.6) + (0.5*87.7*12.3^2) + (0.5*5.9*2.1^2) = 207.4*d

(2234.59) + (6634.06) + (13.00) = 207.5*d

d = 42.80 m

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