The tables have turned and a skinny Johnny chases after the

The tables have turned and a skinny Johnny chases after the wolf-monster with a giant powered unicycle having a mass of 200kg and diameter of 8m. If the radius of gyration is 2m and the center of gravity is at G, determine the maximum moment he can apply such that the unicycle does not slip when accelerating. Assume the coefficient of static friction is 0.9. Next, how fast will the unicycle accelerate if the wheel just slips as he applies that maximum moment, assume coefficient of kinetic friction 0.5? Bonus: they are separated by 5m when Johnny carefully applies the maximum moment, how long will it take for him to catch the wolf if it runs with a constant speed of 5m/s and his wheel docs not slip?

Solution

a) The maximum moment to move the unicycle is the net force required for the unicycle to rotate,

F - Fn = ma -------- eqn

Also, linear and angular momentun is related as,

ma = m x r x aplha

Where Fn - coefficient of friction, F - applied force, a - llinear acceleration

Fn = mu_static x W = 0.9 x 200 x 9.81 = 1765.8 N

The angular acceleration is given in terms of torque as,

T = I x alpha where T = F x r, and I = 1/2mr^2 (mass moment)

F x r - Fn x r = I x alpha

T - Fn x r = I x a / r ----- since alpha is ma / I

F x r - Fn x r = 1/2 m r2 x a / r

F x r - Fn x r = 1/2 mr / a

F - Fn = 1/2 x m / a

using eqn 1

ma = 1/2 m / a   

a = sqrt 1/2 = 0.707 m/s2

F = ma + Fn = 1765.8 + 200 x 0.707 = 1907.22 N

The maximum torque required is T = F x r = 1907.22 x 12 = 22.8 kN-m

b) If the kinetic friction coefficient is considered mu = 0.5

Since there is slip in the motion for the applied force of 1907.22 N the acceleration is given by,

F - Fn = ma

Fn = 0.5 x 200 x 9.8 = 980 N

The acceleration is given by,

F - Fn = 1907.22 - 980 = 927.22

a = 927.22 / 200 = 4.63 m/s2

c) Time taken to travel 5m from rest

1) At no slip acceleration

a = 0.707 m/s2

t = (v-u) / a = 5 / 0.707 = 7sec

2) slip acceleration

a = 4.63m/s2

t = 5 / 4.63 = 1.07 sec