Please help me to solve this question. A unicyclist is driving down a hill, with

Question

Please help me to solve this question.

A unicyclist is driving down a hill, without pedalling. His wheel has mass m = 10.0 kg and radius r = 30.0 cm, and can be considered a ring, for the purpose of its moment of inertia (ignore the spokes of the wheel). The combined mass of the cyclist and the unicycle is M = 90.0 kg. You may assume that the centre of gravity/center of mass is immediately above the axis of the wheel at all times. Note also that the gravitational force acts on the wheel via the axis of the wheel, and so does not give torque to the wheel. If the slope of the hill is 10.0 degrees from horizontal, what is the acceleration of the centre of mass? And what is the angular acceleration of the wheel? What is the minimum necessary coefficient of static friction, to prevent the wheel from sliding? The hill is d = 100 m long, and immediately after reaching the bottom, the road goes back up. What is the speed of the centre of mass when reaching the bottom, and what is the total kinetic energy? You may assume that the only friction present is the static friction responsible for the wheel not slipping. You may also assume that the unicyclist started from rest at the top of the hill. The unicyclist now starts pedalling uphill, providing a constant torque of = 30.0 Nm (he is pedalling forwards, obviously). What is his acceleration, if the hill is again sloping (this time upwards) at an angle of 10.0 degrees? How far up the hill does he get before coming to a halt, assuming that he started out with the speed from part c)?

Explanation / Answer

a) Force due to gravity = F = mgcos80

frictional force = f

F - f = Mgcos80 - f = Ma

f*R = I*alpha = I*a/R

f = Ia/R^2

Mgcos80 - I*a/R^2 = Ma

a = Mgcos80/(M + I/R^2) = 90*9.8*cos80/(90 + 10) = 1.531m/s^2

b) f = Ia/R^2 = u*Mg*cos10

u = ma/(Mgcos10) = 10*1.531/(90*9.8*cos10) = 0.0176 = coeff of friction

c) Mgh = 0.5*M*v^2 + 0.5*I*w^2 + f*d

90*9.8*100sin10 = 0.5*9.8*v^2 + 0.5*10*v^2 + 0.0176*90*9.8*cos10*100

v^2 = 1392.62

v = 37.3m/s

d) similar equations add torque in f*R = I*a/R

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