Assume that the wheel\'s geometry can be approximated by a thin ring and that th
ID: 1819104 • Letter: A
Question
Assume that the wheel's geometry can be approximated by a thin ring and that the unicyclist is able to keep the unicycle in a vertical position while climbing the hill. Consider the system of interest to be the unicycle's wheel and the weight of the unicycle and the unicyclist applied to a point located at the center of the wheel.
Explanation / Answer
consider (unicycle + uni-cyclist) as the system m=mass of (unicycle + uni-cyclist)=72.5 Work done by gravity = - m*g*h =- 72.5*9.8*h = - 710.5h (-ve sign indicates work done against gravity force) Work done by uni-cyclist =M*(s/r) ( torque multiplied by the angle of rotation of wheel) s= h/sin15 (from figure) => Work done by uni-cyclist = M*(h/(r*sin15)) =489.4*h change in kinetic energy of wheel = -0.5*I*(w^2) I = moment of inertia of wheel = (mass of wheel)*r^2 = 6.5*0.45^2 =1.31625 kg-m^2 w=angular velocity at beginning = v/r = 13.5/0.45 = 30 rad/s =>change in kinetic energy of wheel = -0.5*I*(w^2) =-592.3125 change in kinetic energy of uni-cyclist = -0.5*(mass of uni-cyclist)*v^2 = -0.5*(72.5-6.5)*(13.5^2) = -6014.25 Energy conservation equation: Work done by gravity + Work done by uni-cyclist = change in kinetic energy of wheel + uni-cyclist. 489.4*h - 710.5h = -592.3125 + (-6014.25) solving we have, h=29.88m NOTE: The answer may be different if g is not taken as 9.8. If the answer is not close 2 the textbook answer, then my solution may have some calculation mistake but follow the same procedure and calculate again. Cheers!!