Assume that the wheel\'s geometry can be approximated by a thin ring and that th
ID: 2075932 • 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.
problems that involve velocities, forces, moments, and displacements. For a rigid body, the principle is 1-2 Figure 2 of 2 Vo Part B As shown, a unicyclist traveling at vo 9.50 m/s approaches a hill inclined at 15.0 degrees (Figure 2). The combined mass of the unicyclist and unicycle is 102 kg; the unicycle's wheel has a mass of mur 5.50 kg, and it has a radius of T 30.55 m If the wheel does not slip, and the unicyclist is able to apply a constant couple moment of magnitude M 48.0 N.m to the wheel, what is the maximum elevation, h, that the unicyclist can climb? Assume that the wheels 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 Express your answer numerically in meters to three significant figures Submit Hints My Answers Give Up Review Part Provide Feedback ContinueExplanation / Answer
Work done = M * theta = M * (s / r ) = M * [ h / sin phi * r]
KEwheel = 1/2 I w2
w = v / r = 9.5 / 0.55 = 17.27 rad/s
I = 5.5 * 0.552 = 1.66
KEwheel = 1/2 * 1.66 * 17.272 = 248
KEcyclist = 1/2 m v2 = 1/2 * 102 * 9.52 = 4602.75
PE = m g h = 102 * 9.81 * h = 1000.62 h
Total energy1 = 248 + 4602.75 = 4850.75
Total energy2 = 48 * [h / sin 15 * 0.55] - 1000.62
Total energy2 = -663.42 h
-663.42 h = -4850.75
h = 7.31 m