1. Energy of a point from a wheel Taylor 4.37 only a) and b) A massless wheel of
ID: 1769987 • Letter: 1
Question
1. Energy of a point from a wheel Taylor 4.37 only a) and b) A massless wheel of radius R, mounted on a frictionless horizontal axle is shown below in Figure1 oCLASSICAL Figure 1: Frictionless wheel, two masses A point mass M is glued on the edge of the wheel, and a mass m hangs from a string wrapped around the perimeter of the wheel. a) Write down the total potential energy of the two masses as a function of the angle b) Use the potential energy to find the values of m and M for which there are any positions of equilibrium. Describe the equilibrium positions, discuss their stability, and explain your answers in terms of torquesExplanation / Answer
A) Total Potential energy E =MgR(1- cos phi) - mg*R*phi
B) it will be in equilibrium if potential energy will be minimum.
dE/dphi =0
MgR*( sin phi) - mgR = 0
M sin phi - m =0
sin phi = m/M
M> or = m
Equilibrium position will be at phi = arcsin(m/M).
It will be stable equilibrium as potential energy has to be minimum, not maximum.
At equilibrium position, torque will be zero, because torque due to M will cancel out that due to m.
Torque = mgR - MgR sin phi = mgR-MgR*m/M = 0
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