s/mrshuggins/Downloads/Phys1114%20HMwKW208.pdf I, A uniform spherical globe(,) h
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s/mrshuggins/Downloads/Phys1114%20HMwKW208.pdf I, A uniform spherical globe(,) having a radius of 15 and a mass of 0.05 kg, rotates about its vertical axis on frictionless bearings. A light chord is wound around the equator of the globe, then run over a frictionless pulley (1,-1Y2and aachod to a moctal weight as shown in the figure above. The pulley is a solid wheel having a mass of 0.03 kg and a radius of 5 em. The metal weight has a mass of O.1 kg. If released froen rest, Let us determine the velocity of the metal weight after it has descended a distance of 0.2 m, using conservation of encrgy a) Write an expression for the gravitational potential cnergy, U, of the metal weight if it is hanging from a height of 0.2m. b) Write an eapression for the translational (linear) kinctic energy, K,, of the metal wcight as it is descending c) Write an expression for the otational kinetic energy, of the pulley. d) Write an expression of the rotational kinetic anergy, K, of the spinning globe. e) Equate the gravitational posential energy.U, of the mctal weight with the three kinctic tenns PHYS1114 Beeson Spring 2018 OSU Physics Dept. 0 Sabstiaae in the appeoprlate expecssions for the s of ieetia of the pulley and the globe and substioue in appropriate valucs of the angular velocities of the palley and the globe in terms of ta ential velocity and radius. Then simplify the algebra. g) Rewrite the equation to solve for velocity, plag in values, and determine the desired vclocity of the descending weightExplanation / Answer
1. given
I = 2mr^2/3
r = 0.15 m
m = 0.05 kg
Ir = 0.5mr*r'^2
mr = 0.03 kg
r' = 0.05 m
M = 0.1 kg
d = 0.2 m
a. U = Mgd = 0.1962 J
b. KE of weight = 0.5Mv^2 = 0.05v^2
c. rotatinoal KE of pulley, KEp = 0.5*Ir*w^2
w = v/r'
KEp = 0.5(0.5mr*r'^2)v^2/r'^2 = 0.25mrv^2 = 0.0075v^2
d. rotational KE of the globe = 0.5*I*w'^2
w' = v/r
KEg = 0.5*2*m*v^2/3 = mv^2/3 = 0.01666v^2
e. hence from KE = PE
0.01666v^2 + 0.0075v^2 + 0.05v^2 = 0.1962
v = 1.6264 m/s
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