A luggage carousel at an airport has the form of a section of a large cone, stea
ID: 1413924 • Letter: A
Question
A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 17.0degree with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 41.5 s. Calculate the force of static friction exerted by the carousel on the bag. The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 31.2 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.Explanation / Answer
Taking +x axis horizontally toward the center and +y axis vertically upward, we apply Newton’s 2nd law. Since we need to calculate the minimum coefficient of friction
mr2 = fcos-nsin = n(cos-sin)
n = mr2/(cos-sin)
0 = fsin+ncos-mg n= mg/(sin+cos)
mr2/(cos-sin) = mg/(sin+cos)
= (gsin+r2cos)/(gcos- r2sin)
w = 2pi/T = 2 pi/41.5 s= 0.151 rad/s
= (9.8sin17º+7.46×0.1512cos17º)/(9.8cos17º-7.46×0.1512sin17º)
=0.324
fs = us N = us mg = 0.324( 30) ( 9.8) = 95.49 N
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(b)
us = fs/n = fs cos theta/ mg- fs sin theta
= cos theta/(mg/fs) - sintheta
= cos theta/ g/ g sin theta+ 4 pi^2 r cos theta/T^2) - sinteta
= cos17/(9.8/9.8 sin17+ 4 pi^2 ( 7.94) cos17/(31.2)^2) - sin 17
=0.341
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