Asked by Joe
A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.8rev/s . The wheel can be considered a uniform disk of mass 4.7kg and diameter 0.34m . The potter then throws a 3.1kg chunk of clay, approximately shaped as a flat disk of radius 13cm , onto the center of the rotating wheel.
What is the frequency of the wheel after the clay sticks to it?
I got 1.64 rev/s and it is not correct. Can anyone explain what I might be doing wrong?
What is the frequency of the wheel after the clay sticks to it?
I got 1.64 rev/s and it is not correct. Can anyone explain what I might be doing wrong?
Answers
Answered by
Damon
P = angular momentum = I omega
I1 = (1/2) m r^2 = (1/2)(4.7)(.17)^2
= .0679
P1 = (.0679) 2 pi (1.8)
I2 = I1 + (1/2)(3.1) (.065)^2
= .0679 + .00655 = .0744
P2 = P1
so
.0744 (2 pi)(rps) = .0679 * 2 pi *1.8
rps = 1.64 revs/s so I agree with you wholeheartedly
I1 = (1/2) m r^2 = (1/2)(4.7)(.17)^2
= .0679
P1 = (.0679) 2 pi (1.8)
I2 = I1 + (1/2)(3.1) (.065)^2
= .0679 + .00655 = .0744
P2 = P1
so
.0744 (2 pi)(rps) = .0679 * 2 pi *1.8
rps = 1.64 revs/s so I agree with you wholeheartedly
Answered by
Ashley
In I2 you halfed the radius of the flat disk as if it were a diameter like pottery wheel. it's not 0.065 it's 0.13
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