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Phy 241
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I have a question about one of the problems on test 2 for PHY 241. It states A uniform disk is growing in such a way that its radius inc by 0.3cm every min. The mass density of the disk is 5g/cm^2 of cross-sectional surface area. If I(r) is the moment of inertia of the disk when its radius is r, then what are dI/dr and dI/dt at the instant the radius is 50cm?
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Ok, I understand that since this is a uniform rod that the moment of inertia will be I=(1/2)MR^2.
disk, not rod; your moment of inertia formula is OK
I think when it's asking for dI/dr, that I would just do: I=0.5(5g/cm^2)*(50cm)^2, since 50cm was the radius given and I got 6250.
M is 5 g/cm^2 * r.
r is 50 cm, and dr/dt = .03 cm/minute.
I and m are both functions of radius, which is in turn a function of clock time t. So I changes with respect to r, and therefore with respect to t. The chain rule is therefore relevant: dI/dt = dI/dr * dr/dt.
As a function of radius, m = 5 g / cm^2 * pi r^2, so I = m r^2 = 5 g / cm^2 * pi r^2 * r^2 = 5 pi g / cm^2 * r^4.
So dI / dr = 5 pi g/cm^2 * 4 r^3.
r is in turn a function of t, so
dI/dt = dI/dr * dr/dt = 5 pi g/cm^2 * 4 r^3 * dr/dt = 5 pi g/cm^2 * 4 * (50 cm)^3 * .03 cm/min = 2.5 * 10^6 g cm * .03 cm/min = 7.5 * 10^4 g cm^2 / min.
I THINK that's what I would do for that one. Now, when it's asking for dI/dt, I think I would use the fact that the disk is growing in a way that the radius increases 0.3cm every minute. So would this mean I would do I=0.5(5g/cm^2)*(0.3cm/min)??? Or would I have to do something with the 0.3cm/min before plugging it into the equation?
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You need to set this up symbolically. Both m and r are functions of t. See my notes.