#$&*
Phy 231
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I'm resubmitting my work regarding some test #2 questions to see if I'm on the right track, thanks for your help.
question form
#$&*
Phy 231
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'm looking over the #2test in preparation for taking the real thing, and I have some questions about 2 of the problems. The 1st is. . .
A uniform disk is growing in such a way that its radius increases by .3 cm every minute. The mass density of the disk is 5 grams per 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 50 cm?
Do you use the equation I=.5mr^2? If so it would be I(50cm)= .5m(50cm)^2, but I'm not sure how to incorporate the .3cm increase in the radius for every minute or how to tie in the 5grams per cm^2 of cross-sectional area.
@& Let t = 0 at the instant the radius is 50 cm.
Then r(t) = 50 cm + .3 cm/min * t.
This could extend into negative times, back to t = -166 2/3 minute (when r would be zero).
The area of the disk is
pi r^2 = pi ( 50 cm + .3 cm/min * t).
Thus the mass of the disk is
m(t) = pi ( 50 cm + .3 cm/min * t) * 5 grams / cm^2
and its moment if inertia I = 1/2 m r^2 is a function of t:
I(t) = m(t) * r^2 (t).
You can easily substitute for m(t) and r(t) then find dI/dt
You can also express I as a function of r.
m is a function of r, with m(r) = 5 grams / cm^2 * pi r^2.
I = m r^2.
Substituting m(r) for m you get an expression whose derivative you can find with respect to r.*@
&&&&I(t)= ½ m(t)*r^2(t)
I(0)= ½ (pi(.5m + .003m/min(0)))^2*.5kg/m^2*(.5m + .003m/min(0))^2
I(0)= .154kgm^2
I(5)= ½ (pi(.5m + .003m/min(5min)))^2*.5kg/m^2*(.5m + .003m/min(5min))^2= .174kgm^2
dI/dt= (.174kgm^2 - .154kgm^2)/(5min - 0min)= .02kgm^2/5min= .004kgm^2/min
I= ½ (.5kg/m^2*pi r^2)r^2
Derivative of I= kg/m^2*pi r^3
I(.5m)= kg/m^2*pi(.5m)^3= .393kgm
I(1m)= kg/m^2*pi(1m)^3= 3.14kgm
dI/dr= (3.14kgm- .393kgm)/(1m - .5m)= 2.747kgm/.5m= 5.494kg&&&&
@& I(0) and I(5) are constant, and their derivatives are zero. You've got an average rate of change, but you don't have an instantaneous rate of change. To get the instantaneous rate you need to take a derivative.
There was an error in the information I gave you above. mass = area * density = pi r^2 * density
so
m(t) = pi ( 50 cm + .3 cm/s * t) ^2 * 5 g / cm^3.
Suppressing units r(t) = 50 + .3 t, with the understanding that when t is in minutes r is in cm.
Supressing units on density we understand that the units of L in the expressions below are gram cm^2:
I(t) = m(t) r^2(t)
= [ pi ( 50 + .3 t)^2 * 5 ] * (50 + .3 t)^2
= 5 pi ( 50 + .3 t)^4.
IThe rate of change of L with respect to t is dI / dt, the derivative of the expression for I(t).
To get the rate of change at t = 5 minutes you find dI/dt and plug in 5, being careful you understand and correctly express the units of your result.
Alternatively dm / dt is easily enough calculated (we get 5 pi * .3 cm/s * 2 * (50 cm + .3 cm/s * t) = 3 pi ( 50 cm + .3 cm/s * t). This derivative was found using the chain rule.
Again using the chain rule,
dI / dt = dI / dm * dm / dt.
I = 1/2 m r^2 so dI / dm = m r.
dm/dt was just calculated.
The expression
dI/dt = dI / dm * dm / dt
should match the expression you obtain for dI/dt by finding the derivative of 5 pi ( 50 + .3 t)^4.*@
My 2nd question is. . .
An Atwood machine consists of masses of 1.2 kg and 1.31 kg on opposite sides of a light frictionless pulley. The system is given an initial velocity of 1.3 m/s in the direction of the 1.2 kg mass.
How much work does gravity do on the system between the initial instant and the instant at which the system comes to rest, and how much work is done by the system against gravity during this time?
How far does the system therefore travel during this time (solve using energy considerations)?
How are the work done by the system and its kinetic energy change related?
.dW= Fave*ds= 24.6N * -1.95m= -47.97Joules by system against gravity
dW by gravity= 24.6N * -.09m= -2.2Joules
.v0=(-1.3m/s), vf= 0m/s, a= .43m/s^2, dt= 3s, ds= -1.95m
This is what I've come up with so far, but I don't think this is correct.
Any help would be very much appreciated, thanks.
&&&&KE= ½ mv^2= ½(2.51kg)(1.3m/s)^2= 2.12Joules
I chose the positive direction to be that of the 1.31kg mass going down.
F= 1.2kg * 9.8m/s= 11.76N, but in the negative direction so -11.76N
F= 1.31kg * 9.8m/s= 12.84N in the positive direction
F_net= 12.84N - 11.76N= 1.08N
dW and dKE values are roughly the same but opposite in signs, added together they should equal 0. But since my values are approximations its not exactly 0 and they are not exactly opposites.
-2.2Joules + 2.12Joules= -.08Joules&&&&
@& Good.*@
self-critique rating
2
@& Your solution is a little out of order (the ds should be found before it is used to find the work), but that's OK.
Part of the system goes up (with an accompanying increase in PE) and part goes down (with accompanying decrease in PE), so the net change in gravitational PE is a couple of Joules.
However you are supposed to solve using energy considerations.
So what is the KE of the system at 1.3 m/s?
The system will come to rest for an instant when this KE has been converted to PE.
What is the net force on the system? Note that whichever direction you choose as forward, one of the masses will move upward and one downward, so the gravitational force on one mass will be in the positive direction and the gravitational force on the other will be in the negative direction.*@
@& I've inserted some notes. See if you can make more progress based on those notes. You're welcome to submit your additional work, and/or questions.*@
@& Check my additional notes on the first question. Your work on the second looks good.
Additional questions and/or revisions are welcome. If you do submit more, use #### to denote insertions.*@