#$&*
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.
** **
These questions are from test 2. I planed on coming down and asking you today but... snow fail!
** **
question 1) I believe I understand the derivation of the velocity function being:
Fg = GMm/r^2
=>mv^2/r = GMm/r^2
=>v_r= sqrt(GM/r)
=>d(v)/dr = sqrt(GM/(2r^2)dr
however what is not as clear is the derivation of the v(r) with respect to time. I can obviously do partial differentiation.
Right, but you want to say that a little differently:
d(v)/dr = sqrt(GM/(2r^2)) so
dv = sqrt( G M / (2 r^2) ) dr.
dv/dt would be dv/dr * dr/dt, by the chain rule.
However you didn't include a copy of the question. I'm not aware of any question where you are asked for dv/dt. In this context that wouldn't make a lot of sense, but of course that doesn't mean I didn't ask it.
The reason it wouldn't make sense is that you wouldn't continuously change from one circular orbit to another.
and the result would be:
sqrt(GM/r)dt, I think this is right but I want to make sure. My intuition says other wise bc of units:
if G is said to be in J*s which I did not understand the units were not right or anything. Then I checked it out on Wikki and found the units for the gravitational constant to be m^3/(kg*s^2) this makes more sense but I wanted to confirm that i am on the right track. if we let gG be in in m^3/(kg*s^2) and r is in meters and M is in kg. It follows that v= the sqrt[m^3/(kg*s^2)*kg/m] = m/s. As a double check if i wanted to go from meters/s to rad/s I would just divide the vr by r to get w (omega) = rad/s
There is a faulty value of G on the test. The correct value is 6.67 * 10^-11 N m^2 / kg^2. In fundamental units that comes out to m^3 / (kg s^2), but that's a very nonintuitive unit. The intuitive unit is easily reasoned out. You have to get Newtons when you multiply G by M m / r^2. M m / r^2 has units kg^2 / m^2. So the units of G have to be N * m^2 / s^2.
question 2)
I am still not comfortable with the moment of inertia for different regions.
If you have a disk I believe the moment of inertia is 1/2mr^2
where m is the mass and r is the radius.
so if given a constant dr and a point of r as well as the mass the the function the moment of inertia is 1/2mr^2
=> dI/dr=2mrdr
and=>dI/dt= 1/2*m*r^2*dt
Once more it would be a good idea to include a copy of the problem.
I believe you are given the rate of change of r with respect to t. That is, you are given dr/dt, not just dr.
Your expression
dI/dr=2mrdr
is not quite right, and I believe this might be misleading you here.
dI/dr=2mr
and by the chain rule
dI / dt = dI / dr * dr / dt.
Thus
dI/dt = 2 m r * dr/dt.
This assumes that the disk has constant mass. If it is the mass density of the disk that is constant, then the increasing radius also implies an increasing mass, and your expression for the moment of inertia would nee to be adjusted accordingly.
then after solving for dI/dr we can multiply that value by dr and have dI then we can say dI = m*r^2*dt^2 => dt= sqrt((dI/mr^2))
if this is exactly right cool please let me know so if not OK I need to know what I missed.
** **
If the college is open tomorrow will you be in so I can review these concepts before I take the test? I do have one more question but I will be submitting it later on today. If I do need to be at highlands at a certain time to ensure I meet with you just let me know when.
Looks possible that we're down until Monday. If we are in session on Friday, I hope to be there all afternoon. However I'm somewhat under the weather (which is slowing me mentally more than physically), and it might be as late as 2:00 before I get there.
I do plan to be at VHCC all afternoon on Monday.
#$&*
q1`1
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.
** **
I am still doing the same test and just want to see what I am not doing.
I was given a disk problem the problem included the radius of the disk the fact that it was frictionless, that there were three masses i will call m1, m2, and m3 as well as the fact that there was a Force acting tangent to the disk I will call F_theta.
the first question(test): if the disk has traveled through a distance d(theta) then what is the `dKE of the unit.
Well it seemed apparent that the only Force was F_theta. If this were true I needed to find the torque it acted with upon the system. I found this by crossing rxF_theta = tue (this is my abbreviation for torque) tau, but I'll know what you mean
so ow I had tue and needed `dKE. tue dotted with d(theta)= work = `dKE = 1/2 Iw^2, where I is the the total moment of inertia and w = omega = rad/s.
omega, not w; w is an especially bad idea when work is involved; however I understand your meaning
`dKE = 1/2 I omega_f^2 - 1/2 I omega_0^2, not just 1/2 I omega^2. This is a common source of error.
Assuming tau and `dTheta are in the same direction, as I believe is the case with this problem, tau = r * F. So the work done is tau * dTheta = r * F * dTheta, and in the absence of nonconservative forces this is equal to `dKE.
If your initial angular velocity is zero then
`dKE = 1/2 I omega_f^2 - 1/2 I omega_0^2 = 1/2 I omega_f^2 - 0 = 1/2 I omega_f^2.
*********************
question 2) I w0 = 0 and we are still using the 5 rad = d(theta)then what is wf. bc we know `dKE = 1/2*Iw^2 it follows
2*`dKE / I = w^2 =>w = sqrt(2*`dKE / I)
we know `dKE so is all we need is I.
To find I, I [hahaha] found the moment of inertia for the disk:
[again I am a little shaky here.] The moment of Inertia (I) = I_disk+I_m1+I_m2+I_m3 = 1/2*m*r^2 + m1 x r1 + m2 x r2 + m3 x r3, where r1, r2.. = the distance from the center [note this would not be true if the rotational point was not about its center; in the test it states that the object is rotating about its center.]
Having the moment of inertia for the entire unit we solve for w.
Right, with the qualifications I mentioned on the first question.
***********
question 3)(test) what is the KE of each of the masses at the finial velocity? this = 1/2*m1(w * r1)^2,...1/2*m3(w * r3)^2. the sum of these KE will be less that the `dKE bc we did not include the `dKE of the disk.
so:1/2*m1(w * r1)^2,...1/2*m3(w * r3)^2 < 1/2 * I * w^2
I think you've got it. Just to be sure,
KE = 1/2 m1 (omega * r1)^2 + 1/2 m2 (omega * r2)^2 + 1/2 m3 (omega * r3)^2 + 1/2 I_disk * omega^2.
*********************
test question 4)if the force being applied is due to a mass suspended form a light string then what is the mass on the string and how far does it travel. We have already found w and it start from rest + we are given d(theta) so we will let alfa = a'
to find a` we can use the eqtn of constant acceleration => a` = w^2/2*`d(theta)
now that we have a`:
a`*r = the acceleration at the end of the vertical disc = a
We are given F so F/a = m
This is good reasoning and will work as long as m is small compared to the mass of the system (more accurately, as long as m r^2 is small compared to the moment of inertia of the system).
A more direct alternative: alpha = tau / I (Newton's Second Law) and a = alpha * r (definition of the radian). If you can see how this is equivalent to your reasoning, you're doing very well.
now we are asked to find `dy:
there are two ways
1: given d(theta) and r if we cross them we get `dx; [simplest way]
d(theta)*r=dy
2:also to check our other work:
we can say: w x r = v and we know a => ds = v^2/(2*a)
we could also say the energy in rotation = the work gravity did so 1/2 I w^2 or tue * d(theta) = F_theta*ds => `ds = tue * d(theta)/ F_theta, 1/2 I w^2 / F_theta = `ds
** **
Note that m r^2 for the descending mass would be part of the moment of inertia. If it's not neglible then you would simply include it in the first part of the problem, when you first calculated I. If you know the force F exerted by gravity on the mass, then F = m g and m = F / g, and m r^2 = F / g * r^2 so I = m1 r1^2 + m2 r2^2 + m3 r3^2 + F / g * r^2 + 1/2 M_disk r^2.
I put above the problem and what I though to do. I never used F = -kx and do not know what scenarios to apply it to. It there is anything correct in my above symbolic soln please let me know as well as what I am doing wrong[of course this is the most important to me :)]
** **
Good questions. Let me know if you have more.