course
David A. Smith
So would we need to use the relationship of k1+u1=k2+u2
The way I would put it:
`dW_noncons_ON = `dKE + `dPE.
No friction, etc., implies that `dW_noncons_ON = 0 so that
`dKE + `dPE = 0.
If you use the book's notation, `dKE = k2 - k1 and `dPE = u2 - u1, so that
k2 - k1 - (u2 - u1) = 0 and
k2 + u2 = k1 + u1.
This is a useful formulation, but in my opinion it's better to start with a single formulation of energy conservation, such as
`dW_noncons_ON = `dKE + `dPE
and then apply it specifically, clearly stating special conditions such as `dW_noncons_ON = 0, so that the situation can be understood in its most general context.
where k1= 1/2Momega(o) ^2 (Am I understanding correctly that the cylinder has a moment of inertia and the rope has a moment of inertia as its being rotated around the cylinder? Then once the rope comes off it has no moment of inertia?)
The rope continues to be part of the system. Its mass remains concentrated at perpendicular distance R from the axis of rotation. So its moment of inertia remains the same throughout.
using mg (since the ropes length is 2piR, will its center of mass be piR?) then mgpiR for the inital gravitational potential energy
The rope is uniform and of length 2 pi R so its center of mass descends through distance R. Relative to the descended position, its initial gravitational potential energy is pi m g R, as you state.
then for K2 =1/2 (I - if at this instant the rope falls off the cylinder, the cylinder and rope are going to have the same angular speed, but will the moment of inertia of just the cylinder be the only thing considered)
just before the rope falls off the cylinder it has an angular speed; the cylinder's angular velocity at this instant is not significantly different from its angular velocity the instant after the rope slides off, if indeed it does slide off (it makes no difference whether it does or not))
omega(F)^2 and then the gravitational potential energy will be mg2piR because the rope has completely came off the cylinder. Will this, then, give the correct answer by solving for omega(f)?
In other words, you lose nothing by assuming that the rope hasn't yet separated from the cylinder
On Tue, Nov 18, 2008 at 10:30 PM, David Smith wrote:
The center of mass of the rope doesn't decrease by 2 pi R. That would be the change in position of the end of the rope, not of its center of mass.
Your equation essentially says that KE0 = KEf - `dPE, which if you think about it is equivalent to saying that -`dKE = -`dPE, which in turn is equivalent to `dKE = `dPE. It's not quite so.
David A. Smith
Associate Professor of Mathematics
Virginia Highlands Community College
dsmith@vhcc.edu
so my error is in using the incorrect moment of inertia for the rope? Are my energy considerations correct with the exception of the moment of inertia of the rope?
On Tue, Nov 18, 2008 at 11:00 AM, David Smith wrote:
The rope is not a rigid rod rotating about an end. Its mass is completely concentrated at a single perpendicular distance from the axis of rotation.
Cylinder- M => I1= 1/2MR^2
Radius=R
Rope mass=m => I2=1/3ML
L=2piR
I=2/3mpi'R it has an initial angular velocity omega(o)
1/2 { I1+I2} omega(o)^2= 1/2{ i1+i2) omega(f)^2+ mg2piR
then solve for omega(f)
On Mon, Nov 17, 2008 at 6:36 PM, David Smith wrote:
See my notes. Then submit a copy of this message using your access code so it can be posted.
David A. Smith
Associate Professor of Mathematics
Virginia Highlands Community College
dsmith@vhcc.edu
From: Breeding Dillon [mailto:dwb2199@email.vccs.edu]
Sent: Monday, November 17, 2008 6:05 PM
To: David Smith
Subject: Rope with cylinder (website down)
M-cylinder; rope- m;
I am considering on whether or not the fact that the moment of inertia of a rope would affect this problem, or whether we can just ignore it and add it to the Mass of the cylinder (M+m), still using 1/2 MR^2 for I.
The mass of the cylinder is not spread out over concentric rings in the manner of the mass of the cylinder. At the beginning the entire mass of the rope is at distance r from the axis of rotation. At the end its distance perpendicular to the moment arm is still r. So its moment of inertia is constant, but it's not ½ M R^2.
As you have said, gravitational potential energy is being in use here: the initial potential energy is 0 and the final is mgh, but we have a rope whose length is equal to the circumference of the circle, so 2(pi)R, but would the center of mass of the cylinder come into play here ( would half the distance be correct (pi' R for h)?
The cylinder is rotating about an axis through its center of mass.
(M/4+ m/2)R^2 (omega0)= (M/4+m/2) R^2omega1+ mg2piR
2 pi r m g would have units of torque. M r^2 omega has units of kg m^3 / s, not units of torque. So the addition is not possible.
I believe the system starts from rest.
What is the moment of inertia of the cylinder?
What is the moment of inertia of the rope?
What is the total kinetic energy at angular velocity omega?
What is the change in the gravitation PE of the system?
Would this be close, correct?
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