Register to reply

Conceptual Problem in rotating cone problem

by WannabeNewton
Tags: conceptual, cone, rotating
Share this thread:
WannabeNewton
#1
Dec11-12, 10:30 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
1. The problem statement, all variables and given/known data
A cone of height h and base radius R is free to rotate about a fixed
vertical axis. It has a thin groove cut in the surface.
The cone is set rotating freely with angular speed ω0 and a small block
of mass m is released in the top of the frictionless groove and allowed to
slide under gravity. Assume that the block stays in the groove. Take the
moment of inertia of the cone about the vertical axis to be I0.
(a) What is the angular velocity of the cone when the block reaches
the bottom?
(b) Find the speed of the block in inertial space when it reaches the
bottom.
3. The attempt at a solution
My issue isn't really with solving the two parts because that part is quite straightforward. Just apply conservation of angular momentum about the axis of rotation for part (a) and then conservation of energy for part (b). My problem is more with this: since there is never a presence of a net torque about the axis of rotation, [itex]\partial _{t}L_{y} = 0[/itex] (taking the y axis to be the vertical axis of rotation) so, since the block and cone become one combined system much like in a perfectly inelastic collision, [itex]I_{0}w_{0} = (I_{0} + mR^{2})w_{f}[/itex] and we can easily solve for the final angular velocity of the combined system which is equivalent to the final angular velocity of the block. My question is simply: where does the angular velocity of the initial cone without the block go? At the end when the block reaches the ground there is obviously a loss in the angular velocity of the system from before the mass was placed on the rotating cone. I'm not sure where the angular velocity actually goes. Using conservation of energy I got that [itex]v_{f} = \sqrt{\frac{I_{0}}{m}(w_{0}^{2} - w_{f}^{2}) + 2gh}[/itex] and since [itex]w_{0} >= w_{f} >= 0[/itex], [itex]w_{0}^{2} >= w_{f}^{2} >= 0[/itex] so unless the rotation was zero to begin with (in which case vf is just sqrt(2gh)), there is a positive term that adds on to the 2gh term so am I right in saying that the loss in angular velocity simply gets added to the translational velocity of the block? It seemed too obvious so I don't know if there is some shady business here (=p). Thanks!
Phys.Org News Partner Science news on Phys.org
New type of solar concentrator desn't block the view
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Asian inventions dominate energy storage systems
mfb
#2
Dec11-12, 10:51 AM
Mentor
P: 11,815
and we can easily solve for the final angular velocity of the combined system which is equivalent to the final angular velocity of the block.
Why?

where does the angular velocity of the initial cone without the block go?
There is a torque acting between block and cone, slowing down the cone.

At the end when the block reaches the ground there is obviously a loss in the angular velocity of the system from before the mass was placed on the rotating cone.
Right, and its moment of inertia increased.
WannabeNewton
#3
Dec11-12, 11:45 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by mfb View Post
Why?
Because once we place the block on the already rotating cone, we have a single system consisting of the combined moments of inertia of the block and cone about the vertical axis and the reduced angular velocity of said combined system about the vertical axis. The block is rotating with the cone as one about said axis. Thanks for the reply.

mfb
#4
Dec11-12, 12:20 PM
Mentor
P: 11,815
Conceptual Problem in rotating cone problem

The block can rotate quicker or slower than the cone. Actually, it has to do this in order to get down.
WannabeNewton
#5
Dec11-12, 12:32 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by mfb View Post
The block can rotate quicker or slower than the cone. Actually, it has to do this in order to get down.
Wait do you mean set up the conservation of angular momentum like [itex]I_{0}w_{0} = I_{0}w_{f,cone} + mR^{2}w_{f,block} [/itex]? This is what I had initially done and it went with my intuition but I had to kind of force my mind to change my initial intuition because the solution itself took the final angular velocity of the block to be that of the system as well so I felt my initial logic was flawed somehow.
mfb
#6
Dec11-12, 04:42 PM
Mentor
P: 11,815
Maybe that solution assumed that the velocity of the block relative to the cone is negligible. In general, it is not.
WannabeNewton
#7
Dec11-12, 04:49 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by mfb View Post
Maybe that solution assumed that the velocity of the block relative to the cone is negligible. In general, it is not.
Weird because the question didn't state it. Now I have another variable to take care of however if that assumption is not made.
WannabeNewton
#8
Dec11-12, 05:20 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Looking again at how it wrote down the energy conservation equation, it totally disregarded the angular velocity of the block and considered only the new angular velocity of the cone after the block is placed on it (it included the translational velocity of the block of course) but I can see no justification for this based on the problem statement. Do you?
haruspex
#9
Dec11-12, 06:18 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,783
Quote Quote by WannabeNewton View Post
Looking again at how it wrote down the energy conservation equation, it totally disregarded the angular velocity of the block
Doesn't look that way to me. mvf2/2 includes the energy due to rotation, i.e. the horizontal component of its KE.
WannabeNewton
#10
Dec11-12, 06:24 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by haruspex View Post
Doesn't look that way to me. mvf2/2 includes the energy due to rotation, i.e. the horizontal component of its KE.
Not sure what you mean by horizontal component of KE since KE is a scalar but usually when one writes down conservation of energy for motion involving both translation and rotation, one includes the 1\2mv^2 term and the 1\2Iw^2 term separately for a given body massive object that is assuming the speed v the question is asking for is that of the orbital motion of the center of mass (the radial component that is but if they just want the overall speed then I agree with you). However this still doesn't address the part about the conservation of angular momentum where it just considers the angular velocity of the whole system.
haruspex
#11
Dec11-12, 06:40 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,783
Quote Quote by WannabeNewton View Post
Not sure what you mean by horizontal component of KE since KE is a scalar
If the horizontal and normal to horizontal components of velocity are vh, vn, then it is reasonable to describe the KE as having components mvh2/2 and mvn2/2, no?
energy for motion involving both translation and rotation, one includes the 1\2mv^2 term and the 1\2Iw^2 term separately for a given body massive object that is assuming the speed the question is asking for is the orbital motion of the center of mass.
It's asking for the speed. The mass is considered to be a point mass, so its 'rotational' KE about the cone's axis will be m(Rωf2)/2 = mvh2/2. I.e. it doesn't matter whether you consider vh to be a speed (transiently) about the cone's axis or just the horizontal component of a linear velocity, you get the same answers for its total KE and for its scalar speed.
WannabeNewton
#12
Dec11-12, 06:54 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by haruspex View Post
If the horizontal and normal to horizontal components of velocity are vh, vn, then it is reasonable to describe the KE as having components mvh2/2 and mvn2/2, no?
Well when you said components I assumed you were talking about an expansion in terms of basis vectors.
It's asking for the speed. The mass is considered to be a point mass, so its 'rotational' KE about the cone's axis will be m(Rωf2)/2 = mvh2/2. I.e. it doesn't matter whether you consider vh to be a speed (transiently) about the cone's axis or just the horizontal component of a linear velocity, you get the same answers for its total KE and for its scalar speed.
I have no problem with this at all if it is indeed asking for the speed itself. I thought it was asking for the radial component. Still unsure about the conservation of angular momentum part though.
haruspex
#13
Dec11-12, 07:02 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,783
Quote Quote by WannabeNewton View Post
I have no problem with this at all if it is indeed asking for the speed itself. I thought it was asking for the radial component.
In the OP it just says speed.
Still unsure about the conservation of angular momentum part though.
What's your doubt there?
WannabeNewton
#14
Dec11-12, 07:05 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by haruspex View Post
What's your doubt there?
As to why it assumes both the block and the cone will have the same final angular velocity (as in why the tangential velocity of the block along the cone will be the same as the angular velocity of the cone itself at the end). Thanks for all the help so far btw.
haruspex
#15
Dec11-12, 07:24 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,783
Quote Quote by WannabeNewton View Post
As to why it assumes both the block and the cone will have the same final angular velocity (as in why the tangential velocity of the block along the cone will be the same as the angular velocity of the cone itself at the end). Thanks for all the help so far btw.
The problem statement omits to say that the groove is a straight line. But you have to assume that to be able to solve it.
WannabeNewton
#16
Dec11-12, 07:34 PM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,562
Quote Quote by haruspex View Post
The problem statement omits to say that the groove is a straight line. But you have to assume that to be able to solve it.
Ah so in that case we can indeed treat the two objects as being one for the purposes of the final angular velocity since their tangential velocities will be directed along the same [itex]\hat{\theta }[/itex] if the block is constrained to a straight line?
haruspex
#17
Dec11-12, 08:33 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,783
Yep.


Register to reply

Related Discussions
Inertia Tensor - Rotating Cone Advanced Physics Homework 1
Cone rotating on the table Advanced Physics Homework 8
Rotating cone filled with water Introductory Physics Homework 2
Rolling Cone - (Rotating Coordinate systems) Advanced Physics Homework 0
Loudspeaker cone problem Introductory Physics Homework 3