# Kleppner:Mechanisc: 7.33 Grooved cone and mass

## Homework Statement

1. A cone of height h and base radius R is free to rotate around a fixed vertical axis. It has a thin groove cut in its surface. The cone is set rotating freely with angular speed ω0, and a small block of massm 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 around the vertical axis to beI0.
h is the height of the cone.

(a) What is the angular speed of the cone when the block reaches the bottom?

(b) Find the speed of the block in inertial space when it reaches the bottom.

Li = Lf

## The Attempt at a Solution

OK on this question for part a I understand that angular momentum is conserved
Li = I0*w0
the final angular momentum should it not be that of the cone plus the marble as the marble is moving downwards with a velocity v which can be calculated from mgh=1/2 m v2 and so
Lf = (I0 + mR2)*wf + m*R* √(2*g*h)*h/√(h2+R2)
the solution does not take into account the speed of the marble at the bottom IE: m*R* √(2*g*h)*h/√(h2+R2)
Last part is sinθ = h/√(h2+R2)
The problems for this chapter are really subtle and confusing me..

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haruspex
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the marble is moving downwards with a velocity v which can be calculated from mgh=1/2 m v2
Can you be more specfic about the direction of that component of its velocity?
Lf = (I0 + mR2)*wf + m*R* √(2*g*h)*h/√(h2+R2)
the solution does not take into account the speed of the marble at the bottom
Not all components of the block's velocity contribute to its angular momentum about the vertical axis.

Why not?
that component should be contributing please see image #### Attachments

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haruspex
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Why not?
that component should be contributing please see imageView attachment 228854
If $\vec v$ is the component in the vertical plane through the axis then $\vec r\times\vec v$ is horizontal. To contribute to the angular momentum about the vertical axis it would need to have a vertical component.

Can you be more specfic about the direction of that component of its velocity?

Not all components of the block's velocity contribute to its angular momentum about the vertical axis.
Why not? That component should be contributing. See image please.
If $\vec v$ is the component in the vertical plane through the axis then $\vec r\times\vec v$ is horizontal. To contribute to the angular momentum about the vertical axis it would need to have a vertical component.
I am sorry I don't understand r×v should be out of the page and hence contributes to L no?ℂ

haruspex
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I am sorry I don't understand r×v should be out of the page and hence contributes to L no?ℂ
Angular momentum about an axis is a vector along that axis.
A velocity out of the page would contribute to angular momentum about the vertical axis, but the component of the velocity vector that you are considering is through the vertical axis, so cannot contribute to angular momentum about it.

Angular momentum about an axis is a vector along that axis.
A velocity out of the page would contribute to angular momentum about the vertical axis, but the component of the velocity vector that you are considering is through the vertical axis, so cannot contribute to angular momentum about it.
Thanks. Yes I completely understand but the velocity is not vertically down it is at an angle as in image and the cross product spans an inverted cone with a horizontal top circle as the cone in question is spinning ie it has a component of angular momentum along that vertical axis that adds up as cone rotates. What am I missing?

haruspex
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it has a component of angular momentum along that vertical axis
Indeed, but that component is not given by the v you calculated from mgh.

Think about the cone+block system. What forces act on it? Which if any have a torque about the axis?

Indeed, but that component is not given by the v you calculated from mgh.

Think about the cone+block system. What forces act on it? Which if any have a torque about the axis?
OK thank you very much. I got it. I wrote the vectors of r and v in cylindrical coordinates and indeed the cross product generates zero momentum along the vertical axis. You are right no external toques anyways act on the system so angular momentum has to be conserved vertically but not necessarily in any direction. Am I right.

haruspex
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OK thank you very much. I got it. I wrote the vectors of r and v in cylindrical coordinates and indeed the cross product generates zero momentum along the vertical axis. You are right no external toques anyways act on the system so angular momentum has to be conserved vertically but not necessarily in any direction. Am I right.
Nearly.
Two external forces on the system are gravity and a vertical force opposing gravity. But those combine to produce a torque tending to topple the cone. So there is also a torque from the axle to keep it upright. But none of those have a torque component about the vertical axis.

Nearly.
Two external forces on the system are gravity and a vertical force opposing gravity. But those combine to produce a torque tending to topple the cone. So there is also a torque from the axle to keep it upright. But none of those have a torque component about the vertical axis.
Gravitation of cone and cube got it
what is the Force opposing gravity?
what is the source of torque from axle?

haruspex
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Gravitation of cone and cube got it
what is the Force opposing gravity?
what is the source of torque from axle?
The cone must be mounted on an axle to keep it in place. The forces from that will be whatever is required to keep the centre line where it is, in altitude, position and orientation.
Since the block and cone (maybe) have weight, this must include a vertical force to stop the cone moving down. It will be more than the weight of the cone but less than the combined weight, since the block is accelerating.
This upward force will be in the same line as the weight of the cone, but the block is pressing down on one side, resulting in a torque. So the axle must provide a countertorque. This will be horizontal and normal to the plane containing the axle and the block.
Also, the block is pushing sideways on the cone, so the axle must provide an opposing horizontal force.
Since the cone is free to turn on the axle, the axle cannot provide any force or torque that has a moment about itself.

The cone must be mounted on an axle to keep it in place. The forces from that will be whatever is required to keep the centre line where it is, in altitude, position and orientation.
Since the block and cone (maybe) have weight, this must include a vertical force to stop the cone moving down. It will be more than the weight of the cone but less than the combined weight, since the block is accelerating.
This upward force will be in the same line as the weight of the cone, but the block is pressing down on one side, resulting in a torque. So the axle must provide a countertorque. This will be horizontal and normal to the plane containing the axle and the block.
Also, the block is pushing sideways on the cone, so the axle must provide an opposing horizontal force.
Since the cone is free to turn on the axle, the axle cannot provide any force or torque that has a moment about itself.
Ok got it thanks very much for this thorough and succinct explanation.