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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 $ω_0$, 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 around the vertical axis to be $I_0$. $h$ 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.

In (a) it is trivial that the angular momentum of the system is conserved because there is no torque along the vertical axis ( but there is torque due to gravity in the perpendicular of vertical axis)! So I use coservation of angular momentum to get the final angular velocity of both the block and cone!

But in (b) the problem is arrising : at first based on my reasoning when the block reaches bottom it has two velocity components , (1) due to the final angular velocity of the combined system and (2) due gravitation that is $\sqrt {2gh}$ along the the groove! So the net velocity should be combined of these two velocity!

But on the other hand in a second reasoning it can be said that the total energy of the system is conserved because there is dissipative force at all! So we can use conservation of energy to get the final velocity. When solving this equation I am getting different result!

And in the solution the second way is preferd; so I really can not understand what is wrong with my first reasoning?

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  • $\begingroup$ It is very difficult to identify where you have gone wrong unless you give more details of your reasoning. I have one idea - since the cone is rotating it is not obvious that the force exerted by the cone on the block will be perpendicular to the block's velocity. If not, then the cone is doing work on the block and the block's final velocity will not be $\sqrt{2gh}$. $\endgroup$ – gandalf61 Sep 22 '20 at 10:12
  • $\begingroup$ @gandalf61 yes I also think that but I really can't find the force which is doing work due to which kinetic energy along the groove is changing! $\endgroup$ – PRITAM the cat of Newton Sep 22 '20 at 10:23
  • $\begingroup$ There is a contact force between the cone and the block which is what keeps the block in its groove. If this contact force is not perpendicular to the block's velocity then it will do work on the block and so affect the block's energy. However, if you use conservation of energy for the total system (cone and block together) then the work done by the contact force nets to zero (because it is an internal force) so you can ignore it. $\endgroup$ – gandalf61 Sep 22 '20 at 11:04
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Your assumption that the final velocity along the groove is $\sqrt {2gh}$ is incorrect. If you analyze in the frame of reference of the cone, you'll see that the centrifugal force is also doing some work on the block, along with gravity. Analyzing it in an inertial frame gives the same result (obviously), but in a more roundabout way. In any case, you'll have to use calculus to solve it this way as the centrifugal force is constantly varying. Instead of going through all this, using conservation of energy is the cleanest method.

NOTE: In a comment on your post, you said that you couldn't find a force other than gravity which could increase the kinetic energy along the groove. As the cone is rotating, the direction along which you are applying the work kinetic theorem, i.e. the 'along the groove' direction, is constantly changing. In fact it is rotating, which means that your frame of reference is non-inertial. This is why you have consider a pseudo force i.e. the centrifugal force, while writing $\Delta K = F \cdot ds$

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  • $\begingroup$ Kinetic energy = (1/2)m$v^2$ = (1/2)m${v_x}^2$ + (1/2)m${v_y}^2$. It looks like you can break kinetic energy into components. $\endgroup$ – R.W. Bird Sep 22 '20 at 15:40
  • $\begingroup$ Yeah sorry, my last statement was incorrect, I have removed it. $\endgroup$ – dnaik Sep 22 '20 at 15:51
  • $\begingroup$ Thanks! Dude! Now I am explaining it (with the help of your idea) in inertial frame: as the block is in a circular motion at every instant there must be some sort of force that is producing centrepetal force and that force is doing work on the block![correct me if I am wrong:) ] $\endgroup$ – PRITAM the cat of Newton Sep 23 '20 at 15:05
  • $\begingroup$ The analysis is correct, but your frame still appears to be non-inertial. Considering an inertial frame isn't really the way to go here, as it will lead to further complicated equations. Explaining in the inertial frame with text without diagrams is very difficult, but i will attempt it. (continuing it another answer because of comment character limit) $\endgroup$ – dnaik Sep 23 '20 at 16:27
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An inertial frame means that even though the cone is rotating continuously, the direction you consider will be stationary (wrt ground). Which means that if you pick the direction along the groove at one time, at the next instant, the groove will move on, but your line of reference is still the same. So the components of the various forces, accelerations and the velocity, along the direction which you are considering, are all changing continuously. I'm not going to attempt to solve it, but I'm pretty sure it will involve unnecessarily difficult calculus. You can certainly try to visualize what is happening from an inertial frame, but trying to actually write the equations is pointless.

Note: The above is for applying the work kinetic energy theorem along a particular direction. If you want to write newtons equations for the block in an inertial frame you can easily do so, and solve it without using any calculus. Using conservation of energy and momentum is still the easiest, however.

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  • $\begingroup$ Yeah! From now I will try to apply energy conservation rather solving each and every equation of motion for rigid body motion (and because most of the time systems are not defined properly in many problem of kleppner; that's also makes it difficult to think what actually it is trying to say about! And it is also hard to identify all internal forces in that system! So it will be better to use conservation laws because internal forces and work (for rigid body) adds to give zero!!😁 $\endgroup$ – PRITAM the cat of Newton Sep 23 '20 at 16:42

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