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I'm confused about energy of a body in circular motion. In particular I'm having trouble to find the correct answer to this question.

Consider the body in the picture that is set in motion from rest. How much energy is needed to set the body in motion? What energy does the body in motion own?

enter image description here

In my answer I would surely include the kinetic energy that the body owns once in motion, that is

$K=\frac{1}{2} m v^2=\frac{1}{2} m (\omega r)^2$

(Where $r$ is the radius of the circular trajectory of the body in motion)

But is that all? I mean the body has somehow rise a bit from the previous position, therefeore it should own potential energy too, is that correct?

$U_{grav}=mg h$

(Where $h$ is the height of the body in motion with respect to the previous position)

Finally should also centrifugal potential energy be included? That would be

$U_{centrifugal} = -\int_{0}^{r} F_{centrifugal} dx= -\int_{0}^{r} m \omega^2 x dx=-\frac{1}{2} m (\omega r)^2$

So adding up all the pieces

$E_{TOT}= K+U_{grav}+U_{centrifugal}=U_{grav} =m g h$

But I'm not convinced of what I tried, am I missing something important?

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  • $\begingroup$ Not sure why you've made the centrifugal kinetic energy negative - K.E. is always positive (or zero). And I suspect that the question wants you to put the answer in terms of the quantities in the diagram, so you'd need to express $h$ and $r$ in terms of $L$ and $d$. $\endgroup$ – PhillS Feb 23 '16 at 13:10
  • $\begingroup$ @PhillS To express everyrthing in terms of $L$ and $d$ is not a problem, I'm only confused about what energies are involved. I' m sorry I meant centrifugal potential energy, which must be negative because centrifugal force does positive work when the distance from the center, $x$ is increasing, so the potential energy must become less $\endgroup$ – Gianolepo Feb 23 '16 at 13:29
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    $\begingroup$ I misread your post slightly. I think you are mixing up different reference frames. If you are looking in the non-rotating reference frame, you only need to worry about the rotational kinetic energy and the gravitational potential energy (there is no centrifugal force in that frame). If you work in a frame rotating with the mass, there is no kinetic energy (the object isn't moving in that frame), but there is gravitational potential energy and the centrifugal potential energy. By including both $K$ and $U_{centr}$ you are counting the same thing twice. $\endgroup$ – PhillS Feb 23 '16 at 13:45
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You don´t have to consider fictitious forces if you set an inertial frame of reference. Suppose the rod at which the mass is tied to by the rope is the z-axis, with z=0 at the lower extreme and raising from bottom to top, and the other x and y axis are in the plane perpendicular to the z-axis, no matter the specific location inside it because the problem has symmetry about z-axis.

It´s obvious that the mass has less energy when idle than when twirling, just because both the potential and kinetic energy have increased but, how much?

Well, if you locate the "zero level" of the potential at z=0 and you consider gravity to be constant, the potential energy is $V_{1}=mg(L-d)$, and mechanic energy too, because the mass is iddle and kinetic energy is zero $T_{1}=0$.

When twirling at the given height, you have that kinetic and potential energy are:

$T_{2}=\dfrac{1}{2}m\omega^{2}r^{2}=\dfrac{1}{2}m\omega^{2}(d^{2}-\dfrac{L^{2}}{4})$

$V_{2}=mg\dfrac{L}{2}$

$r$ means the same than in your exposition, computing its value only requires to apply pythagorean theorem.

With this, the problem is practically solved. The energy of the mass when twirling is:

$E_{2}=T_{2}+V_{2}=\dfrac{1}{2}m\omega^{2}(d^{2}-\dfrac{L^{2}}{4})+mg\dfrac{L}{2}$

And the energy increment is:

$\Delta E=E_{2}-E_{1}=E_{2}-V_{1}=\dfrac{1}{2}m\omega^{2}(d^{2}-\dfrac{L^{2}}{4})-mg\dfrac{L}{2}+mgd$

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  • $\begingroup$ So, reading again your question, you don´t have to include "centrifugal potential energy" terms or things like that if you consider an inertial system as I´ve said before (Well, it´s not exactly inertial, but the aproximation is fine). If you calculate the energy from a non inertial frame, remember that the energy will change, because velocity vector is different. In fact, you have computed the total energy from a frame of reference with same z-axis as mine but x y axis twirling with the mass, because the velocity is zero from this frame. $\endgroup$ – amejmar Feb 23 '16 at 15:35
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I think one important point is missing -the observer who is measuring the energy of a rotating body - the person is on earth an approximate Inertial frame and he has devised a physical toy by which he can rotate a body of mass m in a circular path of radius r .

therefore the rotational kinetic energy will be there. If the body was earlier at a point below the plane of rotation say length h .then gravitational potential energy m.g.h is to be added to to the above . And no third type of energy ,which is being termed as centrifugal potential energy should be there as for our observer he does not experience any centrifugal force acting on the body- apart from centripetal acceleration provided by him through the string to maintain the rotational motion at constant speed. However for another observer sitting on the body i.e. in non-inertial frame he can measure the kinetic energy of the body which will be motion of the body going radially outward .

the experimenter can pull the string to increase centripetal force and that work done can lead to increase in kinetic energy.

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