If an ice skater is spinning on a frictionless surface

[shortydeb]

..and he brings in his arms to decrease his moment of inertia in order to increase his angular velocity, that means the rotational kinetic energy increases. But that means there's a change in rotational kinetic energy, right? But wouldn't there be no external forces acting on him (assuming no air resistance or friction from the ice), so how come energy isn't conserved? Since the gravitational potential energy doesn't change, and there is no change in translational kinetic energy (though i am really not sure on these two). It looks like the final total mechanical energy is different from the initial total mechanical energy.

Any help at all would be terrific...

 

[jablonsky27]

of course the energy is conserved!
the rotational kinetic energy is I.(w^2)/2
when I reduces, w increases correspondingly to maintain the above and vice-versa..

 

[robertm]

Exactly. Here it is a little more in depth:

When the arms are brought inward we know what is changing: Moment of inertia and Angular Speed

So to show that E is conserved we must look at the equation for rotating E:

KErot = 1/2 * moment of inertia * (angular speed)^2

Now since we know that his A.speed increased we know that his MoI must have decreased by an amount so that if the A.speed doubled then the MoI must have decreased by a factor of 1/4:

KE = [2*(2^2)]/2 = 4

Double the speed:

KE = [1/2*(4^2)]/2 = 4

So 2 (the MoI in the first equation) is multiplied by 1/4 to equal 1/2 when the A.speed is doubled (and squared) and so the total Rotational Kenetic E. is conserved.

 

[Doc Al]

Mechanical energy is not conserved!

It looks like the final total mechanical energy is different from the initial total mechanical energy.

It is different. Mechanical energy is not conserved! (But angular momentum is conserved.) The skater must do work in order to bring his arms in towards his body. Essentially what is happening is that internal energy--chemical energy in the skater's muscles--is converted into mechanical energy.

Just because all the forces are internal does not imply that rotational kinetic energy cannot change. It does imply that there's no external torque on the system and thus angular momentum is conserved. It also implies that the translational momentum and translational KE of the center of mass does not change.

 

[Danger]

In reference to the OP, I just want to mention that one could not actually skate on frictionless surface.

 

[shortydeb]

thanks robertm and jablonsky27, i forgot to take into account that the moment of inertia would decrease since the radius of the arm mass would be much less...


so if the skater has to do work to bring in his arms, that would mean the force he exerts to do so is nonconservative? My book has the equation W (by nonconservative forces) = (change in KE) + (change in Rot. KE) + (change in PE) . I'm not sure if the nonconservative forces have to be external or not. Would the force the skater exerts to bring in his arms be internal? (Thanks btw!)

 

[rohanprabhu]

Would the force the skater exerts to bring in his arms be internal? (Thanks btw!)

Yes, the force is internal. Whether a force is internal or not has nothing to do with it being conservative or non-conservative. A force is internal or external depending on the system you consider.

Take the case of a bullet hitting a wooden block. If you consider your bullet as a system, then any force exerted by any other object than the bullet is external. Since the wooden block is not a part of our system, the force exerted by the wooden block is external and hence, there is a net external force on our system and the momentum of our system i.e. the bullet will change. Now, if we consider the bullet + the block to be our system, then the force exerted by the block on the bullet is internal. Hence there is no external force and momentum will not change. This is due to the fact that, if the bullet exerts a force on the block, then the block will exert a equal and opposite force on the bullet as well. Hence, there is no net change in momentum. But here, the momentum of the bullet + block system will remain same, not the bullet or the block alone.

A non-conservative force is a force which does no work in a cycle. Take the case of gravitation. If we take a stone to a height 'h', it's energy will be 'U' [let us take the GPE at h = 0 to be 0]. When it is dropped and it reaches h = 0, just before hitting the ground it's KE will be 'U'. We can again take the stone to a height 'h' and the same amount of energy will build up. As such, we can reclaim the energy in a sense.

Take the case of frictional heating. The energy is not 'destroyed' in any sense, but if we take a block on a frictional surface from s = 0, to s = d then back to s = 0, the work done in this case is twice as that for one trip. Hence, the work done in a cycle is not zero and hence it is not a conservative force.

Also, conservation of momentum always applies in all cases no matter the force being conservative or non-conservative as it follows from Newton's Third Law.

 

[jablonsky27]

A non-conservative force is a force which does no work in a cycle. Take the case of gravitation. If we take a stone to a height 'h', it's energy will be 'U' [let us take the GPE at h = 0 to be 0]. When it is dropped and it reaches h = 0, just before hitting the ground it's KE will be 'U'. We can again take the stone to a height 'h' and the same amount of energy will build up. As such, we can reclaim the energy in a sense.

its the other way around - conservative forces do no work around a closed loop. or alternatively, the work done by a conservative force is independant of the path.

Take the case of frictional heating. The energy is not 'destroyed' in any sense, but if we take a block on a frictional surface from s = 0, to s = d then back to s = 0, the work done in this case is twice as that for one trip. Hence, the work done in a cycle is not zero and hence it is not a conservative force.

ya, correct. i guess the previous thing was a typos.

so if the skater has to do work to bring in his arms, that would mean the force he exerts to do so is nonconservative? My book has the equation W (by nonconservative forces) = (change in KE) + (change in Rot. KE) + (change in PE) . I'm not sure if the nonconservative forces have to be external or not. Would the force the skater exerts to bring in his arms be internal? (Thanks btw!)

no, it the internal force he exerts is not nonconservative. in general, the work done on a system is given by W = change in its mechanical energy + change in its thermal energy + change in its internal energy.