Does the Work-Energy Theorem hold true for objects in rotational motion?

[JSGandora]

Consider a solid sphere and a cube of equal mass, both on a frictionless table. Now, you apply a force to both objects at the point of contact between the object and the table. Then the linear accelerations of both objects will be the same (since the same force is applied to the two objects of equal mass). Therefore, at some point in time, the velocities of both objects will be the same so their translational kinetic energies will also be the same.

Additionally, since you are applying a nonzero torque to the sphere, the sphere will rotate and therefore have a nonzero rotational kinetic energy at the same point in time. Therefore, the total kinetic energy of the sphere is greater than the total kinetic energy of the cube (since it does not rotate).

However, by the Work-Energy Theorem, shouldn't both objects have the same total kinetic energy since they both traverse the same distance (same accelerations) and both have the same force applied to them? Can someone explain this apparent paradox?

 

[jbriggs444]

The work done on the sphere by the force is equal to the product of the force times the distance moved by the point of contact. If the sphere is allowed to rotate as a result of the applied force, the distance moved by the point of contact will be greater than the distance moved by the sphere.

Strictly speaking, it's not quite a product. It's an integral of the instantaneous force applied times the incremental distance moved.

 

[dauto]

The work applied on the ball is in fact larger because even though the two objects move at the same speed the ball rotates while the cube doesn't which causes the point where the force is applied to move faster (compounded translation plus rotation). The formula for the power P = Fv uses the speed v of the point of contact, not the speed of the center of mass.

 

[pikpobedy]

There is no paradox.
You started off with the erroneous premise that the force is the same. The force on the sphere is greater to obtain the results that you stated.

A common junior physcs problem is as follows:
Consider two identical spheres. One rolls down an inclined plane that has a typical coefficient of friction, say 0.3 so there is no slippage. The other slides down a frictionless inclined plane of the same angle.
The rolling sphere will be slower.

 

[JSGandora]

Thanks jbriggs444 and dauto! That makes sense. After some algebra I see that $W=\frac{F^2 t^2}{2\beta m}$ while translational kinetic energy is $\frac{F^2 t^2}{2m}$.

There is no paradox.
You started off with the erroneous premise that the force is the same. The force on the sphere is greater to obtain the results that you stated.

I think you misunderstood my question. The point is that the force applied on both objects is the same; though, in order to obtain the same total kinetic energy in both the sphere and the cube, a smaller force must be applied to the sphere for the same amount of time (by a factor of $\sqrt{\beta}$).

 

[pikpobedy]

OK
More work was done on the sphere. Jbriggs and dauto are fine.

 

[Fantasist]

Consider a solid sphere and a cube of equal mass, both on a frictionless table. Now, you apply a force to both objects at the point of contact between the object and the table.

Maybe I am missing something, but on a frictionless table there shouldn't be any force at the contact point with the object other than the vertical gravitational force (for a cube it would be a contact surface anyway).

 

[JSGandora]

I mean the force is applied, not naturally there. For example, a string is attached to the bottom of the cube and you pull the string parallel to the table. You apply that force yourself in some way.

 

[Fantasist]

It all depends how you apply the force at the sphere. If you apply it radially (vertically to the surface) it will just translate, if you apply it tangentially it will just rotate. In general you have a mixture between the two, with the total energy the same in all cases.

 

[jbriggs444]

It all depends how you apply the force at the sphere. If you apply it radially (vertically to the surface) it will just translate, if you apply it tangentially it will just rotate. In general you have a mixture between the two, with the total energy the same in all cases.

If you apply force and the sphere is not restrained, it will always translate. This remains true, even if the force is applied tangent to the sphere. Newton's second law still applies. F=ma

The total energy is not the same in all cases, as has been explained already.

 

[sophiecentaur]

There are no paradoxes in the ideal world of School Mechanics.

 

[A.T.]

Related Puzzle:

https://www.youtube.com/watch?v=vWVZ6APXM4w

 

[russ_watters]

Note that the OP says nothing about the distances through which the force is applied, so it must be making the erroneous assumption that they are equal.

 

[voko]

Related Puzzle:

https://www.youtube.com/watch?v=vWVZ6APXM4w

The puzzle is not that interesting; it is interesting why this is a puzzle to begin with. The answer should matter a lot to teachers of physics.

 

[sophiecentaur]

The puzzle is not that interesting; it is interesting why this is a puzzle to begin with. The answer should matter a lot to teachers of physics.

It may be interesting to you but it's no surprise to me that people who do not 'know' Physics come to the most amazing conclusions about the way the World works. Just read a few of the posts we get on PF - and they are from people who are, at least. a bit interested in the topic, even if they may not know a lot.

 

[sophiecentaur]

They do not show you close-ups of the bullet holes.. . . . . .

 

[voko]

It may be interesting to you but it's no surprise to me that people who do not 'know' Physics come to the most amazing conclusions about the way the World works.

I understood the puzzle was being offered to some fellow maintainers of science related channels. What was remarkable is that they were easily going to consider conservation of energy, but not of momentum.

 

[Fantasist]

If you apply force and the sphere is not restrained, it will always translate. This remains true, even if the force is applied tangent to the sphere. Newton's second law still applies. F=ma

The total energy is not the same in all cases, as has been explained already.

If the force is applied to a fixed point it has to rotate with the sphere in order to stay tangential. So there can not be a net translation.

As for F=ma, let me invert your argument: if there is no force (but only a torque), then there is no acceleration.

 

[Dale]

If the force is applied to a fixed point it has to rotate with the sphere in order to stay tangential. So there can not be a net translation.

If there is a net force then it will translate, regardless of whether the force is tangential or not and regardless of whether it is applied to a fixed point or not.

 

[Dale]

Newton's second law still applies. F=ma

Newtons second law is better stated as ∑F=ma. Otherwise you invite comments like Fantasist's.

 

[Fantasist]

If there is a net force then it will translate, regardless of whether the force is tangential or not and regardless of whether it is applied to a fixed point or not.

There is no net force if it is tangential at all times to a fixed point on a rotating sphere.

 

[sophiecentaur]

There is no net force if it is tangential at all times to a fixed point on a rotating sphere.

"At all times"? It is instantaneous and at one point at a time.

 

[pikpobedy]

There is no net force if it is tangential at all times to a fixed point on a rotating sphere.

Are you are referring to a rotary lawn sprinkler or Hero's steam turbine type of gadget.

 

[Dale]

There is no net force if it is tangential at all times to a fixed point on a rotating sphere.

This doesn't follow. Consider a spherical spaceship ("That's not a moon ...") located far from any gravitating bodies and having an exhaust nozzle pointed tangent to the surface. The force would be tangential at all times to a fixed point on a rotating sphere and yet there would be a net force at all times causing an acceleration of the center of mass of the ship.

Again, if there is a net force then it will translate, regardless of whether the force is tangential or not and regardless of whether it is applied to a fixed point or not. The details like "tangential" and "fixed point" are simply irrelevant for the correct application of Newton's 2nd law.

 

[A.T.]

There is no net force if it is tangential at all times to a fixed point on a rotating sphere.

Of course there is a net force, if a single tangential force is applied to sphere. If the tangential force roates with the sphere, so does the net force. Then the translation of the mass center is not linear, but a curved path.

 

[Fantasist]

"At all times"? It is instantaneous and at one point at a time.

Even if it instantaneous, I can not see how an impulse transfer at a point where the normal to the surface is vertical to the direction to the center of mass should lead to a translational motion (in contrast, a perfect sphere always has a normal in direction of the center of mass, so it will never rotate but always translate on interaction with other bodies (in the absence of friction)).

Try it e.g. with a pencil at home. Put it on a smooth table and hit it at one end. It will practically only rotate (at least it will stay on the table). But hit it with the same force in the center, and it will fly through the room.

 

[Fantasist]

This doesn't follow. Consider a spherical spaceship ("That's not a moon ...") located far from any gravitating bodies and having an exhaust nozzle pointed tangent to the surface. The force would be tangential at all times to a fixed point on a rotating sphere and yet there would be a net force at all times causing an acceleration of the center of mass of the ship.

Any reference to support this claim (I mean experimental data)?

Again, if there is a net force then it will translate, regardless of whether the force is tangential or not and regardless of whether it is applied to a fixed point or not. The details like "tangential" and "fixed point" are simply irrelevant for the correct application of Newton's 2nd law.

There seems to be a kind of circular argument in place here: from a kinematical point of view (according to F=ma), you know the force only if you know the acceleration. But the acceleration is the unknown here (we are trying to figure out whether the sphere will translate or not in a given situation)

 

[Doc Al]

Any reference to support this claim (I mean experimental data)?



There seems to be a kind of circular argument in place here: from a kinematical point of view (according to F=ma), you know the force only if you know the acceleration. But the acceleration is the unknown here (we are trying to figure out whether the sphere will translate or not in a given situation)

This is basic Newtonian physics. Nothing controversial here.

If there is a net force on an object, then the center of mass of that object will accelerate according to Newton's 2nd law.

 

[Fantasist]

If there is a net force on an object, then the center of mass of that object will accelerate according to Newton's 2nd law.

The problem is to define the force.

 

[Doc Al]

The problem is to define the force.

What do you mean?

 

[Dale]

Any reference to support this claim (I mean experimental data)?

This is standard material in any introductory statics class. My textbook was Meriam and Kraige, Statics, V1, 3rd Ed, p. 52-53, but I cannot imagine that any statics text would skip such a basic concept.

As far as experiments go, you can do this one yourself. Just get an air hockey table, apply a tangential force to the puck, and note that the puck translates as well as rotates.

There seems to be a kind of circular argument in place here: from a kinematical point of view (according to F=ma), you know the force only if you know the acceleration. But the acceleration is the unknown here (we are trying to figure out whether the sphere will translate or not in a given situation)

According to Newton and everyone since him, the puck will translate and rotate due to an isolated tangential force. According to you it will only rotate.

I fail to see the circular argument, other than the fact that you argue incessantly about anything involving circular motion.