I'm calculating more energy out than I put in

[jbriggs444]

I think I'll leave it the way it is.

OK. I will try to follow your analysis and see where it goes wrong.

The force being constant produces a constant translational acceleration a = F/M and a constant rotational acceleration α. = L / I where L (torque) =F⋅R and I is the moment of inertia. The work required to do this has two parts one rotating it and one moving the center of mass horizontally. The force is applied for a distance of its length x causing the sphere to rotate θ radians where X=R⋅θ.

You appear to be assuming that although there is rolling without slipping that the frictional force of the sphere on the platform imparts no momentum, provides no torque and does no work. That is, that the frictional force turns out to be zero.

I disagree with this, but will accept it for the purposes of argument.

Let us go ahead and apply the force $F$. The claim is that the resulting acceleration is $a = F/M$.

We can go ahead and apply the torque as well. The claim is that the resulting angular acceleration $\alpha = \frac{L}{I} = \frac{F \cdot R}{I}$. The moment of inertia of a sphere is $\frac{2}{5}MR^2$. So we end up with $\alpha = \frac{5F}{2MR}$. If we multiply by $R$ to convert this angular acceleration rate to a rate of horizontal acceleration, we get $a = \frac{5}{2}F/M$.

The angular acceleration is 2.5 times too fast. We would not actually maintain rolling without slipping under these conditions.

Non-zero friction with the platform has to provide enough retarding torque and advancing horizontal force so that the rotational acceleration and the horizontal acceleration match.

 

[A.T.]

The sphere moves without slipping horizontally on a flat surface as shown below.

Note that the OP doesn't specify this. And as @jbriggs444 points out, you cannot assume zero friction for accelerated rolling without slipping in general.

 

[gleem]

You appear to be assuming that although there is rolling without slipping that the frictional force of the sphere on the platform imparts no momentum, provides no torque and does no work. That is, that the frictional force turns out to be zero.

I disagree with this, but will accept it for the purposes of argument.

The OP did not specify a coefficient of friction so I think my assumption of not slipping is justified. If the frictional force is zero, there would be no rotation, so there must be friction. The frictional force is not moving so it does no work (neglecting rolling friction) and does not affect the final kinetic energy.

My post was to demonstrate that the work done was equal to the final kinetic energy. that is

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FaX + L\alpha =FaX+I\alpha \theta $$
$$v^{2}=2aX $$ and $$ \omega^{2} =2\alpha \theta$$
$$KE_{final}=maX+I\alpha \theta =mXv^{2}/2X +I\theta \omega ^{2}/2\theta $$




Now I did find this analysis surprisingly simple since when I considered using V² = 2Xa, I could almost see where it was going to take me.

 

[PeterDonis]

@gleem please use the LaTeX feature for equations. Your equations are not very readable.

 

[jbriggs444]

The OP did not specify a coefficient of friction so I think my assumption of not slipping is justified. If the frictional force is zero, there would be no rotation, so there must be friction. The frictional force is not moving so it does no work (neglecting rolling friction) and does not affect the final kinetic energy.

My post was to demonstrate that the work done was equal to the final kinetic energy. that is

½ mV² +½ Iω² = FX + Lθ = maX + Iαθ

What is $F$ here? The applied force? If so, that is incorrect.
What is $L$ here? The torque from the applied force? If so, that is incorrect.

 

[gleem]

What is F here? The applied force? If so, that is incorrect.
What is L here? The torque from the applied force? If so, that is incorrect.

Yes to both. Perhaps I was sucked into an error by my approach. Can you see it?

 

[jbriggs444]

Yes to both. Perhaps I was sucked into an error by my approach. Can you see it?

$F \ne ma$
$\sum F = ma$

One can use an energy approach. But then should treat $a$ and $\alpha$ as unknowns to be determined.

You have $F$ (the applied force) and $X$ (the displacement of the center of mass) to work with. And you have the assumption of rolling without slipping.

If the center of mass is displaced by $X$ then the cumulative distance moved by the rod that is supplying force $F$ is $2X$. The mechanical work supplied by $F$ is then obviously: $W=2\vec{F} \cdot \vec{X}$.

If you like, you can interpret the work done as center of mass work $W_\text{com} = FX$ and mechanical work in the center of mass frame $W_\text{mech} = \tau \theta$. Torque $\tau$ is given by $FR$ while rotation angle $\theta$ is given by $\frac{X}{R}.$ It follows that $W_\text{mech} = FX$ and by no coincidence, $W_\text{tot} = W_\text{com} + W_\text{mech} = FX + FX = 2FX$

By the work-energy theorem, the final kinetic energy is equal to the initial kinetic energy plus the work done:$KE_\text{final} = 2FX$.

Final kinetic energy is also given by $\frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$

But rotation rate $\omega$ is $\frac{v}{R}$ and moment of inertia $I$ is $\frac{2}{5}mR^2$. So we can substitute in and rewrite the previous equation as $\frac{1}{2}mv^2 + \frac{1}{2} \frac{2}{5}mv^2$ for a total of: $$KE = 2FX = 0.7 mv^2$$We can solve for $v$ and get:$$v = \sqrt{\frac{20FX}{7m}}$$Given the total distance covered and the final velocity we can do the boring work of determining the horizontal acceleration $a$ and the rotational acceleration $\alpha$.

Let us go ahead and do that. Elapsed time $t$ is given by distance $X$ divided by average velocity $\frac{v}{2}$. That gives us:$$t = \frac{X}{0.5\sqrt{\frac{20FX}{7m}}} = \sqrt{\frac{7mX}{5F}}$$Acceleration is then given by the change in velocity divided by the elapsed time. That gives us:$$a = \frac{v}{t} = \frac{\sqrt{\frac{20FX}{7m}}}{\sqrt{\frac{7mX}{5F}}} = \frac{10F}{7m}$$More directly, we could have simply used the fact that the "effective mass" of a rolling sphere is $\frac{7}{5} = 1.4$ times its regular mass and the fact that we have a 2:1 mechanical advantage.

Edit: Massive edit to add the equations.

 

[gleem]

This is my thought. The forces are the applied force and the frictional force I assumed. But the frictional force does no work so it was not included in the equation for work. One might think the frictional force might contribute to the rotational kinetic energy. But consider this if the sphere were in free space the applied tangential force would cause a rotation which if placed in contact with a frictional stationary surface underneath would not change the motion, would it? So no additional torques would be present.

Now if the force were applied to the center of the sphere there would be no rotation until it was in contact with a stationary surface which would produce a torque.

 

[jbriggs444]

This is my thought. The forces are the applied force and the frictional force I assumed. But the frictional force does no work so it was not included in the equation for work. One might think the frictional force might contribute to the rotational kinetic energy. But consider this if the sphere were in free space the applied tangential force would cause a rotation which if placed in contact with a frictional stationary surface underneath would not change the motion, would it? So no additional torques would be present.

Now if the force were applied to the center of the sphere there would be no rotation until it was in contact with a stationary surface which would produce a torque.

See the work editted into #77 above. Both acceleration $a$ and angular acceleration $\alpha$ have contributions from the frictional force on the bottom. Although this does not contribute to energy, it does contribute to acceleration.

Think about it this way if you like... The frictional force on the bottom takes away from horizontal acceleration and adds to rotational acceleration. There is no net change in energy since this force does zero work. But it does change the balance between the kinetic energy of linear motion and the kinetic energy of rotation.

 

[gleem]

If the center of mass is displaced by X then the cumulative distance moved by the rod that is supplying force F is 2X.

I just rolled a cylinder without slipping with a ruler on top as in the OP. The center of the cylinder moved the same distance as the ruler.

 

[jbriggs444]

I just rolled a cylinder without slipping with a ruler on top as in the OP. The center of the cylinder moved the same distance as the ruler.

What, exactly, did you measure?

If you measured the forward progress of the center of the cylinder and separately measured the forward progress of one end of the ruler then you should clearly witness a 2:1 ratio, ruler:cylinder.

If, however, you measured the forward progress of the center of the cylinder and separately measured the rearward progress of the point of contact where the cylinder rolls on the ruler, you should clearly witness a 1:1 ratio. (The cylinder rolls forward 1 unit, the ruler forward by 2 units and the delta covered by the point of contact relative to the ruler is -1 unit).