I I'm calculating more energy out than I put in

[kuruman]

@Dale is just stating the work energy theorem for the scenario in question. At any time, the work done (which must come from whatever external energy source is providing the applied force) must equal the kinetic energy produced (as you note, there is no potential energy so the kinetic energy produced is the total energy produced).

Thanks. That clarifies it.

 

[A.T.]

Most of the time when I ask about it online I get the following derivation of the power force velocity equation $work = force * displacement$ and we divide both sides by time and we get $power = force * velocity$

Also what is the time in the formula, is the time equal to the time the force is applied and velocity is a function of time that increases as the force is applied?

Rather than "dividing by time" you take the derivative w.r.t. time on both sides. You only need to specify a time interval if you want to compute the work done (integrate).

Does this formula mean that at very high velocities a car generates more power for the same force or torque it applies to the wheels? Or does it mean that at higher speeds a car requires more power for a given force?

Yes.

 

[annabrown]

Hi, You've done a detailed analysis, but let's clarify where the discrepancy might lie.

1. Work Done by Force:
   Work = Force × Distance = 5 N × 1 m = 5 J
2. Moment of Inertia:
   For a solid sphere, I = (2/5) × M × R²
   I = (2/5) × 5 kg × (1 m)² = 2 kg·m²
3. Torque and Angular Acceleration:
   Torque (τ) = Force × Radius = 5 N × 1 m = 5 N·m
   Angular acceleration (α) = τ / I = 5 / 2 = 2.5 rad/s²
4. Time and Angular Velocity:
   θ = (1/2) × α × t² → t = √(2 / 2.5) ≈ 0.894 s
   Angular velocity (ω) = α × t = 2.5 × 0.894 ≈ 2.23 rad/s
5. Linear Velocity:
   Linear impulse = 5 N × 0.894 s
   v = Linear impulse / Mass = 4.47 / 5 = 0.894 m/s
6. Kinetic Energy:
   Translational KE = (1/2) × Mass × v² = (1/2) × 5 × (0.894)² ≈ 2 J
   Rotational KE = (1/2) × I × ω² = (1/2) × 2 × (2.23)² ≈ 4.97 J
   Total KE = 2 J + 4.97 J = 6.97 J

The discrepancy arises because the total kinetic energy includes both translational and rotational components. The work done (5 J) is converted into both forms of energy, resulting in the higher total kinetic energy calculated (6.97 J).

Hope this helps!

 

[A.T.]

The work done (5 J) is converted into both forms of energy, resulting in the higher total kinetic energy calculated (6.97 J).

Hope this helps!

No, this doesn't help, because it makes no sense to get 6.97 J total energy if you did just 5 J of work.

The error of the OP was already correctly identified by @Dale:

The force was not applied at the center of mass. It was applied at the edge of the sphere, right? Therefore the displacement of the edge will be larger than the displacement at the center of mass. That is where your missing work is.

Then see my answer above. That is exactly what I assumed.

Since you are getting confused with the displacements it will be better to calculate power rather than work. Use $P=\vec F \cdot \vec v$ and integrate over time.

 

[kuruman]

Just to settle this issue in case of future revivals of this thread, here is a derivation for the work-energy theorem in the case of a circular body of radius $R$ which rolls without slipping on a horizontal surface as it is pulled by constant force $F$ applied at its axis.

Let the moment of inertia of the body about its CM be $~I_{\text{CM}}=qmR^2~$ where $~0\leq q\leq 1.$ Constant $q$ is zero for a non-rotating body, $\frac{1}{2}$ for a cylinder, $1$ for a ring, etc.

We write Newton's second law for rotations about the instantaneous point of contact P on the surface, $$\begin{align}
& FR=I_P~ \alpha=(I_{\text{CM}}+mR^2)\alpha=(q+1)mR^2\alpha=(q+1)mR^2\frac{a_{\text{cm}}}{R} \nonumber \\
& \implies F=(q+1)ma_{\text{cm}}. \nonumber
\end{align}$$ The work done by the force on the object rolling over a displacement $\Delta x_{\text{cm}}$ is $$W=F\Delta x_{\text{cm}}=(q+1)ma_{\text{cm}}\Delta x_{\text{cm}}.$$From the kinematic equation under constant acceleration, $a_{\text{cm}}\Delta x_{\text{cm}}=\frac{1}{2}\left(v_{\text{cm,f}}^2- v_{\text{cm,i}}^2 \right)$ so that $$F\Delta x_{\text{cm}}=(q+1)\frac{1}{2}m\left(v_{\text{cm,f}}^2- v_{\text{cm,i}}^2 \right)
=(q+1)\Delta K_{\text{cm}}.$$ Thus, the work-energy theorem in the case of a circular body which rolls without slipping on a horizontal surface as it is pulled by constant force $\mathbf F$ applied at its axis, $$\boxed{ \Delta K_{\text{cm}}=\frac{1}{q+1}\mathbf F\cdot \Delta \mathbf x_{\text{cm}} }$$Arguments involving mechanical energy conservation and/or work done by external forces can start here. For example, a cylinder rolling down an incline starting from rest after dropping vertical distance $h$ will have speed $v$ given by $$\frac{1}{2}mv^2=\frac{1}{\frac{1}{2}+1}mgh.$$

 

[gleem]

Just to settle this issue in case of future revivals of this thread

EDIT: This post has been replaced by Post #91 based on suggestions by jbriggs444 and other subsequent posts.

Me too.

A sphere or radius R and mass M is pushed tangentially with constant force F for X meters as with a flat bar of length x moving across the top of the sphere. The sphere moves without slipping horizontally on a flat surface as shown below.

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.

Thus the translational work is W_t= F⋅X and the rotational work is W_r = L⋅θ.

After the force has been applied the sphere moves with constant velocity V and rotates with an angular velocity ω. The two which are related by V = R⋅ω.

This results in the kinetic energy of translation KE_t=½⋅M⋅V² and rotation KE_r=½⋅I⋅ω²,

Because of the constant acceleration, the translational and rotational velocities are related to their respective accelerations and the distance of the applied force by similar equations. V ² = 2⋅ a ⋅x and ω ²= 2⋅ α ⋅θ. (Avoids the consideration of the time the force is applied.)

Using the above relationships, it is easily shown that the total work done is equal to the total resulting kinetic energies.

 

[jbriggs444]

A sphere or radius R and mass M is pushed tangentially with constant force F for X meters as with a flat bar of length x moving across the top of the sphere. The sphere moves without slipping horizontally on a flat surface as shown below.

View attachment 348281

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.

Are you sure that you are counting all of the forces?

 

[gleem]

I think as long as it is not sliding we should be all set. suggestions?

 

[jbriggs444]

I think as long as it is not sliding we should be all set. suggestions?

You could do an energy analysis. Only one of the two forces provides any energy input into the system. This is the way that I would likely proceed.

You could let the frictional force be an unknown and solve the simultaneous equations with the constraint that the rolling rate will match the translation rate.

You might make some progress by creatively choosing an axis of rotation so that one of the two forces exerts zero torque and proceed from there. Be careful that you properly include all of the places where angular momentum can show up.

 

[gleem]

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

 

[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

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FX+ L\theta=maX+I\alpha \theta $$
$$v^{2}=2aX $$ 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).

 

[A.T.]

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.

It should be obvious that the top of a rolling cylinder moves twice as fast as it's center. So if you don't have any slippage (neither at the ground, nor at the ruler), then a horizontal ruler touching the top will also move twice as fast as the center.

From: https://www.miniphysics.com/uy1-rolling-motion.html

 

[A.T.]

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FaX + L\alpha =FaX+I\alpha \theta $$

The dimensions in ths equation don't match.

 

[gleem]

First, @jbriggs444
I see said the blind man as he picked up his hammer and saw that the speed at the top of the sphere is moving twice as fast as the center so the bar moves twice the distance compared to the center. Thanks for your patience.

Second: @A.T.

It should be obvious that the top of a rolling cylinder moves twice as fast as it's center

Actually, I always knew that but somehow did not see its relevancy until @jbriggs444 kept my attention on the actual work that was being performed.

The dimensions in ths equation don't match.

As for this, I was requested to use LaTex for my equations and in a rush accidentally forgot to replace F with m. but thanks anyway.

I will edit my previous posts to show appropriate corrections and errors suitably struck out.

 

[jbriggs444]

$$KE_{tot}= \frac{1}{2}mv^2 +\frac{1}{2}I\omega ^2 = FaX + L\alpha =FaX+I\alpha \theta $$

The dimensions in ths equation don't match.

As I reconstruct the intent we have several typos and some idiosyncratic usage.

@gleem has been using $ma$ as a synonym for $F$. One $FaX$ should read $FX$ and the other should read $maX$.
@gleem has been using $L$ for torque. $L\alpha$ should have read $L\theta$. I will use $\tau$ for torque instead.

With those corrections in mind we arrive at:$$KE_\text{tot} = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 = FX + \tau \theta = maX + I \alpha \theta$$That equation is, at least, dimensionally consistent.

Although it is wrongly motivated, the resulting equation is even correct.

While $F \ne ma$ and $\tau \ne I \alpha$, the sums of the two terms: $FX + \tau \theta$ and $maX + I \alpha \theta$ do indeed happen to match. The discrepancy in the one exactly cancels with the discrepancy in the other. Because the unaccounted for force does zero work.

 

[A.T.]

With those corrections in mind we arrive at:$$KE_\text{tot} = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2 = FX + \tau \theta = maX + I \alpha \theta$$That equation is, at least, dimensionally consistent.

Although it is wrongly motivated, the resulting equation is even correct.

While $F \ne ma$ and $\tau \ne I \alpha$, the sums of the two terms: $FX + \tau \theta$ and $maX + I \alpha \theta$ do indeed happen to match. The discrepancy in the one exactly cancels with the discrepancy in the other. Because the unaccounted for force does zero work.

OK, but $KE_\text{tot} = FX$, not $FX + \tau \theta$.

 

[jbriggs444]

OK, but $KE_\text{tot} = FX$, not $FX + \tau \theta$.

If $X$ is the distance that the center of mass moves then the mechanical work done is $2FX$. And since $\tau = FR$ and $\theta = \frac{X}{R}$, it turns out that $\tau \theta = FX$.

 

[A.T.]

If $X$ is the distance that the center of mass moves then the mechanical work done is $2FX$.

I went by diagram by @gleem where $X$ is the distance of the upper bar applying the force $F$, so $FX$ would be the total work done and thus the total change in $KE$.

 

[jbriggs444]

I went by diagram by @gleem where $X$ is the distance of the upper bar applying the force $F$, so $FX$ would be the total work done and thus the total change in $KE$.


View attachment 348445

OK, I agree with that.

I had not interpreted the drawing that way. But it makes perfect sense, now that you've drawn my attention to it. It does mean that evaluating the work done by the same force twice would be double dipping, just as you've said.

 

[gleem]

OK, I agree with that.

Wait. I am currently working on revising my post per your suggestions and have changed my diagram to

It would have been posted by now but overnight I lost a lot of my work for some reason and will have to redo it.