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

[Chenkel]

Hello everyone,

I'm currently working on a physics problem involving the rotation of a 5 kilogram $M=5$ solid sphere subjected to a force of 5 newtons $F=5$, and I've encountered an inconsistency in my calculations. I'm seeking guidance or insights into where I might have gone wrong.

My initial goal was to apply a force of 5 Newtons tangentially to a solid sphere with a 1-meter radius through a distance of 1 meter. The objective was to examine whether the total kinetic energy (sum of linear and angular kinetic energy) of the system equals the energy I put into it. I postulate that I put in 5 joules of energy but my calculations say I put in 6.97 joules.

I calculated with a solid sphere that has a mass of 5 kilograms, and I used the following formula to find the moment of inertia about the center of mass of a solid sphere $I=(2/5)*M*R^2$, $M=5$ and $R=1$ so I plugged my values in and I got a moment of inertia of $I=2$

Then I calculated the work I put into the system as 5 joules or 5 Newtons applied through a distance of 1 meter.

But the force also creates a 5 Newton meter torque about the center of mass during the impulse.

I postulated that if the force acts through a distance of 1 meter tangentially to the sphere then the sphere will rotate by an angular displacement of 1 radian during the impulse (because the sphere has a radius of 1 meter).

So now I wanted to calculate how long the impulse was so I calculated the torque, and divided the torque by the moment of inertia to find the angular acceleration and then integrated the angular acceleration to find the angular displacement and I solved for t. So ${\theta} = (1/2)*{\alpha}*t^2$ where $\theta$ is the angular displacement, and $\alpha$ is the angular acceleration.

To find the angular acceleration I divide the torque $\tau$ of 5 Newton meters by the moment of inertia $I$ which is 2 and I get an acceleration of 5/2 which is an $\alpha$ value of 2.5 radians per second squared.

Now I can solve for t when $\theta$ (angular displacement) is one radian (the angular displacement during the impulse) and I get $t=sqrt(2/{\alpha})=sqrt(2/2.5) = .894$ seconds

Then I calculated an angular impulse of time_of_impulse*applied_torque or
sqrt(2/2.5)*(5 Newton meters) = 4.47 Newton meter seconds.

I divide this value by the moment of inertia to get the angular velocity after the impulse and I get an angular velocity of 4.47/2 or 2.23 radians per second.

Now I need to calculate the linear impulse and divide by the total mass of the object to get the linear velocity and I get (sqrt(2/2.5)*5)/5 = .894 meters per second

Now I plugged these values into the total kinematic energy formula (1/2)*5*(.894^2) + (1/2)*2*(2.23^2) and I get 6.97 joules instead of what I postulated I put in which was 5 joules.

Any assistance or suggestions regarding the approach or calculations would be highly appreciated. Specifically, I'm looking for insights into the discrepancy between the expected and calculated total kinetic energy.

My apologies if my analysis was tedious and and thanks to anyone who read through it.

If you can shed any light on this matter I would appreciate it and I thank you!

 

[kuruman]

Can you describe what you did with equations rather than numbers? Use $R$ for the radius of the sphere, $M$ for its mass and so on. It will be much easier for us to figure out what you did.

If I understand correctly, you have a sphere suspended in space (no gravity) and you apply a tangential force $F$ duration that gets it spinning and translating until it moves by distance $L$. You want to calculate its rotational plus translational kinetic energy at that point. Is that right?

 

[Chenkel]

Can you describe what you did with equations rather than numbers? Use $R$ for the radius of the sphere, $M$ for its mass and so on. It will be much easier for us to figure out what you did.

If I understand correctly, you have a sphere suspended in space (no gravity) and you apply a tangential force $F$ duration that gets it spinning and translating until it moves by distance $L$. You want to calculate its rotational plus translational kinetic energy at that point. Is that right?

That's correct, thanks again for reading, I'll try to fix up the equations with latex.

 

[Hill]

Then I calculated the work I put into the system as 5 joules or 5 Newtons applied through a distance of 1 meter.

I didn't follow the details of your calculations, but I think that the work you put into the system should be more than this. Even if the sphere only rotates in place, you need to put in energy, but in your calculation, it'd be "5 Newtons through a distance of 0 meters" = 0 joules.

 

[Chenkel]

I didn't follow the details of your calculations, but I think that the work you put into the system should be more than this. Even if the sphere only rotates in place, you need to put in energy, but in your calculation, it'd be "5 Newtons through a distance of 0 meters" = 0 joules.

The sphere gets a horizontal force parallel to its top edge and the sphere rotates counter clockwise 1 radian to move forward a distance of one meter, therefore the 5 Newton force is applied through a distance of one meter horizontally creating a linear impulse and a rotational impulse but 5 Newton's acting through a distance of 1 meter is 5 joules.

 

[Hill]

The sphere gets a horizontal force parallel to its top edge and the sphere rotates counter clockwise 1 radian to move forward a distance of one meter, therefore the 5 Newton force is applied through a distance of one meter horizontally creating a linear impulse and a rotational impulse but 5 Newton's acting through a distance of 1 meter is 5 joules.

I don't think it answers my objection. If the force of 5 Newton were applied to the center of the sphere, the sphere would move 1 m without rotating, and the work done would be 5 joules. But in your case, the force makes the sphere move and rotate. So, the work should be more then 5 joules.

 

[Chenkel]

I don't think it answers my objection. If the force of 5 Newton were applied to the center of the sphere, the sphere would move 1 m without rotating, and the work done would be 5 joules. But in your case, the force makes the sphere move and rotate. So, the work should be more then 5 joules.

That makes some sense, but the impulse happened over .894 seconds in order to rotate the unbounded sphere one meter forward.

 

[Chenkel]

I'm starting to think I didn't understand this problem well enough before starting on it.

 

[kuruman]

Actually, this is a simple problem. Just calculate the work done by the torque and by the force and add.

 

[Vanadium 50]

Can you describe what you did with equations rather than numbers?

This is the key. Numbers should be the last thing that goes in.

We might catch that a square should be a square root or vice versa. There is no way we will figure out that a 2.563 should actuually be a 5.992.

 

[kuruman]

That makes some sense, but the impulse happened over .894 seconds in order to rotate the unbounded sphere one meter forward.

I get $\sqrt{2}~$s. If $L=\frac{1}{2}\frac{F}{M}t^2$, what is $t$? That's why you should work with symbols.

 

[Chenkel]

Actually, this is a simple problem. Just calculate the work done by the torque and by the force and add.

I'm trying to work the problem again with a 1 second impulse to make things more clear for me.

 

[kuruman]

I'm trying to work the problem again with a 1 second impulse to make things more clear for me.

Are you changing the numbers? One second impulse will not give you 1 meter distance traveled.

Use symbols. Then it will be clear.

 

[Vanadium 50]

I'm trying to work the problem again with a 1 second impulse to make things more clear for me.

Fine. Ignore our advice to work with symbols. Just don't blame us.

 

[Chenkel]

Are you changing the numbers? One second impulse will not give you 1 meter distance traveled.

Use symbols. Then it will be clear.

I'm just making the impulse $F*\Delta t$ where $\Delta t$ is 1 second and the ball will rotate more than 1 radian.

 

[Dullard]

If I followed:

The applied force results in both angular and linear acceleration. Your analysis of the input work seems to ignore the linear displacement.

 

[jbriggs444]

Now I need to calculate the linear impulse and divide by the total mass of the object to get the linear velocity and I get (sqrt(2/2.5)*5)/5 = .894 meters per second

If I follow the scenario, the applied force rotates through an angle of one radian as it is applied. You cannot simply multiply the magnitude of the force by the time it was applied to get total impulse. Force is a vector. You have to treat it as one. Part of the impulse in one direction will cancel with part of the impulse in a different direction.

The required integration is extra nasty because the rotation angle is not changing smoothly. Instead, it changes at an accelerating pace. So you can't just multiply by an integral of cosine theta over some range.

 

[Chenkel]

If I just ignore the linear translation of the ball and just use the the angular velocity based on the angular impulse to calculate the kinetic energy I get 5 joules exactly, with the force perfectly tangential to the the sphere does it cause any change in the velocity of the center of mass or does it cause only rotation?

 

[Chenkel]

If I follow the scenario, the applied force rotates through an angle of one radian as it is applied. You cannot simply multiply the magnitude of the force by the time it was applied to get total impulse. Force is a vector. You have to treat it as one. Part of the impulse in one direction will cancel with part of the impulse in a different direction.

The required integration is extra nasty because the rotation angle is not changing smoothly. Instead, it changes at an accelerating pace. So you can't just multiply by an integral of cosine theta over some range.

But I was thinking of the force as parallel with the ground but at the top of the ball, I wasn't thinking of a rotating force vector.

 

[Chenkel]

Fine. Ignore our advice to work with symbols. Just don't blame us.

Not ignoring your advice.

 

[Chenkel]

I get $\sqrt{2}~$s. If $L=\frac{1}{2}\frac{F}{M}t^2$, what is $t$? That's why you should work with symbols.

I'm trying to figure out how you got that.

 

[Chenkel]

Actually, this is a simple problem. Just calculate the work done by the torque and by the force and add.

Maybe it's simple for you, not for me!

 

[Chenkel]

Fine. Ignore our advice to work with symbols. Just don't blame us.

I'm actually thinking I'm going to stick to the original problem and not change it, I don't need a 1 second impulse, all I need is an impulse through 1 radian.

 

[Chenkel]

Are you changing the numbers? One second impulse will not give you 1 meter distance traveled.

Use symbols. Then it will be clear.

I'm going to stick to the original problem and not change things. And I'm going to use symbols.

 

[Chenkel]

I get $\sqrt{2}~$s. If $L=\frac{1}{2}\frac{F}{M}t^2$, what is $t$? That's why you should work with symbols.

It seems like you set L equal to one meter and you a are looking at the motion of the center of mass where $\frac {F}{M}$ is the acceleration of the center of mass

 

[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.

 

[PeterDonis]

I postulated that if the force acts through a distance of 1 meter tangentially to the sphere then the sphere will rotate by an angular displacement of 1 radian during the impulse (because the sphere has a radius of 1 meter).

If this is true, the force is not applied in a straight line. But your calculation of the final linear velocity assumed a force that was applied in a straight line. The actual final linear velocity will be smaller than what you calculated--I believe you would need to do an integral to find it. Ironically, it would probably be simpler to just subtract the rotational energy you calculated from the total energy of 5 joules and then calculate the magnitude of the final linear velocity from that! (But that won't give you the direction of the final linear velocity.)

 

[PeterDonis]

If you have a constant force acting anywhere on the sphere, the acceleration is constant and equal to $F/M$.

The magnitude of the linear acceleration will be constant. But its direction will not be, because the OP specified that the force is always applied tangentially to the sphere. That requires the force to change direction.

 

[Chenkel]

Let L be the distance through which the force acts

$L = 1$

$L = (1/2)(\frac F M)*t^2$

$1 = (1/2)(\frac 5 5)t^2$

$t = \sqrt{2}$.

Let $\tau$ be the torque around the center of mass

$\tau = FR=(5)(1)=5$

Let P be the linear impulse and Q be the angular impulse

$P = F * t = F * \sqrt{2} = 5 * \sqrt {2}$
$Q = \tau * t = \tau * \sqrt{2} = 5 * \sqrt {2}$.

$v = \frac P M = \frac {5 \sqrt {2}} {5} = \sqrt {2}$

$\omega = \frac Q I = \frac {5 \sqrt {2}} {2} = \frac {5} {\sqrt{2}}$

The kinetic energy E is
$E = {\frac 1 2} M v^2 + {\frac 1 2} I {\omega}^2$
$E = {\frac 1 2} * 5 * {(\sqrt{2})}^2+ {\frac 1 2} * 2 * {(\frac {5} {\sqrt{2}})}^2$
$E = {\frac 1 2} * 5 * 2+ {\frac 1 2} * 2 * {\frac {25} {2}}$
$E = 5 + {\frac {25} {2}}$
$E = 17.5$

So some of the work went into increasing the angular velocity and some of the work went into changing the linear velocity

But I still don't see how the energy put into the system is 17.5 joules.

 

[kuruman]

The magnitude of the linear acceleration will be constant. But its direction will not be, because the OP specified that the force is always applied tangentially to the sphere. That requires the force to change direction.

Interesting. Yes, the force is specified as tangential but there is no specification that it is applied at a fixed point on the sphere. The picture I had in mind was a tangential force that does not change direction, e.g. a thin jet of air at a glancing angle or putting the sphere on an accelerating conveyor belt.

I deleted the post you quoted to avoid confusion. A drawing would have been a good thing to have.

 

[Chenkel]

The magnitude of the linear acceleration will be constant. But its direction will not be, because the OP specified that the force is always applied tangentially to the sphere. That requires the force to change direction.

If the force is on the top horizontal tangent then I don't see how the force needs to change direction.

 

[Chenkel]

Interesting. Yes, the force is specified as tangential but there is no specification that it is applied at a fixed point on the sphere. The picture I had in mind was a tangential force that does not change direction, e.g. a thin jet of air at a glancing angle or putting the sphere on an accelerating conveyor belt.

I deleted the post you quoted to avoid confusion. A drawing would have been a good thing to have.

Yes, it's something like that, a force applied to the top horizontal tangent of the sphere.

 

[Chenkel]

But I still don't see how the energy put into the system is 17.5 joules.

Maybe the force through a distance idea changes a little for energy transfer in rotating systems.

 

[PeterDonis]

Yes, it's something like that, a force applied to the top horizontal tangent of the sphere.

Even if you're doing it that way, the linear distance through which the force is acting won't be 5 meters (which was the underlying point I was trying to make when I said the force would change direction--I was thinking of a different method of applying the force, putting it at a single point on the sphere with no slippage). As @Dale pointed out, the center of mass motion is less than the motion of the tangent point at which the force acts. So the final linear velocity will be smaller than what you are calculating. It should be obvious that the final linear velocity can't be what you would get from 5 joules of linear kinetic energy, since some of the work done is not going into linear motion.

 

[PeterDonis]

Maybe the force through a distance idea changes a little for energy transfer in rotating systems.

Of course it does. If some of the work is going into rotational energy, it's not going into linear kinetic energy.

 

[kuruman]

I cannot make sense of your numbers. The 17.5 Joules is meaningless to me.
For the linear motion
$$L=\frac{1}{2}\frac{F}{M}t^2\implies t^2=\frac{2ML}{F}$$
The translational kinetic energy is
$$K_{trans.}=\frac{1}{2}Mv^2=\frac{1}{2}Ma^2t^2.$$The rotational kinetic energy is $$K_{rot.}=\frac{1}{2}I\omega^2=\frac{1}{2}I\alpha^2 t^2.$$
Put in the appropriate expressions for the accelerations and the moment of inertia. Simplify the expressions and add them.

 

[Dale]

Yes, it's something like that, a force applied to the top horizontal tangent of the sphere.

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.

 

[PeterDonis]

I cannot make sense of your numbers. The 17.5 Joules is meaningless to me.
For the linear motion
$$L=\frac{1}{2}\frac{F}{M}t^2\implies t^2=\frac{2ML}{F}$$
The translational kinetic energy is
$$K_{trans.}=\frac{1}{2}Mv^2=\frac{1}{2}Ma^2t^2.$$

This is already enough to show the issue. We have $a = F / M$, so $L = a t^2 / 2$ and $K_{trans.} = M a L = F L$. But that means nothing is left over for rotational kinetic energy at all!

So the problem is not what equations to use. The problem is what numbers to plug into them. The OP was assuming that the correct numbers for the linear motion are $L = 1$, $F = 5$, and $M = 5$. But that cannot be right if there is any rotation at all. If we assume that $M$ and $F$ can be controlled by the experimenter and hence are known, then $L = 1$ must be wrong for the linear part of the motion. The actual linear $L$ must be smaller than that.

 

[PeterDonis]

it will be better to calculate power rather than work. Use $P=\vec F \cdot \vec v$ and integrate over time.

But what $\vec{v}$ should be used?

It seems to me that the first task is to figure out how to divide up the work being done (or power being expended, either way it's the same problem) between linear and rotational motion. I'm not sure if the problem specification is sufficient to do that.

 

[Dale]

But what v→ should be used?

The only relevant $\vec v$ is the material at the point of application of the force. That is the velocity that is always used in this context.

It seems to me that the first task is to figure out how to divide up the work being done (or power being expended, either way it's the same problem) between linear and rotational motion.

That happens naturally as you calculate the velocity of the material at the point of application of the force

 

[jbriggs444]

But I was thinking of the force as parallel with the ground but at the top of the ball, I wasn't thinking of a rotating force vector.

OK, that makes sense. Then the force is applied over a larger distance since [the material at] its point of application is both rotating and translating.

 

[kuruman]

But that means nothing is left over for rotational kinetic energy at all!

Left over from what? In post #36, I outlined the calculation for translational and rotational energy hoping that OP would finish it. Here it is anyway.$$K_{trans.}=\frac{1}{2}Mv^2=\frac{1}{2}M\left(at\right)^2=Ma\left(\frac{1}{2}at^2\right)=FL.$$ We have seen that the time needed to translate by $L$ under constant acceleration is given by $t^2=\dfrac{2ML}{F}.$ In that time the sphere acquires rotational kinetic energy at constant torque $(\tau=FR=I\alpha)$ about its CM. Thus,
$$\begin{align} K_{rot.}& = \frac{1}{2}I\omega^2=\frac{1}{2}I(\alpha t)^2=I\alpha\left(\frac{1}{2}\alpha t^2\right)=FR\left(\frac{1}{\cancel{2}}\frac{\cancel{F}R}{I}\times\frac{\cancel{2}ML}{\cancel{F}}\right) \nonumber \\
& =FL\frac{MR^2}{I}=\frac{5}{2}FL. \nonumber
\end{align}$$The total mechanical energy acquired is $$ME=\frac{7}{2}FL.$$

 

[A.T.]

Maybe the force through a distance idea changes a little for energy transfer in rotating systems.

No it doesn't. But calculating the distance might be more complicated. As @Dale wrote, you have to integrate the dot-product of force and velocity of the material to which the force is applied (power) to get work.

 

[Chenkel]

Of course it does. If some of the work is going into rotational energy, it's not going into linear kinetic energy.

I'd like to learn more how that works.

 

[Dale]

So always in mechanical problems the power is $P=\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material at the point of application of the force $\vec F$. One simple rule, always.

So here $\vec F$ and $\vec v$ are in the same direction so $P=Fv$. And we have $v=v_c+\omega r$ where $v_c$ is the velocity of the center of mass, $\omega$ is the angular velocity, and $r$ is the radius.

By Newton’s 2nd law $F=m \dot v_c$ which has the solution $v_c=t F/m$. And the rotational analog $\tau=rF=I \dot \omega$ which has the solution $\omega=t \ r F/I$.

Now we just integrate $$W=\int_0^t P \ dt =\frac{t^2 F^2}{2m}+\frac{t^2 r^2 F^2}{2 I}=\frac{1}{2}mv_c^2+\frac{1}{2}I\omega^2$$

 

[kuruman]

Now we just integrate $$W=\int_0^t P \ dt =\frac{t^2 F^2}{2m}+\frac{t^2 r^2 F^2}{2 I}=\frac{1}{2}mv_c^2+\frac{1}{2}I\omega^2$$

Which is what I have in post #42 in terms of $F$ and $L$.

Actually, more generally one can say that when impulse $J$ is delivered tangentially to the sphere, it acquires momentum $$Mv=J\implies v=\frac{J}{M}$$ and angular momentum about the CM $$I\omega=RJ\implies \omega=\frac{RJ}{I}.$$ Then the ratio of rotational to translational kinetic energy is $$ \frac{K_{rot.}}{K_{trans.}}=\frac{I\omega^2}{Mv^2}=\frac{MR^2}{I}=\frac{5}{2}.$$

 

[PeterDonis]

$$K_{trans.}=\frac{1}{2}Mv^2=\frac{1}{2}M\left(at\right)^2=Ma\left(\frac{1}{2}at^2\right)=FL.$$

What is the value of $L$?

 

[PeterDonis]

By Newton’s 2nd law $F=m \dot v_c$ which has the solution $v_c=t F/m$. And the rotational analog $\tau=rF=I \dot \omega$ which has the solution $\omega=t \ r F/I$.

What is the value of $t$?

 

[PeterDonis]

Then the ratio of rotational to translational kinetic energy is $$ \frac{K_{rot.}}{K_{trans.}}=\frac{I\omega^2}{Mv^2}=\frac{MR^2}{I}=\frac{5}{2}.$$

This would indeed be the missing piece that I described earlier.

So if this is true, and if it is also true that $L$ for the translational force is 1 meter, as stated in the OP, then you have just figured out where the OP's figure of 17.5 joules comes from: 5 joules of translational kinetic energy (5 newtons times 1 meter), plus 5/2 times 5 = 12.5 joules of rotational kinetic energy, equals 17.5 joules of total kinetic energy.

However, I don't think that $L$, the total translational distance through which the force acts, is 1 meter. More precisely, that statement is inconsistent with the total rotational distance through which the force acts (angle times radius) being 1 meter (total angle 1 radian). In fact we would expect the ratio of distances to be the same as the ratio of kinetic energies (because the ratio of distances is the ratio of work done), so if the rotational distance is 1 meter, the translational distance is 2/5 of a meter.

That still means the total kinetic energy is more than 5 joules; now it's 5 joules of rotational kinetic energy, plus 2 joules of translational kinetic energy, for a total of 7 joules.

 

[Chenkel]

I don't think it answers my objection. If the force of 5 Newton were applied to the center of the sphere, the sphere would move 1 m without rotating, and the work done would be 5 joules. But in your case, the force makes the sphere move and rotate. So, the work should be more then 5 joules.

If I have a sphere and I apply a force to the top part to rotate it forward the top part of the sphere moves with velocity $v_{top} = v_c + (R){\omega}$ where $\omega$ is the angular velocity in radians per second about the center of mass and R is the distance between the center of mass and the top of the sphere where the force is applied, and $v_c$ is the velocity of the center of mass.

So if the top part is moving with a velocity $v_{top}$ the center of mass will be moving with a velocity of $v_{top} - (R){\omega}$ so the top part of the tire will move more than the center of mass meaning more work will be applied than I initially thought.