I'm calculating more energy out than I put in

[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.

 

[PeterDonis]

more work will be applied than I initially thought

Assuming that the distance $L$ through which the force is applied for one of the motions is 1 meter, yes. But you have to decide whether that distance is for the rotational motion (i.e., the sphere rotates through an angle of 1 radian while the force is being applied) or the translational motion (i.e., the center of mass moves 1 meter while the force is being applied). Those two conditions give different answers for the total work done.

 

[Dale]

What is the value of $t$?

Any real number. It shows that energy is conserved at all times.

 

[PeterDonis]

Any real number. It shows that energy is conserved at all times.

I meant, what value of $t$ would be used to calculate the total kinetic energy for this problem. I agree that energy conservation holds for all $t$, not just that particular one.

 

[Dale]

I meant, what value of $t$ would be used to calculate the total kinetic energy for this problem. I agree that energy conservation holds for all $t$, not just that particular one.

I don’t think it matters because it resolves the concern for all values of $t$. But if you really need one then you can integrate either $v$ or $v_c$ to get the distance and solve for $t$.

 

[Chenkel]

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

I understood that derivation of kinetic energy, for the sphere situation but I'm a little unfamiliar with the equation $P=Fv$

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 $power = force * velocity$

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?

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?

Any help is appreciated, thank you!

 

[PeterDonis]

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?

The latter. Unless by "generates more power" you just mean that the car's engine produces more power, in which case both of your questions are saying the same thing.

 

[kuruman]

Any real number. It shows that energy is conserved at all times.

I don't understand what you mean by this. What kind of energy conservation is this when there is no potential energy and the kinetic energy is a function of time? Mechanical energy conservation means (to me) that $K+U$ at point A is the same as $K+U$ at some other point B. This is not the case here. The sphere starts moving from rest and there is no potential energy.

 

[PeterDonis]

I don't understand what you mean by this.

@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).

 

[Dale]

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?

Yes.

Or does it mean that at higher speeds a car requires more power for a given force?

Yes, it is the same thing as above. For a fixed force more power is required and therefore generated. If the car cannot produce sufficient power then it cannot produce that force at higher speeds.

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?

Yes.

 

[Dale]

What kind of energy conservation is this when there is no potential energy and the kinetic energy is a function of time?

The kind where the change in energy is equal to the work done.

 

[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.