Moment of inertia where mass and torque are at a different positions

In summary: That's because torque relates to work, while Newton's second law doesn't. The moment of inertia depends on both concepts. For Newton's second law it doesn't matter where you apply the force. For work it does, because you can do more work if you use a lever, for example. And when you have a lever, you can put the mass (which can be a point mass, for simplicity's sake) at the end of the lever - where you apply the torque, or closer to the pivot point of the...
  • #1
ynbo0813
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The formula for moment of inertia is:

I=mr^2

A common derivation for this is:
1. F=ma
2. τ=rma
3. τ=rmrα = r^2 mα

This is a rotational version of Newton’s second law, where torque replaces force, moment of inertia replaces mass, and angular acceleration replaces tangential acceleration.

What bothers me is that the two rs in the formula refer to two different things: the r derived from torque (line 2) refers to the position where the force is acting; the r derived from angular acceleration (line 3) refers to the position of the mass. But what if the force is not acting on the same place as the mass? For example, what if a mass is placed on at point r from the pivot point, but the torque is applied at 2r from the pivot point. Does it have the same formula for its moment of inertia?
 
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  • #2
chananyag said:
what if a mass is placed on at point r from the pivot point, but the torque is applied at 2r from the pivot point.
The derivation models the rigid body as a collection of point masses, which all have the force needed for their tangential acceleration applied directly to them (for example via internal forces of the body). That's what the F stands for here. Applying F somewhere else doesn't make sense in this model, because then equation 1 cannot be used to relate F and the tangential acceleration.
 
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  • #3
The formula $$I=mr^2$$ gives the moment of inertia (MoI) of a point particle of mass m. If the body we are trying to find its MoI is not a point particle then the formula for its MoI can be quite complicated. It is actually given by a volume integral $$I=\iiint_V\rho(x,y,z)r^2(x,y,z)dxdydz$$ where ##\rho(x,y,z)## is the mass density of the body at point x,y,z and ##r(x,y,z)## is the distance of the point (x,y,z) from the axis of rotation of the body and ## V## is the region of space that the body occupies. From this formula we can see that ##I## can be different , depending on ##\rho##, V and which axis of rotation we choose(which choice will affect how we calculate the function ##r(x,y,z)##).
 
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  • #4
Delta2 said:
The formula $$I=mr^2$$ gives the moment of inertia (MoI) of a point particle of mass m. If the body we are trying to find its MoI is not a point particle then the formula for its MoI can be quite complicated. It is actually given by a volume integral $$I=\iiint_V\rho(x,y,z)r^2(x,y,z)dxdydz$$ where ##\rho(x,y,z)## is the mass density of the body at point x,y,z and ##r(x,y,z)## is the distance of the point (x,y,z) from the axis of rotation of the body and ## V## is the region of space that the body occupies. From this formula we can see that ##I## can be different , depending on ##\rho##, V and which axis of rotation we choose(which choice will affect how we calculate the function ##r(x,y,z)##).
A.T. said:
The derivation models the rigid body as a collection of point masses, which all have the force needed for their tangential acceleration applied directly to them (for example via internal forces of the body). That's what the F stands for here. Applying F somewhere else doesn't make sense in this model, because then equation 1 cannot be used to relate F and the tangential acceleration.
A.T. said:
The derivation models the rigid body as a collection of point masses, which all have the force needed for their tangential acceleration applied directly to them (for example via internal forces of the body). That's what the F stands for here. Applying F somewhere else doesn't make sense in this model, because then equation 1 cannot be used to relate F and the tangential acceleration.
A.T. said:
The derivation models the rigid body as a collection of point masses, which all have the force needed for their tangential acceleration applied directly to them (for example via internal forces of the body). That's what the F stands for here. Applying F somewhere else doesn't make sense in this model, because then equation 1 cannot be used to relate F and the tangential acceleration.
That's because torque relates to work, while Newton's second law doesn't. The moment of inertia depends on both concepts. For Newton's second law it doesn't matter where you apply the force. For work it does, because you can do more work if you use a lever, for example. And when you have a lever, you can put the mass (which can be a point mass, for simplicity's sake) at the end of the lever - where you apply the torque, or closer to the pivot point of the lever.
 
  • #5
chananyag said:
The moment of inertia depends on both concepts. .
I am not sure which two concepts you mean here (Torque and Work??) but the MoI of a body doesn't depend on anything except the density of the body, the geometry of the body and where we put the axis of rotation (and all these should be clear from the formula I gave). For a point particle of mass m, its MoI is always ##I=mr^2## where r is the distance from the axis of rotation.

chananyag said:
For work it does, because you can do more work if you use a lever, for example. And when you have a lever, you can put the mass (which can be a point mass, for simplicity's sake) at the end of the lever - where you apply the torque, or closer to the pivot point of the lever.
Here you introduce the concept of a lever. I am sure you have in your mind an ideal lever that has no mass and no moment of inertia, but real levers have both and we must take those into consideration when we consider the dynamics of a system with a lever.
The moment of inertia of a point mass will still be ##I=mr^2## even if we put the point mass on an ideal lever. The moment of inertia of the system (point mass+lever) will be different depending on where is the axis of rotation and where we put the point mass.
 
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  • #6
chananyag said:
That's because torque relates to work, while Newton's second law doesn't. The moment of inertia depends on both concepts.
As you see in the derivation, the moment of inertia is derived directly from Newton's 2nd Law applied to the tangential acceleration of each point mass. The concept of work doesn't enter the derivation.
 
  • #7
I thought it might help to see a more complete derivation. I will assume you are restricting your attention to rigid body dynamics. Let us also suppose for simplicity that the rigid body is undergoing pure rotation. The angular velocity and angular acceleration of the rigid body are ##\omega## and ##\alpha## respectively.

Consider a mass element ##m_i## at position ##\vec{r}_i## w.r.t. the axis of rotation (which I take to be the ##\hat{z}## axis). The force on any individual mass element can be decomposed into a resultant due to internal forces plus a resultant due to external forces, $$\vec{F}_i= {\vec{F}_i}^{ext} + {\vec{F}_i}^{int} = m_i\ddot{\vec{r}}_i$$Now left multiply by ##\vec{r}_i##, $$\vec{\tau}_i = {\vec{\tau}_i}^{int} + {\vec{\tau}_i}^{ext}= \vec{r}_i \times m_i \ddot{\vec{r}}_i = \frac{d}{dt} (\vec{r}_i \times m_i \dot{\vec{r}}_i) = \frac{d}{dt} (\vec{r}_i \times \vec{p}_i) = \frac{d\vec{L}_i}{dt}$$If we sum over all particles ##i## in the rigid body, the internal torques cancel (assuming all forces are central). But more important for your question is to notice that ##\dot{\vec{r}_i} = \omega \times \vec{r}_i## since all particles are undergoing circular motion about the axis of rotation. Hence ##\vec{L}_i = \vec{r}_i \times m_i (\omega \times \vec{r}_i) = m_i {r_i}^2 \omega \hat{z}## so that ##\vec{L} = \sum \vec{L}_i = \sum m_i {r_i}^2 \omega \hat{z}= I\omega \hat{z}##, where ##I## is defined as the moment of inertia ##I = \sum m_i {r_i}^2##. Putting it all together, $${\vec{\tau}}^{ext} = \frac{d\vec{L}}{dt} = I\frac{d \omega}{dt} \hat{z} = I\alpha \hat{z}$$Like @A.T. mentioned, there is no need to bring the concept of work into any of this. All points in the rigid body move with the same angular velocity and angular acceleration, and any external torques end up affecting all of the particles in the rigid body since the effects of external interactions are "propagated" by internal forces.
 
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  • #8
chananyag said:
The formula for moment of inertia is:

I=mr^2

A common derivation for this is:
1. F=ma
2. τ=rma
3. τ=rmrα = r^2 mα

This is a rotational version of Newton’s second law, where torque replaces force, moment of inertia replaces mass, and angular acceleration replaces tangential acceleration.

What bothers me is that the two rs in the formula refer to two different things: the r derived from torque (line 2) refers to the position where the force is acting; the r derived from angular acceleration (line 3) refers to the position of the mass. But what if the force is not acting on the same place as the mass? For example, what if a mass is placed on at point r from the pivot point, but the torque is applied at 2r from the pivot point. Does it have the same formula for its moment of inertia?
Perhaps the following example illustrating what @A.T. said in post #2 will help you understand how this works by addressing your "what if" question.

Two masses ##m_1## and ##m_2## are connected by a light rod. A second light rod of length ##R## is welded at right angles to the first rod and the assembly is pivoted at point O as shown in the figure. Force ##F## is applied. Find the torque acting on each mass.

TorqueOnMasses.png

Solution
The moment of inertia of the assembly about O is ##I=(m_1r_1^2+m_2 r_2^2)##.
The torque on the assembly about O is ##\tau=FR##.
Newton's Second law says ##\tau=I\alpha##.
Therefore the common angular acceleration of the two masses is ##\alpha =\dfrac{\tau}{I}=\dfrac{FR}{(m_1r_1^2+m_2 r_2^2)}##.

The linear acceleration of mass 1 is ##a_1=\alpha r_1=\dfrac{FRr_1}{(m_1r_1^2+m_2 r_2^2)}##.

The net force on mass 1 is ##F_{1}=m_1a_1=\dfrac{FRm_1r_1}{(m_1r_1^2+m_2 r_2^2)}##.

The net torque on mass 1 about O is ##\tau_1=F_{1}r_1=\dfrac{FRm_1r_1^2}{(m_1r_1^2+m_2 r_2^2)}##.

Likewise, the net force on mass 2 is ##F_{2}=m_2a_2=\dfrac{FRm_2r_2}{(m_1r_1^2+m_2 r_2^2)}## and the net torque is
##\tau_2=F_{2}r_2=\dfrac{FRm_2r_2^2}{(m_1r_1^2+m_2 r_2^2)}##.

Note that ##\tau_1+\tau_2=FR## as it should. It should be clear now how the different position vectors enter the calculation and how the external torque on the rigid body is divided between individual torques acting on each mass. These ideas can be extended the same way if you add a third, fourth or any number of masses.
 
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  • #9
kuruman said:
Note that ##\tau_1+\tau_2=FR## as it should. It should be clear now how the different position vectors enter the calculation and how the external torque on the rigid body is divided between individual torques acting on each mass. These ideas can be extended the same way if you add a third, fourth or any number of masses.

This example is great. I thought I might add one extra thing just to put it in context with what I wrote above. The torques ##\tau_1## and ##\tau_2## are internal torques exerted by the frame on the two masses respectively, and by Newton III the two masses exert opposite torques of magnitude ##\tau_1## and ##\tau_2## back on the frame. This is also justification for why ##\tau_1 + \tau_2 = FR##, since it is a statement of balancing torques on the the system consisting of the frame without the two masses.
 
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  • #10
chananyag said:
... What bothers me is that the two rs in the formula refer to two different things: the r derived from torque (line 2) refers to the position where the force is acting...
Welcome! :cool:
I would like you to consider that the torque of the formula has a value that corresponds to many possible combinations of tangential force(s) and distance(s) to the axis of rotation of the mass.

A torque steadily applied ("how" is not important, "where" does not make sense) in a perpendicular way to the axis of a rotational mass (punctual, multiple or distributed) will induce a steady angular acceleration/deceleration.

For convenience and simplicity, we apply the tangential force onto the mass for a single-mass rotational system.
As you have described, the tangential force could be applied anywhere along the imaginary line that connects the mass and its axis of rotation.

If we want to use the radial location of the mass, then we assume that an imaginary force applied there induces the same value of torque than the actual force that is applied somewhere along our imaginary line.
Note that neither the actual nor the imaginary force appears in the formula.

Besides the square of the radius of gyration of the mass, what is important for the value of the resistance of the system to accelerate or decelerate is the magnitude of the resultant torque value, not the magnitude of the tangential force(s) and distance(s) of application.

Here you can see a practical case, where the force than induces the torque is applied at a radius from the axis of rotation that has nothing to do with the radius in the formula of moment of inertia of a body with a peculiar shape:
https://docs.google.com/document/d/1VRpUxal8UKHfk8NEpNWBVVv5O4Wz48FDtiFVp5WypVE/edit
 
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  • #11
etotheipi said:
Like @A.T. mentioned, there is no need to bring the concept of work into any of this.
But isn't the formula of torque itself derived from the formula for work? In other words, when a force is applied further from the axis, then the circumference of the circle it acts along is greater than if it were applied closer to the axis - meaning that the Fd is greater?
 
  • #12
chananyag said:
But isn't the formula of torque itself derived from the formula for work?
Torque applies to static equilibrium situations as well, where no work is done.
 
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  • #13
kuruman said:
Perhaps the following example illustrating what @A.T. said in post #2 will help you understand how this works by addressing your "what if" question.

I'm a bit confused as to whether you're proposing a different 'what if' question or not. My 'what if' question was this:
1590382902280.png
 
  • #14
A.T. said:
Torque applies to static equilibrium situations as well, where no work is done.
Interesting point. But if it isn't derived from the concept of work, where is it derived from? Can it even be derived from first principles (i.e. Newton's Laws, work-energy theorem) at all, or is only derived from experiment?
 
  • #15
chananyag said:
Interesting point. But if it isn't derived from the concept of work, where is it derived from? Can it even be derived from first principles (i.e. Newton's Laws, work-energy theorem) at all, or is only derived from experiment?

Let's take a simpler example. A single particle is moving w.r.t. a coordinate system. The net force on it is ##\vec{F}##. Newton's second law tells us that $$\vec{F} = m\ddot{\vec{r}}$$We left multiply by ##\vec{r}##, $$\vec{r} \times \vec{F} = m\vec{r} \times \ddot{\vec{r}}$$ and define a new quantity, torque, ##\vec{\tau} = \vec{r} \times \vec{F}##. We now rearrange the RHS, $$\vec{\tau} = \frac{d}{dt} (\vec{r} \times m\dot{\vec{r}}) =\frac{d}{dt} (\vec{r} \times \vec{p}) = \frac{d\vec{L}}{dt}$$ where ##\vec{L}## is defined as ##\vec{r} \times \vec{p}##. As has been shown already, if the distance to a certain origin is fixed (as in the case of rigid body about a fixed axis, or a body fixed frame on the rigid body), you can usefully define a moment of inertia also.

So the concept arises naturally from Newton's laws.
 
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  • #16
etotheipi said:
Let's take a simpler example. A single particle is moving w.r.t. a coordinate system. The net force on it is ##\vec{F}##. Newton's second law tells us that $$\vec{F} = m\ddot{\vec{r}}$$We left multiply by ##\vec{r}##, $$\vec{r} \times \vec{F} = m\vec{r} \times \ddot{\vec{r}}$$ and define a new quantity, torque, ##\vec{\tau} = \vec{r} \times \vec{F}##. We now rearrange the RHS, $$\vec{\tau} = \frac{d}{dt} (\vec{r} \times m\dot{\vec{r}}) =\frac{d}{dt} (\vec{r} \times \vec{p}) = \frac{d\vec{L}}{dt}$$ where ##\vec{L}## is defined as ##\vec{r} \times \vec{p}##. As has been shown already, if the distance to a certain origin is fixed (as in the case of rigid body about a fixed axis, or a body fixed frame on the rigid body), you can usefully define a moment of inertia also.

So the concept arises naturally from Newton's laws.
It's a neat derivation, but I think you just introduced the concept of torque rather than derive it. Then, once you introduced it, you showed that it's equivalent to the change in angular momentum. All it shows is that there are many ways to define torque.

Another way to put it is this: you need to explain why levers work. Why should applying a force further along the lever make a difference? Newton's laws don't seem to explain it, but its seems the concept of work does. Take the two examples illustrated below. The second one has twice as much torque, twice as much rotational energy (so it seems to me), and therefore (so it seems it me) twice as much moment of inertia. The angular momentum, however, is the same for both.
1590400341683.png
 
  • #17
chananyag said:
IThe second one has twice as much torque, twice as much rotational energy (so it seems to me), and therefore (so it seems it me) twice as much moment of inertia. The angular momentum, however, is the same for both.
View attachment 263466
The moment of inertia of this system remains the same (as long as we don't change the position of the point mass or the pivot point) regardless of where we apply the torque.
 
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  • #18
chananyag said:
It's a neat derivation, but I think you just introduced the concept of torque rather than derive it.
The quantity of torque is defined as ##\vec{\tau} = \vec{r} \times \vec{F}##.
chananyag said:
Another way to put it is this: you need to explain why levers work. Why should applying a force further along the lever make a difference?
Because then the force results in a larger angular acceleration, as we have shown above.
chananyag said:
Newton's laws don't seem to explain it, but its seems the concept of work does. Take the two examples illustrated below. The second one has twice as much torque, twice as much rotational energy (so it seems to me), and therefore (so it seems it me) twice as much moment of inertia. The angular momentum, however, is the same for both.
The moment of inertia is constant (as @Delta2 points out also). That quantity only depends on the sum of ##m_i r_i^2##'s to the axis of rotation.

You can perhaps rationalise some behaviour in terms of work, but it's not any more fundamental. About a fixed angular displacement, the arc length along a path with twice the radius will be twice as long, so double the work is done by a tangential force moving along that path. If it is applied about a fixed angle, it will result in twice the amount of kinetic energy to the structure compared to if the force were applied at the centre through this fixed angle.

That is to say ##W = \tau \Delta \theta = \frac{1}{2} I \Delta (\omega^2)## That doesn't mean that the moment of inertia changes, but instead the change in the square of the angular velocity doubles if the torque is doubled. (I assume no gravitational field here).
 
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  • #19
etotheipi said:
The quantity of torque is defined as ##\vec{\tau} = \vec{r} \times \vec{F}##. Because then the force results in a larger angular acceleration, as we have shown above. The moment of inertia is constant (as @Delta2 points out also). That quantity only depends on the sum of ##m_i r_i^2##'s to the axis of rotation.

You can perhaps rationalise some behaviour in terms of work, but it's not any more fundamental. About a fixed angular displacement, the arc length along a path with twice the radius will be twice as long, so double the work is done by a tangential force moving along that path. If it is applied about a fixed angle, it will result in twice the amount of kinetic energy to the structure compared to if the force were applied at the centre through this fixed angle.

That is to say ##W = \tau \Delta \theta = \frac{1}{2} I \Delta (\omega^2)## That doesn't mean that the moment of inertia changes, but instead the change in the square of the angular velocity doubles if the torque is doubled. (I assume no gravitational field here).
Would you make the same argument in this case? I'm not disagreeing with you, I'm just trying to flesh it out. Hopefully it'll make it clearer for me.
1590402818503.png
 
  • #20
chananyag said:
Would you make the same argument in this case? I'm not disagreeing with you, I'm just trying to flesh it out. Hopefully it'll make it clearer for me.

Assuming the plank is massless, then you have now quadrupled the moment of inertia of the system. You can work everything else out from that :smile:
 
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  • #21
A.T. said:
Torque applies to static equilibrium situations as well, where no work is done.
I think @chananyag is referring to virtual work applied to static problems. See Feynman's treatment near the end of section 4.2.
 
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  • #22
kuruman said:
I think @chananyag is referring to virtual work applied to static problems. See Feynman's treatment near the end of section 4.2.
Yes, but is this really how torque is derived in general, or just a confirmation that the definition of torque is consistent with the definition of work and energy conservation in the special case of rigid bodies? Isn't the conservation of angular momentum the fundamental principle that torque is based on, as the rate of angular momentum transfer?
 
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  • #23
I think the OP has some misconception about moment of Inertia depending on where we apply the torque or how much torque or work is done. Moment of Inertia does not depend on torque or work done, it depends on the geometry of the body, its mass distribution along that geometry and where we put the axis of rotation.
 
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  • #24
chananyag said:
... Take the two examples illustrated below. The second one has twice as much torque, twice as much rotational energy (so it seems to me), and therefore (so it seems it me) twice as much moment of inertia. The angular momentum, however, is the same for both.
View attachment 263466
Can you see what is incorrect about those statements in this example?
Consider that the angular velocity has been reduced in half for the second case.

If your force is going to produce the same amount of work for both systems (as it should be in order to do a proper comparison) and if it remains perpendicular to the lever, the lengths of the arcs each describes must be the same.

Copied from
https://en.wikipedia.org/wiki/Rotational_energy

"In the rotating system, the moment of inertia, I, takes the role of the mass, m, and the angular velocity,
\omega
, takes the role of the linear velocity, v.
The rotational energy of a rolling cylinder varies from one half of the translational energy (if it is massive) to the same as the translational energy (if it is hollow)."

For a hollow cylinder with thick wall, you have this equation:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Returning to your original question:
What is the radius in these equations, the radius of application of the tangential force that induces the torque or the radius at which the concentrated mass is located?

 
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  • #25
etotheipi said:
Let's take a simpler example. A single particle is moving w.r.t. a coordinate system. The net force on it is ##\vec{F}##. Newton's second law tells us that $$\vec{F} = m\ddot{\vec{r}}$$We left multiply by ##\vec{r}##, $$\vec{r} \times \vec{F} = m\vec{r} \times \ddot{\vec{r}}$$ and define a new quantity, torque, ##\vec{\tau} = \vec{r} \times \vec{F}##. We now rearrange the RHS, $$\vec{\tau} = \frac{d}{dt} (\vec{r} \times m\dot{\vec{r}}) =\frac{d}{dt} (\vec{r} \times \vec{p}) = \frac{d\vec{L}}{dt}$$ where ##\vec{L}## is defined as ##\vec{r} \times \vec{p}##. As has been shown already, if the distance to a certain origin is fixed (as in the case of rigid body about a fixed axis, or a body fixed frame on the rigid body), you can usefully define a moment of inertia also.

So the concept arises naturally from Newton's laws.
Let me phrase it a more intuitive way. If two forces are equal they should produce the same tangential acceleration. The position of the force shouldn't matter for tangential acceleration. However, it does matter for angular acceleration. For example, take two cars accelerating at the same rate, but one has smaller wheels. If I'm not mistaken, the cars will both be accelerating at the same rate, but the smaller wheels will be spinning faster (i.e. it will take them quicker to make a full circle). But in that case, the force is closer to the axis! The force of the friction from the road is closer to the axis because the wheels are smaller. This is why I still don't understand, intuitively at least, how torque arises from Newton's laws. I'm guessing I made a misstep somewhere, just trying to find it.
 
  • #26
chananyag said:
For example, take two cars accelerating at the same rate, but one has smaller wheels. If I'm not mistaken, the cars will both be accelerating at the same rate, but the smaller wheels will be spinning faster (i.e. it will take them quicker to make a full circle). But in that case, the force is closer to the axis! The force of the friction from the road is closer to the axis because the wheels are smaller. This is why I still don't understand, intuitively at least, how torque arises from Newton's laws. I'm guessing I made a misstep somewhere, just trying to find it.

I think everything you said about the wheels is correct. For the same magnitude of acceleration of the car, the two friction forces must be of the same size (if the cars are of equal mass). The torque of the friction force about the axis for the larger wheel is also larger, for the same value of this acceleration. However, note that with a wheel being driven by an axle, it is the torque from the axle in the opposite direction to the torque of the friction force that is driving the angular acceleration in the same sense as the acceleration of the car (so the example is slightly more complicated than it seems!).

You could try working through the maths. By the rolling condition, you can first write ##r\alpha_1 = R\alpha_2##, where ##R>r## and go from there!

chananyag said:
Let me phrase it a more intuitive way. If two forces are equal they should produce the same tangential acceleration. The position of the force shouldn't matter for tangential acceleration. However, it does matter for angular acceleration.

That's right too! ##\tau_z = I_z\alpha_z## whilst ##F_{t} = ma_{t}##. It might help to think about angular velocities first; consider a particle moving in a circle of radius ##r##. In a time ##\delta t##, its path subtends an angle of ##\omega \delta t##, and the actual distance moved is ##v\delta t = r\omega \delta t##. For a fixed ##\omega##, the latter depends on the radius! If you hold one end of a rod 4.4 light years away from me at the other end and I turn the rod by 1 degree, you'll move a lot further than if the rod is 4.4 metres long! The same applies for angular accelerations; the same ##\alpha## at different radii results in different tangential accelerations.

You still have to be careful. If ##F_t = ma_t = mr\alpha##, we see the same tangential force applied to a particle at a larger radius produces a smaller angular acceleration, even though its torque is larger. This is not a contradiction; the moment of inertia ##I = mr^2## is larger too, so ##\alpha = \frac{\tau_z}{mr^2} = \frac{F_t r}{mr^2} = \frac{F_t}{mr}## is inversely proportional to ##r##.
 
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  • #27
etotheipi said:
If ##F_t = ma_t = mr\alpha##, we see the same tangential force applied to a particle at a larger radius produces a smaller angular acceleration, even though its torque is larger. This is not a contradiction; the moment of inertia ##I = mr^2## is larger too, so ##\alpha = \frac{\tau_z}{mr^2} = \frac{F_t r}{mr^2} = \frac{F_t}{mr}## is inversely proportional to ##r##.
But if τ=dL/dt shouldn't the angular acceleration be larger if the torque is larger?
 
  • #28
chananyag said:
But if τ=dL/dt shouldn't the angular acceleration be larger if the torque is larger?

If you fix the moment of inertia then you're right, ##\tau = I\alpha## so a larger torque yields a larger angular acceleration.

If we instead fix the tangential force, then the angular acceleration is greater for smaller radii (even though the torque of the force at the larger radius is larger!).
 
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  • #29
etotheipi said:
If we instead fix the tangential force, then the angular acceleration is greater for smaller radii (even though the torque of the force at the larger radius is larger!).
But then how can you derive torque from F=ma? Your derivation has become circular ...
 
  • #30
chananyag said:
But then how can you derive torque from F=ma? Your derivation has become circular ...

I showed how the concept of torque is defined in posts #7 and #15; nothing is circular, ##\vec{\tau}## is defined as ##\vec{r}\times \vec{F}##. With that you can come up with lots of other nice formulae. In #28 we're just looking at some consequences.
 
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  • #31
etotheipi said:
I showed how the concept of torque is defined in posts #7 and #15; nothing is circular, ##\vec{\tau}## is defined as ##\vec{r}\times \vec{F}##. With that you can come up with lots of other nice formulae. In #28 we're just looking at some consequences.
Sorry let me rephrase that. The derivation from F=ma doesn't explain why levers work. But then again, levers are useful because they perform more work with less force. So to explain levers I think you need the concept of work, but that doesn't mean you can't derive torque from F=ma.
 
  • #32
chananyag said:
Sorry let me rephrase that. The derivation from F=ma doesn't explain why levers work. But then again, levers are useful because they perform more work with less force. So to explain levers I think you need the concept of work, but that doesn't mean you can't derive torque from F=ma.

Hmm, suppose I have a massless see-saw type contraption with a hinge in the middle, and I put a mass on the end. I balance the torque of the mass on that end by exerting a force on the opposite end.

Now suppose I always exert a tiny bit more force than is required for equilibrium, so that the see-saw tilts and the mass rises, but with negligible acceleration, until the height of the mass increases by ##h##.

No matter where I apply this force on the other side of the see-saw, when the mass reaches the height ##h## I will have done the same amount of work on the system.

The reason why levers are useful is that I get to exert a lower force over a longer arc length if I push near the end. The total amount of work I do in this case to raise the height by ##h## is the same. Levers don’t create energy!
 
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  • #33
chananyag said:
Sorry let me rephrase that. The derivation from F=ma doesn't explain why levers work.

Side-note:
Analyse a lever in static equilibrium, so that ##\alpha=0##. You then know ##\tau=0## on the system. That is sufficient to start to reason as to why a lever multiplies force...
 
  • #34
chananyag said:
... But then again, levers are useful because they perform more work with less force...
Levers do not perform more work: work in = work out
They can only multiply force/torque at the expense of distance/angle.

Please, see:
https://en.wikipedia.org/wiki/Simple_machine
 
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  • #35
chananyag said:
... levers are useful because they perform more work with less force.
What do you mean by "more work" here?

chananyag said:
So to explain levers I think you need the concept of work,
I don't think you should need it, because levers function in equilibrium too, when no work is done.

The simplest possible rigid body (you need that for a lever) is a truss with 3 nodes (triangle), you should be able to solve the linear force equilibrium equations for all nodes. You don't need to assume the concept of torque here, it should follow from the external forces that give you the force equilibrium for all nodes.

As noted earlier, you can also "explain" levers by invoking the principle of energy conservation and the concept of virtual work. But these are additional assumptions, and a more indirect proof based on energy rather than forces.
 
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