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The acceleration a of the center of mass of a system is equal to F/M, where F is the net external force and M is the total mass of all the particles of the system, or at least that is my current understanding of it

A uniform rod in space is subjected to a force F. Now, this force is the net external force and the acceleration a of the center of mass should be equal to F/M, regardless of the distance of application x from the centre of mass.

However, the net external torque, T does depend on x, and a greater T increases angular velocity, which in turn increases the total kinetic energy of the object.

My question essentially is, why is a, the translational acceleration of the center of mass of the object, dependent on the point of application of the force? And if it is not dependent on that, and a is in fact the same no matter what the x, how is total kinetic energy different for the same force applied over the same small distance?

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5 Answers 5

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Kinetic energy of a rigid body is always expressed as sum of energy of center of mass(COM) and the rotational energy in frame of reference of COM . Its the latter energy which depends on torque.

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The first point is that the acceleration of $G$ do not depend of the point of application of the force$\overrightarrow{T}$.

If the force is not applicated to the center of mass , the rotating speed can increase or decrease.

But there is no paradox : if the rod is rotating, the power associated to the force $\overrightarrow{T}\cdot \overrightarrow{v}$ is greater (or smaller) that if it is not rotating with $\overrightarrow{T}$ applied in $G$ : $\overrightarrow{T}\cdot \overrightarrow{{{v}_{G}}}$.

Sorry for my english !

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first of all every force can provide torque as well as push or pull to the body.

Kinematical part

Now centre of mass is defined because the study we did is for particle(small objects) system but in reality the world consists of bodies of multiparticles so to ease our study we have defined the concept of centre of mass. Here is a link for complete derivations like how is it defined and why we show forces on centre of mass.

Rotational part

Now the point of application of the force provides a torque to the body about its centre of mass.Torque provides it an angular accleration.

Now these motions occur simultaneously and the quantity that connects them is time(youcan define distance travelled even, but that becomes a frame dependant quantity).

So these both motions carry energy. And the net mechanical energy of the body is the sum of these energies.

how is total kinetic energy different for the same force applied over the same small distance Now as you are harnessing more work from the force so the work done by the force would obviously be greater.

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If the given force is not directed toward the center of mass, the time required for the force to act through a given distance is less. Then the linear impulse is smaller and the resulting velocity of the center of mass is smaller. The rod gains rotational energy at the expense of translational energy.

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The answer to your question lies in the definition of translational momentum, and the special status the center of mass has.

Here are some things to consider:

  1. Rigid Body Motion if the distance between any two arbitrary points on the body is fixed, then the motion of the body as a whole can be decomposed as the translation of a single point, and a rotation about this point. The choice of point does not matter at this point. This is often called Chasle's Theorem.

    You can use the law of velocity transform $$\boldsymbol{v}_B = \boldsymbol{v}_A + \boldsymbol{\omega} \times ( \boldsymbol{r}_B - \boldsymbol{r}_A) \tag{1}$$ to find the transitional velocity of point B from the translational velocity of point A and the rotational velocity $\boldsymbol{\omega}$ of the body. Above $\boldsymbol{r}_A$ and $\boldsymbol{r}_B$ are the position vectors of these points.

  2. Translational Momentum The net translational momentum of a rigid body is defined as the sum of the individual momenta of each particle on the body. $$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i \tag{2}$$ where $m_i$ is the mass of each particle and $\boldsymbol{v}_i$ is the translational velocity of the particle.

    Given the rigid body motion, and the choice of an arbitrary point A on the body to describe the motion, the motion of each particle is given by (2) by $$\boldsymbol{v}_i = \boldsymbol{v}_A + \boldsymbol{\omega} \times \boldsymbol{d}_i$$ where $\boldsymbol{d}_i = \boldsymbol{r}_i - \boldsymbol{r}_A$ is the relative position of each particle to point A.

    Use that to find the net translational momentum of the body from (2) as $$ \begin{aligned}\boldsymbol{p}_{\rm net} & =\sum_{i}m_{i}\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{d}_{i}\right)\\ \boldsymbol{p}_{\rm net} & =\left(\sum_{i}m_{i}\right)\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\left(\sum_{i}m_{i}\boldsymbol{d}_{i}\right) \end{aligned} \tag{3} $$

  3. Center of Mass Now imagine for A we find a special point C such that $$\sum_i m_i \boldsymbol{d}_i = 0 \tag{4}$$, then the net translational momentum (3) reduces to

    $$\boldsymbol{p}_{\rm net}=m\,\boldsymbol{v}_{C} \tag{5}$$ where $m$ is the total mass of the body, and $\boldsymbol{v}_C$ is the translational velocity of this point.

    Such a point can be always be found by choosing the arbitrary point A from $$\boldsymbol{r}_A = \frac{ \sum_i m_i \boldsymbol{r}_i}{\sum m_i} \tag{6}$$

    You can prove the above with $$\small \begin{gathered}\sum_{i}m_{i}\left(\boldsymbol{r}_{i}-\boldsymbol{r}_{A}\right)=\\ =\sum_{i}m_{i}\left(\boldsymbol{r}_{i}-\frac{\sum_{j}m_{j}\boldsymbol{r}_{j}}{\sum_{j}m_{j}}\right)=\\ =\sum_{i}m_{i}\boldsymbol{r}_{i}-\sum_{i}m_{i}\frac{\sum_{j}m_{j}\boldsymbol{r}_{j}}{\sum_{j}m_{j}}=\\ =\sum_{i}m_{i}\boldsymbol{r}_{i}-\sum_{j}m_{j}\boldsymbol{r}_{j}=0 \end{gathered}$$

    Note that equation (4) becomes the definition of center of mass.

  4. Newton's 2nd Law Motion states that net force $\boldsymbol{F}$ is defined as the change in net momentum $\boldsymbol{p}$ over time $$\boldsymbol{F}_{\rm net} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p}_{\rm net} = m \tfrac{\rm d}{{\rm d}t} \boldsymbol{v}_C = m\,\boldsymbol{a}_C \tag{7}$$

    See how the acceleration of the center of mass has entered into the equations of motion. It is because we choose to decompose motion about the center of mass which simplifies the equations. This means that Newton's second law describes the motion of the center of mass only.

    But it does not have to be like so. You can keep using the motion of point the arbitrary point A which is not the center of mass point C.

    $$\begin{aligned}\boldsymbol{p}_{{\rm net}} & =m\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{c}\right)\\ \boldsymbol{F}_{{\rm net}} & =m\tfrac{{\rm d}}{{\rm d}t}\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{c}\right)\\ & =m\left(\boldsymbol{a}_{A}+\boldsymbol{\alpha}\times\boldsymbol{c}+\boldsymbol{\omega}\times\left(\boldsymbol{v}_{C}-\boldsymbol{v}_{A}\right)\right) \end{aligned} \tag{8}$$

    where $\boldsymbol{c} = \boldsymbol{r}_{C}-\boldsymbol{r}_{A}$. The above is equally as valid as equation (7), but a lot more complex.

    You are concerned of why the spatial location an applied force does not enter into the laws of motion, and the answer is the forces only describe the motion of the center of mass and not of the entire body. The spatial relationship missing is that the motion of each particle on the body differs by location, as opposed to a pure rotation where all the particles on the body share the same rotational motion, but the spatial position of an applied force drives this rotation.

  5. Rotational Momentum is defined again by summing up the individual rotational momentum of each particle, similar to (3), but about the center of mass (for simplification).

    $$\small \begin{aligned}\boldsymbol{L}_{C} & =\sum_{i}\left(\boldsymbol{d}_{i}\times m_{i}\,\boldsymbol{v}_{i}\right)\\ & =\sum_{i}\left(\boldsymbol{d}_{i}\times m_{i}\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{d}_{i}\right)\right)\\ & =\left(\sum_{i}m_{i}\boldsymbol{d}_{i}\right)\times\boldsymbol{v}_{A}+\sum_{i}m_{i}\left(\boldsymbol{d}_{i}\times\left(\boldsymbol{\omega}\times\boldsymbol{d}_{i}\right)\right)\\ & =\sum_{i}m_{i}\left(\boldsymbol{\omega}\left(\boldsymbol{d}_{i}\cdot\boldsymbol{d}_{i}\right)-\boldsymbol{d}_{i}\left(\boldsymbol{d}_{i}\cdot\boldsymbol{\omega}\right)\right)\\ & =\sum_{i}m_{i}\left(\boldsymbol{d}_{i}\cdot\boldsymbol{d}_{i}-\boldsymbol{d}_{i}\odot\boldsymbol{d}_{i}\right)\boldsymbol{\omega}\\ & ={\rm I}_{C}\,\boldsymbol{\omega} \end{aligned} \tag{9}$$ which leads to the definition of the mass moment of inertia tensor about the center of mass as $${\rm I}_C = \sum_{i}m_{i}\left(\boldsymbol{d}_{i}\cdot\boldsymbol{d}_{i}-\boldsymbol{d}_{i}\odot\boldsymbol{d}_{i}\right) \tag{10} $$ where $\cdot$ is the vector dot product, and $\odot$ the vector outer product. There is an 3×3 identity matrix converting the scalar dot product into a matrix in there, that I am omitting for clarity.

  6. Euler Law of Rotation describes the motion of a rigid body about the center of mass by equating the net torque to the change in rotational momentum.

    $$\boldsymbol{\tau}_{C}=\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{C}={\rm I}_{C}\boldsymbol{\alpha}+\boldsymbol{\omega}\times\boldsymbol{L}_{C} \tag{11}$$

    Of course the choice of which point to decompose motion over is ours, and similarly with (8) we can describe the net torque about an arbitrary point A in terms of the motion of the body

    $$ \small \begin{aligned}\boldsymbol{\tau}_{A} & =\boldsymbol{\tau}_{C}+\boldsymbol{c}\times\boldsymbol{F}_{{\rm net}}\\ & ={\rm I}_{C}\boldsymbol{\alpha}+\boldsymbol{\omega}\times\boldsymbol{L}_{C}+\boldsymbol{c}\times m\left(\boldsymbol{a}_{A}+\boldsymbol{\alpha}\times\boldsymbol{c}+\boldsymbol{\omega}\times\left(\boldsymbol{v}_{C}-\boldsymbol{v}_{A}\right)\right)\\ & =m\boldsymbol{c}\times\boldsymbol{a}_{A}+{\rm I}_{C}\boldsymbol{\alpha}-m\,\boldsymbol{c}\times\left(\boldsymbol{c}\times\boldsymbol{\alpha}\right)+\boldsymbol{\omega}\times\boldsymbol{L}_{C}+\boldsymbol{c}\times\left(\boldsymbol{\omega}\times\left(\boldsymbol{p}-m\boldsymbol{v}_{A}\right)\right)\\ & =m\boldsymbol{c}\times\left(\boldsymbol{a}_{A}-\boldsymbol{\omega}\times\boldsymbol{v}_{A}\right)+{\rm I}_{C}\boldsymbol{\alpha}-m\,\boldsymbol{c}\times\left(\boldsymbol{c}\times\boldsymbol{\alpha}\right)+\boldsymbol{\omega}\times\boldsymbol{L}_{A}+\boldsymbol{v}_{A}\times\boldsymbol{p} \end{aligned} \tag{12}$$

    which is some much more complex than (11).

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