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The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply a force to the wooden ball plus an extra force to $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply a force to the wooden ball plus an extra force to $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. But the force on the first and second ball is the same throughout the process. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply the force to the wooden ball plus an extra mass $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply a force to the wooden ball plus an extra force to $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron and lift it to the same height. The second process involves more work, since the second ball is heavier. But the force on the first and second ball is the same throughout the process. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply the force to the wooden ball plus an extra mass $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body

The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.

Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usally means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$. However, according to Newton's third law, for every internal force that a particle applies to its neighboors, there is a counter-force of same magnitude but opposite direction. Therefore one has $$ \sum_i \left( \sum \vec{F}_{i,\text{int}} \right) = 0 $$ such that the equation of motion for the center of mass becomes: $$ M~\vec{a}_c = \sum_i \left( \sum \vec{F}_{i,\text{ext}} \right) $$

So concerning your problem: If the two forces, are equal in magnitude and direction, there will be no difference in the acceleration of the center of mass, no matter what the point is on which the forces are applied (e.g. at the center of mass or tangential).

However, the energies of both systems will differ, as the second system will not only move linearly but also spin. This seems odd in the first place, since the magnitude of force was the same, but think about it like this:

Take a ball of wood an lift it in the air, then take a ball of iron with equal size and lift it to the same height. The second process involves more work, since the second ball is heavier. But the force on the first and second ball is the same throughout the process. So the extra work can be explained by the mass difference $\Delta m$ that has to be lifted in the second process. It's like you apply the force to the wooden ball plus an extra mass $\Delta m$.

This reasoning can be applied to your problem, too. If the force acts tangentially, it does not only act on the center of mass $M$ but also on the moment of inertia, because there is a torque due to this force.

edit according to: Equation of motion for the center of mass of a rigid body