4
$\begingroup$

I have seen a lot of questions here which ask why a free rigid body always rotates about it's centre of mass. The answer in most cases is like a "thought experiment". First, we prove that when a force is applied to a rigid body, it behaves like a point object where the entire mass of the object is concentrated to one point called the "centre of mass". Then we transfer out attention to a co-ordinate system at the centre of mass (so that the centre of mass is at rest, relatively). Then we say that the definition of a rigid body is that the distance between the particles of the rigid body always remain constant. This means that the distance between the centre of mass, and any point in the rigid body also remains constant. So, the only possible motion of any point will be a circular path around the centre of mass : hence, the only possible motion of a rigid body about the centre of mass is a rotation. Also, since the distance between any points in the rigid body must be constant, particles inside the rigid body cannot rotate in opposite directions or different axes, as this would change the distances.

Now, I have also been taught this way. In school and university, even in our Dynamics text book (Meriam & Kraige), the concept of "rotation" and "moment" was just introduced..like its common sense. There was no "mathematical proof" that rotation is the motion around the centre of mass (CM). Rotation and translation are always treated differently, even though its taught that the net motion will be a sum of the two.

I have been wondering whether you can prove that the motion of a particle in a rigid body with respect to the centre of mass is a rotation. I have come up with a kind of half baked derivation below :

First, as always we consider a rigid body as a system of particles connected with massless rigid rods. For simplicity I have considered only the 2D case. In the figure below, I have considered a 3 particle system, with all relevant variables marked.

3 particle system of particles with applied force

The red point is the centre of mass (CM) of the system. Here a force $\vec f$ is applied to mass $m_1$ which does not pass through the CM. So, this system would rotate.

To apply the principles of dynamics, we first isolate all masses and draw the free body diagram

free body diagram of mass 1

Here $\vec f_{12}$ and $\vec f_{13}$ are the reaction forces on $m_1$ from $m_2$ and $m_3$. Applying newtons second law to $m_1$ we have $$\vec f + \vec f_{12} + \vec f_{13} = m_1\ddot{\vec r_1}$$

For mass $m_3$

free body diagram of mass 3

we have $$\vec f_{31} + \vec f_{32} = m_3\ddot{\vec r_3}$$

and for mass $m_2$

free body diagram of mass 2

we have $$\vec f_{21} + \vec f_{23} = m_2\ddot{\vec r_2}$$

Now adding up all the above equations and noting that $\vec f_{12}=-\vec f_{21}$ and $\vec f_{13}=-\vec f_{31}$ and $\vec f_{32}=-\vec f_{23}$, we have $$\vec f=m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$$ Introducing the position of centre of mass as $$\vec r_{cm}=\frac{m_1\vec r_1+m_2\vec r_2+m_3\vec r_3}{m_1+m_2+m_3}$$ and differentiating $$\ddot{\vec r_{cm}}=\frac{m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}}{m_1+m_2+m_3}$$ Now we can substitute for $m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$ in the dynamic equation to get $$\vec f=(m_1+m_2+m_3)\ddot{\vec r_{cm}}$$ This is nothing but the equation of motion of a point particle whose mass is $m_1+m_2+m_3$ situated at position $\vec r_{cm}$. Thus, the rigid body behaves like the entire mass is concentrated at the centre of mass. Now we turn our attention to the centre of mass co-ordinate system $x_{cm} - y_{cm}$. To do this we note that $\vec r_1=\vec r_{cm}+\vec r_{1c}$ and $\vec r_2=\vec r_{cm}+\vec r_{2c}$ and $\vec r_3=\vec r_{cm}+\vec r_{3c}$ Substituting for $\vec r_1$, $\vec r_2$ and $\vec r_3$ in the dynamic equation for each mass, we have $$\vec f + \vec f_{12} + \vec f_{13} - m_1\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}\\\vec f_{31} + \vec f_{32} - m_3\ddot{\vec r_{cm}}= m_3\ddot{\vec r_{3c}}\\\vec f_{21} + \vec f_{23} - m_2\ddot{\vec r_{cm}}= m_2\ddot{\vec r_{2c}}$$ Again adding all of the above up, we have $$\vec f-(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}+m_2\ddot{\vec r_{2c}}+m_3\ddot{\vec r_{3c}}$$ Now we invoke the definition of rigid body. This means that the distance between any 2 masses is constant. This may be written for our case as $$\frac {d}{dt}\left(\vec r_{12}\cdot\vec r_{12}\right)=0$$ since the magnitude of the vector between any 2 masses is constant. However $\vec r_{12}=\vec r_{2c}-\vec r_{1c}$. So we have $$\frac {d}{dt}\left[\left(\vec r_{2c}-\vec r_{1c}\right)\cdot\left(\vec r_{2c}-\vec r_{1c}\right)\right]=0\\\frac {d}{dt}\left[{\vert\vec r_{2c}\vert}^2+{\vert\vec r_{1c}\vert}^2-2\vec r_{2c}\cdot\vec r_{1c}\right]=0$$ This essentially means that $$\frac {d}{dt}\left[\vec r_{2c}\cdot\vec r_{1c}\right]=0$$ Applying product rule $$\vec r_{2c}\cdot\dot{\vec r_{1c}}+\vec r_{1c}\cdot\dot{\vec r_{2c}}=0$$ Differentiating once again, $$\vec r_{2c}\cdot\ddot{\vec r_{1c}}+\vec r_{1c}\cdot\ddot{\vec r_{2c}}+2\dot{\vec r_{1c}}\cdot\dot{\vec r_{2c}}=0$$ Since the last term is a product of derivatives, we say that it is infinitesimally small, and ignore it. This gives $$\vec r_{2c}\cdot\ddot{\vec r_{1c}}+\vec r_{1c}\cdot\ddot{\vec r_{2c}}=0$$ Applying same treatment for $\vec r_{13}$, we have $$\vec r_{1c}\cdot\ddot{\vec r_{3c}}+\vec r_{3c}\cdot\ddot{\vec r_{1c}}=0$$ From the above 2 equations, we can write $$\ddot{\vec r_{2c}}=\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}\\\ddot{\vec r_{3c}}=\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Substituting for $\ddot{\vec r_{2c}}$ and $\ddot{\vec r_{3c}}$ in the summed up dynamic equation, we get $$\vec f-(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Now we focus on the term $(m_1+m_2+m_3)\ddot{\vec r_{cm}}$. From the definition of the centre of mass, we have $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_1}+m_2\ddot{\vec r_2}+m_3\ddot{\vec r_3}$$ We will now proceed to invoke the rigid body condition the same way we did above, by noting that $\vec r_{12}=\vec r_2-\vec r_1$ and that $\vec r_{13}=\vec r_1-\vec r_3$. After applying the same treatment as above, we get $$\ddot{\vec r_2}=\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}\\\ddot{\vec r_3}=\frac{-\vec r_1\cdot\vec r_3\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}$$ Substituting these into the centre of mass definition above, we have $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=m_1\ddot{\vec r_1}+m_2\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}+m_3\frac{-\vec r_1\cdot\vec r_3\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}$$. Now, if we take the common term $\ddot{\vec r_1}$ apart, all the other terms on RHS are scalar products. So we may write $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=K_1\ddot{\vec r_1}$$ where $$K_1=\frac{m_1{\vert\vec r_1\vert}^2-m_2\vec r_1\cdot\vec r_2-m_3\vec r_1\cdot\vec r_3}{{\vert\vec r_1\vert}^2}$$ Now we make the observation $$\vec r_{2c}-\vec r_{1c}=\vec r_2-\vec r_1=\vec r_{12}$$Differentiating twice, we have $$\ddot{\vec r_{2c}}-\ddot{\vec r_{1c}}=\ddot{\vec r_2}-\ddot{\vec r_1}$$ Substituting for $\ddot{\vec r_{2c}}$ in terms of $\ddot{\vec r_{1c}}$ and $\ddot{\vec r_2}$ in terms of $\ddot{\vec r_1}$ as derived above, we have $$\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}-\ddot{\vec r_{1c}}=\frac{-\vec r_1\cdot\vec r_2\cdot\ddot{\vec r_1}}{{\vert\vec r_1\vert}^2}-\ddot{\vec r_1}$$ Again, we can notice that after taking term $\ddot{\vec r_{1c}}$ on LHS as common and taking term $\ddot{\vec r_1}$ on RHS as common, what left inside the brackets will be a scalar term. So we write $$\ddot{\vec r_1}=K_2\ddot{\vec r_{1c}}$$ So finally we may write $$(m_1+m_2+m_3)\ddot{\vec r_{cm}}=K_3\ddot{\vec r_{1c}}$$ where $$K_3=K_1K_2$$ Now we can substitute for the term $(m_1+m_2+m_3)\ddot{\vec r_{cm}}$ in the summed dynamic equation which becomes $$\vec f-K_3\ddot{\vec r_{1c}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\vec r_{2c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\vec r_{3c}\cdot\ddot{\vec r_{1c}}}{{\vert\vec r_{1c}\vert}^2}$$ Now I am going to do what is called a "pro gamer move". Since scalar product is commutative, I will group the terms in RHS as $$\vec f-K_3\ddot{\vec r_{1c}}=m_1\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\cdot\left(\vec r_{2c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\cdot\left(\vec r_{3c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}$$ Now, the terms in the brackets are scalar products; which means that the second and third terms in the RHS are vectors in the direction of $\vec r_{1c}$ Now to remove those additional terms, I take a cross product with $\vec r_{1c}$ on both LHS and RHS. $$\vec r_{1c}\times\vec f-K_3\vec r_{1c}\times\ddot{\vec r_{1c}}=m_1\vec r_{1c}\times\ddot{\vec r_{1c}}+m_2\frac{-\vec r_{1c}\times\vec r_{1c}\cdot\left(\vec r_{2c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}+m_3\frac{-\vec r_{1c}\times\vec r_{1c}\cdot\left(\vec r_{3c}\cdot\ddot{\vec r_{1c}}\right)}{{\vert\vec r_{1c}\vert}^2}$$ In this case, since the second and third terms in RHS before cross product where vectors in the direction of $\vec r_{1c}$, taking cross product means that these terms will be $0$. Thus finally we have $$\vec r_{1c}\times\vec f=\left(m_1+K_3\right)\left(\vec r_{1c}\times\ddot{\vec r_{1c}}\right)$$ Which is nothing but $$\tau=I\alpha$$ where I call the bracketed term as $I$ (moment of inertia). So I have obtained the moment equation in the centre of mass co-ordinate system. I have the following questions :

  1. Even though I set out to prove that the motion of $m_1$ will be circular, I didn't quite reach there. Does the moment equation prove that $m_1$ will have circular motion ?
  2. Is what I have done correct ?

The answer that satisfied me is written by Claudio Saspinski below (Thankyou very much). Rotation is a basic type of motion like translation, where one point of the rigid body is fixed (relatively). So there is no need to "prove" that the motion of a rigid body is rotation, all bodies rotate about some point on it. As stated in the comment, any point on the rigid body can be taken as the centre of rotation of the body. It is proved that no matter what this point is, the angular velocity is the same, by applying the conditions of a rigid body. So we can take any point other than the centre of mass as the rotation reference. The reason we take the centre of mass as the centre of rotation in my current understanding, is for ease of analysis of motion. This is because the motion of the CM is just like a point object under applied force and is easy to handle. If we take the centre of rotation as another point, then to get the configuration of the body at a later time t, we need the position of this point at that time, which is more involved, as this point does not behave like a point body under applied force.

So in conclusion, my question is kind of not correct or not valid. Centre of rotation of a body is any point you choose to be.

$\endgroup$
5
  • $\begingroup$ Finding the momentum expressions for a rigid body as a collection of particles is far simpler than your effort, and is also standard course material for college level dynamics class. $\endgroup$ Sep 20, 2020 at 3:38
  • $\begingroup$ Thanks for the reference, but, as I said, it doesn't "prove" that the motion of a rigid body about the centre of mass is a rotation. It just introduces the equations as a "given" entity without any consideration of the philosophy behind it $\endgroup$
    – S_holmes
    Sep 23, 2020 at 18:28
  • $\begingroup$ I disagree. The equations are not just given but derived from 1st principles and they prove exactly what you ask for. I am not sure what the difficulty here is, but this is nothing new here. $\endgroup$ Sep 23, 2020 at 21:16
  • $\begingroup$ In the material, rotation and translation are treated differently..just that. If I say that I define angular momentum in a certain way, then I can go ahead and differentiate..that's not the issue here. I was just saying that there is no "proof" that the only motion possible about the centre of mass is a rotation $\endgroup$
    – S_holmes
    Oct 1, 2020 at 18:31
  • $\begingroup$ See physics.stackexchange.com/questions/311606/… $\endgroup$
    – mike stone
    Nov 30, 2021 at 21:14

3 Answers 3

1
$\begingroup$

The point is not that a unconstrained rigid body rotates around its COM. In reality, for any time $t$, a rigid body (constrained or not) always rotates around any of its points with an angular velocity $\boldsymbol \omega (t)$.

Suppose $3$ generic points $P_0, P_1, P_2$ that belongs to the rigid body. And lets $P_0$ be by hypothesis the center of rotation. Let's $\mathbf {r_1}$ and $\mathbf {r_2}$ be the position vectors of $P_1$ and $P_2$ in a non rotating frame where $P_0$ is the origin.

As the distances between the points doesn't change, The modulus of the position vectors are constant. That leads to: $\mathbf {r_1.v_1} = \mathbf {r_2.v_2} = 0$

The distance between points $P_1$ and $P_2$ also doesn't change: $(\mathbf {r_1} - \mathbf {r_2})\mathbf .(\mathbf {v_1} - \mathbf {v_2}) = 0$

From that equations we get: $\mathbf {r_1.v_2} + \mathbf {r_2.v_1} = 0$

We can define $3$ vectors $\boldsymbol \omega_1$, $\boldsymbol \omega_2$, and $\boldsymbol \omega_3$ such that: $\mathbf {v_1} = \boldsymbol \omega_1 \times \mathbf {r_1}$, $\mathbf {v_2} = \boldsymbol \omega_2 \times \mathbf {r_2}$, and $\mathbf {v_1} - \mathbf {v_2} = \boldsymbol \omega_3 \times (\mathbf {r_1} - \mathbf {r_2})$

$\mathbf {r_2.v_1} = \mathbf{r_2.}(\boldsymbol \omega_1 \times \mathbf {r_1}) = \boldsymbol \omega_1 \times \mathbf {r_1.r_2}$

$\mathbf {r_1.v_2} = \mathbf{r_1.}(\boldsymbol \omega_2 \times \mathbf {r_2}) = \boldsymbol \omega_2 \times \mathbf {r_2.r_1}$

$\mathbf {r_2.v_1} + \mathbf {r_1.v_2} = 0 \implies \boldsymbol \omega_1 \times \mathbf {r_1.r_2} + \boldsymbol \omega_2 \times \mathbf {r_2.r_1} = 0 \implies \boldsymbol \omega_1 = \boldsymbol \omega_2$

That equality leads to:

$\mathbf {v_1} - \mathbf {v_2} = \boldsymbol \omega_1 \times \mathbf {r_1} - \boldsymbol \omega_2 \times \mathbf {r_2} = \boldsymbol \omega_1 \times \mathbf {r_1} - \boldsymbol \omega_1 \times \mathbf {r_2} = \boldsymbol \omega_1 \times (\mathbf {r_1} - \mathbf {r_2})$

So, $\boldsymbol \omega_1 = \boldsymbol \omega_2 = \boldsymbol \omega_3 = \boldsymbol \omega$

The choice of $P_1$ and $P_2$ was arbitrary, so the conclusion is valid for all the rigid body. The existence of an unique angular velocity implies that for each time $t$, all points of the rigid body rotates around a generic point of it, chosen arbitrarily as origin of a non rotating frame.

If $P_0$ is the center of mass: $m_1\mathbf {r_1} + m_2\mathbf {r_2} + ... + m_n\mathbf {r_n} = 0$

Taking the derivative with respect to time: $m_1\mathbf {v_1} + m_2\mathbf {v_2} + ... + m_n\mathbf {v_n} = 0$. So the momentum of the body for a non rotating frame attached to the COM is always zero. If the body is unconstrained, that means without external forces applied, its total momentum must be constant for any inertial frame:

$\mathbf p_{tot} = m_1(\mathbf {v_1 + u}) + m_2(\mathbf {v_2 + u}) + ... + m_n(\mathbf {v_n + u}) = \sum m_i\mathbf v_i + \sum m_i\mathbf u = 0 + M\mathbf u$

where $\mathbf u$ is the velocity of the COM with respect to the inertial frame, and $M$ the total mass of the body.

Conclusion: in that case the momentum of the rigid body can be taken as the product of its mass by the velocity of the COM.

$\endgroup$
10
  • $\begingroup$ Thanks! So it means that we do not have to prove that the motion around centre of mass is a rotation right? We can take any point on the rigid body as the centre of rotation, But we stick to the centre of mass as we can decouple the translation of the body from its rotation if we do so. $\endgroup$
    – S_holmes
    Jan 23 at 11:39
  • $\begingroup$ @S_holmes Exactly. $\endgroup$ Jan 23 at 13:08
  • $\begingroup$ How does $[(\boldsymbol{\omega_{2}} - \boldsymbol{\omega_{3}})\times \mathbf{r_{2}}]\cdot \mathbf{r_{1}} = 0$ imply that $\boldsymbol{\omega_{2}} = \boldsymbol{\omega_{3}}$? Using the scalar triple product identity, I see that the expression can be simplified to $(\boldsymbol{\omega_{2}} - \boldsymbol{\omega_{3}})\cdot(\mathbf{r_{2}} \times \mathbf{r_{1}})$ So the omegas are equal iff no three particles of the rigid body are collinear( which is obviously false) and the two operands of the dot product are not orthogonal for all $P_{0}, P_{1} , P_{2}$ in the rigid body. What am I missing here? $\endgroup$
    – Curious
    Feb 22 at 17:56
  • $\begingroup$ @Curious I edited the answer to be more clear. $\endgroup$ Feb 22 at 22:31
  • $\begingroup$ @ClaudioSaspinski Where is the edit? I have a problem with the line “Making the dot product with $\mathbf{r_{1}}$, the LHS is zero which implies that $\boldsymbol{\omega_{2}} = \boldsymbol{\omega_{3}}$” Can you explicitly write the edit at the bottom of the answer as I can’t see any edits regarding that? $\endgroup$
    – Curious
    Feb 23 at 5:59
0
$\begingroup$

For the rotation i will take the sum of the torques about the center of mass, you obtain:

$$\vec{r}_{1c}\times \sum \vec F_1+\vec{r}_{2c}\times \sum \vec F_2+\vec{r}_{3c}\times \sum \vec F_3=\frac{d}{dt}\left(I\,\vec{{\omega}}\right)$$

with

$$\sum \vec F_1=\vec f+\vec f_{12}+\vec{f}_{13}$$ $$\sum \vec F_2=-\vec f_{12}+\vec{f}_{23}$$ $$\sum \vec F_3=-\vec f_{23}-\vec{f}_{13}$$

for rigid body is $~\vec{f}_{ij}$ equal zero

you obtain

$$\vec{r}_{1c}\times \vec{f}=\vec\tau_{\text{CM}}=\frac{d}{dt}\left(I\,\vec{{\omega}}\right)$$

where $I$ is the inertia tensor of the rigid body taken at the COM and $\vec\omega$ the angular velocity of the COM

$\endgroup$
5
  • $\begingroup$ I don't understand how you combined the terms in LHS to eliminate other force values. Like how can you combine cross products ? $\endgroup$
    – S_holmes
    Sep 20, 2020 at 17:51
  • $\begingroup$ I think that $a\times (b+c)=a\times b+a\times c$ $\endgroup$
    – Eli
    Sep 20, 2020 at 19:21
  • $\begingroup$ but how did you apply it..i know I seem dumb..but I dont get it..like..the cross products are with $\vec r_{2c}$ and $\vec r_{3c}$ right ? how did those terms disappear ? $\endgroup$
    – S_holmes
    Sep 23, 2020 at 18:21
  • $\begingroup$ They disappear because $~\vec{f}_{ij}$ equal zero for a rigid body $\endgroup$
    – Eli
    Sep 23, 2020 at 18:31
  • $\begingroup$ But how did you combine the LHS terms to get $\vec f_{ij}$ in the LHS..could you please show me the intermediate steps? $\endgroup$
    – S_holmes
    Oct 1, 2020 at 18:32
0
$\begingroup$

In some sense, the center of mass is defined as the point in which a pure torque will force a body to rotate about, just as a force through the center of mass (and hence no net torque) forces the body to purely translate. You can see that those two statements are equivalent to each other, and proving one proves the other.

The root of all of this is the definition of momentum and angular momentum of a rigid body as a collection of particles that are fixed to each other. The center of mass is exactly the only point in space which de-couples the linear from the rotational momentum such that momentum describes the motion of the center of mass, and angular momentum the motion about the center of mass.

In this answer to Why does a body not rotate if force is applied on the centre of mass? I describe how the decomposition of the position (and hence motion) of each particle $i$ into the position of the center of mass $\boldsymbol{r}_{\rm COM}$ plus the relative position from the center of mass $\boldsymbol{d}_i$ allows us to use the simplification $\sum_i m_i \boldsymbol{d}_i = \boldsymbol{0}$ as the definition for the center of mass, and how this leads to the following expressions for momenta

  • Linear Momentum $$\boldsymbol{p} = m \, \boldsymbol{v}_{\rm COM} \tag{1}$$
  • Angular Momentum $$\boldsymbol{L}_{\rm COM} = \mathbf{I}_{\rm COM} \boldsymbol{\omega} \tag{2}$$

The important point from the above is that they are completely decoupled in the sense that momentum $\boldsymbol{p}$ does not depend on the rotation $\boldsymbol{\omega}$ and that angular momentum $\boldsymbol{L}_{\rm COM}$ does not depend on the motion of the center of mass $\boldsymbol{v}_{\rm COM}$.

Now forces and torques are the time derivatives of momentum and angular momentum, are are also completely decoupled between linear and rotational motion only when expressed at the center of mass.

To see the mathematically, consider a short-lived strong force that causes an impulse in vector form $\boldsymbol{J}= \int \boldsymbol{F} \, {\rm d}t$ applied at some location $\boldsymbol{r}$ not at the center of mass. The effect is going to be an instanteneous change in motion in terms of $\Delta \boldsymbol{v}_{\rm COM}$ and $\Delta \boldsymbol{\omega}$ as a result of this impulse directy changing the momenta of the body.

  • Linear motion $$ \Delta \boldsymbol{v}_{\rm COM} = \tfrac{1}{m} \boldsymbol{J} \tag{3}$$
  • Rotational motion $$ \Delta \boldsymbol{\omega} = \mathbf{I}_{\rm COM}^{-1} (\boldsymbol{r} \times \boldsymbol{J}) \tag{4}$$

Notice that (3) is the inverse of (1) and (4) is the inverse of (2) since $(\boldsymbol{r} \times \boldsymbol{J})$ is the net moment of impulse on the center of mass due to impulse being away from the center of mass.

So to answer your question, when a force is applied away from the center of mass is causes a change in both linear and rotational motion, but if the same force goes through the center of mass (and hence $\boldsymbol{r}=\boldsymbol{0}$) then only the linear motion is affected.

Now consider a different case where the forces is zero $\boldsymbol{J}=\boldsymbol{0}$, but there is still non-zero a net moment of impulse $\boldsymbol{\Gamma} \neq \boldsymbol{0}$ causing (3) to be $\Delta \boldsymbol{v}_{\rm COM} = \boldsymbol{0}$ and (4) to be $\Delta \boldsymbol{\omega} = \mathbf{I}_{\rm COM}^{-1} \boldsymbol{\Gamma} \neq \boldsymbol{0}$.

This is the case where the body starts to rotate, but the center of mass does not change motion. This is the only case where this can happen. Only when the net force is zero and the net torque is not zero.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.