# Confusion regarding rotational motion!

Let us assume I have a rod of some mass m, moment of inertia I, length l and center C.

If I apply a force F on C for a duration of time t, it will accelerate forward. If I apply it elsewhere, the body will rotate. Now, my confusion arises here:

• Will the body still accelerate as much as it did when F was applied on C?
• If so, why?
• How can it rotate and yet accelerate with the same velocity as that of without rotation?
• Where is it getting this extra energy from?
• How can I calculate the respective velocities of linear and angular motion given F and t?

Please cite any reliable sources if possible and state all the formulas.

• Apr 6, 2015 at 13:06

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from $F = \mathrm{d}p/\mathrm{d}t$) and its angular momentum will become approximately $$rF\delta t$$ (from $\tau = rF\sin\theta = \mathrm{d}L/\mathrm{d}t$).

You can then calculate the rod's kinetic energy after this process:

$$K = K_\text{lin} + K_\text{rot} = \frac{p^2}{2m} + \frac{L^2}{2I} = \frac{F^2 \delta t^2}{2m} + \frac{r^2 F^2 \delta t^2}{2mL^2/12} = \frac{F^2 \delta t^2}{2m} + \frac{6r^2F^2}{mL^2}\delta t^2\tag{1}$$

where $L$ is the length of the rod. If you hit the rod away from the center of mass, it acquires more energy.

Now, why is that? Well, let's think about what happens if we do consider the fact that the rod's orientation changes as the force is being applied to it, if the force is off center. We know the rod acquires an angular momentum $rF\delta t$ in time $\delta t$, which corresponds to angular velocity

$$\omega = \frac{L}{I} = \frac{rF\delta t}{mL^2/12} = \frac{12rF}{mL^2}\delta t$$

Assuming the torque is constant, the angular acceleration will also be constant,

$$\alpha = \frac{\Delta\omega}{\Delta t} = \frac{12rF}{mL^2}$$

and for constant angular acceleration we can figure out the total angular displacement as

$$\Delta \phi = \frac{1}{2}\alpha\Delta t^2 = \frac{6rF}{mL^2}\delta t^2$$

The linear displacement corresponding to this angular displacement is just

$$r\Delta\phi = \frac{6r^2 F}{mL^2}\delta t^2$$

which means that when you apply the force off center, the point at which you are applying the force moves $\frac{6r^2 F}{mL^2}\delta t^2$ further than when you apply the force at the center. (Of note: this is zero when $r=0$, as it should be.)

Work is force times distance, when the force is constant, so this means the off-center force does a small amount of extra work relative to the on-center force because of the increased distance:

$$W = \frac{6r^2 F^2}{mL^2}\delta t^2$$

which is exactly the same as the extra amount of energy that the rod acquires when you apply the force off center, from equation (1). This is the origin of that extra energy: the extra distance that the point of application of the force moves.

I'll make an example, to make things clear.

Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$| \vec{r}_1 - \vec{r}_2 | = d$$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus 1). Let's assume that we wanna look at this system when it is initially at rest and the applied external forces will always point perpendicular to the axis of rotation. This means that the system's movement will be constrained to a plane. Let's take a coordinate system such that this movement is in the x-y-plane.

We have therefore 3 effective degrees of freedom (4 minus 1 constraint). One is allowed to choose arbitrary independent variables $q_j$ corresponding to these degrees of freedome.

Take $q_1 := x$ the x-coordinate of the center of mass, $q_2 := y$ the y-coordinate and $q_3 := \phi$ the angle of rotation of the vector joining the two particles relative to the x-axis $$\cos\phi = \frac{\vec{d}\cdot\vec{e}_x}{d}$$ with $\vec{d} = \vec{r}_1 - \vec{r}_2$ and $\vec{e}_x$ the unit basis vector pointing along the x-axis. Then we have: \begin{align} \vec{r}_1 & = \underbrace{\left( \matrix{x\\y} \right)}_{:=\vec{r}_c} + \underbrace{\frac{d}{2} \cdot \left( \matrix{\cos\phi \\ \sin\phi} \right)}_{=\vec{d}/2} \\ \\ \vec{r}_2 & = \left( \matrix{x\\y} \right) - \frac{d}{2} \cdot \left( \matrix{\cos\phi \\ \sin\phi} \right) \end{align}

The equations of motion are given by d'Alembert's principle, which states that the internal forces of constraints do no net work on the body: $$\sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \cdot \frac{\partial \vec{r}_i}{\partial q_j} = 0$$ which gives for every $j=1,2,3$ one equation of motion. For $q_1 = x$ we have: \begin{align} & \sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \cdot \left( \matrix{1\\0} \right) = \left( \sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \right) \cdot \left( \matrix{1\\0} \right) \\ & = \left( m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} + m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{2,\text{ext}} + m \frac{d^2(\vec{d}/2)}{dt^2} - m \frac{d^2(\vec{d}/2)}{dt^2} \right) \cdot \left( \matrix{1\\0} \right) \\ & = \left( 2m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{1\\0} \right) = 0 \end{align}

And similar for $q_2 = y$: $$\left( 2m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{0\\1} \right) = 0$$

Or equivalently to accord for both equations: $$M \frac{d^2\vec{r}_c}{dt^2} = \sum_i \vec{F}_{i,\text{ext}} \tag1$$

the equation of motion of the center of mass $\vec{r}_c$ with $M = 2m$. This shows, that the motion of the center of mass is not affected by the point on which the force is applied (e.g $\vec{F}_{1,\text{ext}} = \vec{F}_0$ and $\vec{F}_{2,\text{ext}} = 0$ leads to the same motion as $\vec{F}_{1,\text{ext}} = 0$ and $\vec{F}_{2,\text{ext}} = \vec{F}_0$). The center of mass moves according to the external forces that are applied. This is a general result, which was for instance answered here:

Equation of motion for the center of mass of a rigid body

Now let's take the last equation for $q_3 = \phi$: \begin{align} & \left( m \frac{d^2\vec{r}_c}{dt^2} + m \frac{d^2(\vec{d}/2)}{dt^2} - \vec{F}_{1,\text{ext}} \right) \cdot \left( \matrix{-\sin\phi \\ \cos\phi} \right) \\ & + \left( m \frac{d^2\vec{r}_c}{dt^2} - m \frac{d^2(\vec{d}/2)}{dt^2} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{\sin\phi \\ -\cos\phi} \right) \\ & = \left( m \frac{d^2(\vec{d})}{dt^2} - \vec{F}_{1,\text{ext}} + \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{-\sin\phi \\ \cos\phi} \right) = 0 \end{align} this is independent of $\vec{r}_c$ and therefore a pure differential equation for $\phi$, which describes the rotational movement. We see that, the rotation around the center of mass is independent of its linear motion, in the sense that it can be described by equations that ignore the movement of the center of mass. One can express that in terms of torque and angular momentum $\vec{L}_r$ relative to the center of mass $$\frac{d \vec{L}_r}{dt} = \sum_i \vec{d}_i \times \vec{F}_{i,\text{ext}} \tag2$$ with $\vec{L}_r = \Theta \cdot \vec{\omega}$ and $\Theta$ the inertia tensor and $\vec{\omega}$ the angular velocity (e.g. $\frac{d\phi}{dt}$ in the above example) at the center of mass and $d_i$ the distances to it.

Regarding your energy problem, I refer you to the last section of this answer:

https://physics.stackexchange.com/a/174208/75518

or think about it like this: Work is defined by the path-integral $\int \vec{F} \cdot d\vec{r}$ or for a many body, rigid system, since the net work of internal forces is zero: \begin{align} \sum_i \int \vec{F}_{i,\text{ext}} \cdot d\vec{r}_i & = \sum_i \int \vec{F}_{i,\text{ext}} \cdot \frac{d\vec{r}_i}{dt} dt \\ & = \int \sum_i \left( \vec{F}_{i,\text{ext}} \cdot \frac{d (\vec{r}_c + \vec{r}_{i,\text{rot}})}{dt} \right) dt \\ & = \int \left( \sum_i\vec{F}_{i,\text{ext}} \right) \cdot \vec{v}_c~dt + \int \left( \sum_i\vec{F}_{i,\text{ext}} \cdot \vec{v}_{i,\text{rot}} \right)~dt \tag3 \end{align} where we have split the motion of a particle into the linear motion of the center of mass, which every particle shares, plus the rotational motion. In the above example this is $\vec{r}_{i,\text{rot}} = \pm \frac{\vec{d}}{2}$. As you can see, the first term corresponds to the work done on the center of mass, while the second corresponds to the work done to let the system rotate. It is this extra displacement $d \vec{r}_{i,\text{rot}}$ that comes with rotation, that causes the system to gain more kinetic-energy in the same ammount of time from a given force of equal magnitude.

Conclusions:

• "Will the body still accelerate as much as it did when F was applied on C?" $\rightarrow$ yes, the acceleration of the center of mass is the same (see equation (1))
• "If so, why?" $\rightarrow$ because the movement of the center of mass is independent of the point where forces are applied. Only the net force matters (see equation (1))
• "How can it rotate and yet accelerate with the same velocity as that of without rotation?" $\rightarrow$ because those two motions are independant of each other
• "Where is it getting this extra energy from?" $\rightarrow$ last section of https://physics.stackexchange.com/a/174208/75518 or see equation (3)
• "How can I calculate the respective velocities of linear and angular motion given F and t?" $\rightarrow$ solve equations (1) and (2) for specific forces.

Note: An impact on a rigid body is, however, a total different situation. See this: Elastic collision of point particle and rod

• @terry: for a force $F$ of constant magnitude $5N$ and constant direction the center of mass will accelerate by $a=F/M$ with M the total mass of the body. That is: $s(t) = 1/2 a t^2 + v_0 t + s_0$, no matter where the force is applied. this is equation (1). However, if an object with a certain momentum collides with the body, this is a total different situation Apr 6, 2015 at 15:14
• The answer for the extra energy can be improved by simply realizing that the work done on the system is given by $\int F \cdot \mathrm d \mathbf s$. When the force is acting on one of the masses, the point where the force acts on moves the distance as if it acted on the center of mass plus the distance it rotates with the mass point in the center-of-mass coordinate frame. So the total distance is longer in the case when the object rotates. The total work can then be written as the sum of the work done on the center of mass plus the work done to rotate the object in the center-of-mass frame. Apr 6, 2015 at 17:24
• @domj33: yeah good point. I added it accordingly Apr 6, 2015 at 18:14
• @svetlana: I derived the equation of motion for the center of mass, showing that only net force, but not the point of application matters. fair enough, right? the OP is not asking about an impact with energy/momentum conservation. just pure forces, which can in principle do as much work as they want to. Apr 7, 2015 at 10:15
• Marcel,I deleted my comments because I am having a discussion with someone who is really an expert: he confirmed you must distinguish between fixed and follower load, and one must consider the value of the effective reduced mass at each point of application/impact. I'll getback to you when I get a definitive word. Try to comply with OP's request and give a concrete example, if you can
– user76816
Apr 9, 2015 at 15:00

Acceleration of the center of mass is always $F/m$, so if force and mass are the same, the center of mass will accelerate the same way, regardless of the point where the force acts.

After the same time of experiencing the same force, the body in the rotating case has greater kinetic energy than in the non-rotating case. This is due to greater work done by the same force in the rotating case - the force is the same, but the velocity of point experiencing the force is greater due to rotation.

• @terry, the question was theoretical and it assumed the same given force $F$ in both cases. There is no reason to complicate the answer with practical issues of how such force is to be accomplished. The mass $m$ in the formula is total mass of the body, idea of effective mass is unnecessary in this case. Apr 5, 2015 at 22:49