I have a question that's been bothering me for years.

Given a rod of uniform mass distribution with total mass $M$ and length $L$ that lies on a horizontal table (with one end fixed to the table around which the rod is free to rotate in the horizontal plane, and a force F applied perpendicular to the rod at the other end), how do you solve for the motion of the rod (and the internal forces) using only Newton's laws and the assumption that the rod is a rotating rigid body? By that I mean only using the most basic conception of Newton's laws and the system's constraints, without the ideas of torque and moment of inertia, energy and momentum, and even without the idea that the net force on the rod gives the acceleration of the center of mass--so only using Newton's laws for point particles, or in this case infinitesimal $dm$ sections of the rod.

I've tried to solve this by breaking the rod down into these small $dm$ components and using an idea I've seen (at least I think I've seen) where you set $F(x+dx)-F(x)=dm(a)$ and then are able to find $F'(x)$ and integrate and then use the boundary conditions on the force. I did this for both tangential and radial components, with radial acceleration equal to $x(\omega(t))^2$ and tangential acceleration equal to $\omega\ '(t)x$, but was unable to obtain the right answer. I used the force at one end of the rod as a boundary condition (is this correct?), but was unable to even solve for the force at the pivot, let alone the angular velocity as a function of time, and have no idea if this technique is even valid. I feel like at a certain point it might be that my force equation switches sign--as the net force that accelerates the infinitesimal mass starts coming from the inward instead of outward side.

I'd also be interested to know more generally how to solve for internal forces and motion of a rigid body using only these most basic assumptions, such as for a free uniform square on a horizontal table with a force applied perpendicularly to one side into one corner.

  • $\begingroup$ Why do you want to find the 'total force'? More importantly, what is specifically implied by total force(given that we are not to use the Center of Mass concept)? $\endgroup$ – Shubham May 22 '14 at 15:04
  • $\begingroup$ Sorry, I obviously have no idea what I'm talking about. I guess I instead mean that I tried to integrate and use the boundary condition on the force at one end, although I have no idea if this is valid. I'll edit. $\endgroup$ – user47050 May 22 '14 at 15:09
  • $\begingroup$ The internal force is the tension in the rod, and the point at the pivot would not need any force(it is not moving). The tangential accelrations would require tangential forces. $\endgroup$ – Shubham May 22 '14 at 15:10
  • $\begingroup$ Yeah, this is part of what confused me. There is obviously no net force on the point of the rod at the pivot, but at least I think there is an internal force tangential force an infinitesimal distance away which exactly counteracts the force of the pivot on the rod. But I'm not sure how all of these internal and external forces come into play when integrating the expression... Are they the boundary values? Do I even have to integrate? $\endgroup$ – user47050 May 22 '14 at 15:23
  • 1
    $\begingroup$ So we can derive the result of torque and moment of inertia and then continue? $\endgroup$ – evil999man May 23 '14 at 6:34

At a general $\theta$,

It is easy to see as the rod does not deform the speeds of particles must be proportional to their distance from centre.

Let $\omega(t)$ be the constant involved here.

$$p(t)=\sum m_i v_i=\sum m_i\omega r_i$$

Note that all particles move in same direction.

$$p(t)=\int_0^l\omega r dm=\int_0^l\omega r mdr/l=\omega ml/2$$

In time $dt$ a particle covers $\omega r dt$ distance. And hence $d\theta=\omega dt$

You can apply Newton's law for both axes and don't forget to consider Force provided by pivot.

You have three unknowns: $N_x,N_y,\theta(t)$

Note that for a particular Force on a particle, $$F=ma=mv\dfrac{dv}{dx}\implies\vec F\cdot d\vec{x}=mvdv$$

Combine this for all particles and dot product of N forces is zero as that point is at rest. This will yield $d(KE)$ on adding.

Hence, $$Fld\theta=d(KE)$$

Compute KE using integration method.

You have three unknowns and three equations.


$\def\ph{\varphi}\def\l{\left}\def\r{\right}\def\nR{\mathbb{R}}\def\hF{\hat{F}}\def\rmL{{\mathrm{L}}}\def\vr{\vec{r}}\def\vF{\vec{F}}\def\m#1{{\mathbf{#1}}}\def\det{\operatorname{det}}\def\vomega{{\vec{\omega}}}\def\vph{{\vec{\varphi}}}\def\vom{{\vec{\omega}}}\def\valpha{{\vec{\alpha}}}\def\vv{{\vec{v}}}\def\ddt{\frac{d}{dt}}$ We assume that the rod can be regarded as a straight line section with the rotation angle $\ph$ as its only degree of freedom. We use the pivot as origin. With these assumptions the bar can be described as $$ z(s,t) = s\exp(i\ph(t)). $$ in the complex plane with $s\in[0,l]$. From this we can easily compute the velocity and the acceleration of the bar points: \begin{align} \dot z(s,t) &= i s \exp(i\ph(t))\dot\ph(t)\\ \ddot z(s,t) &= s \exp(i\ph(t))\bigl(-\dot\ph(t)^2 + i\ddot\ph(t)\bigr) \end{align}

As you say, we know the force applied to the outer bar end. Let us name it $F_l(t)$.

Since you do not want to deal with elasticities you have constraints and you need to apply one of the principles of mechanics for constrained systems. This is as close as you get to Newton's principles for a system with constraints. It has the interpretation that Newton's law must be valid in the direction of the degree of freedom. In all the other directions there are constraint forces keeping the system within the constraints. If you do not want to apply one of the principles of mechanics you have to consider the full 3d-beam with elastic forces and maybe take the limit to a rigid bar. You find a good description about that procedure in Arnolds book for point mass systems. I apply d'Alembert's principle here. Luckily, we do not need to consider the pivot force since there the virtual displacement $\delta z(s,\ph) = s\exp(i\ph(t))i\delta\ph$ is zero because of $s=0$ there. \begin{align*} 0&=\int_{s=0}^l \l\langle\delta z(s,\ph(t))\mid -\ddot z(s,\ph(t))\r\rangle\frac{m}{l}\,ds + \l\langle\delta z(l,\ph(t))\mid F_l(t)\r\rangle \end{align*} Thereby, $\langle\bullet\mid\bullet\rangle$ is the normal scalar product of $\nR^2$ which can be calculated as $\langle a\mid b\rangle=\Re(a^*\cdot b)=a_x b_x + a_y b_y$. \begin{align} 0&= \Re\l(\int_0^l \l(s\exp(-i\ph(t))(-i)\delta\ph\cdot \l(-s\exp(i\ph(t))(-\dot\ph(t)^2+i\ddot\ph(t))\r)\r)\frac ml\,ds + \l( l\exp(-i\ph(t))(-i)\delta\ph\cdot F_l(t)\r)\r)\\ 0&= \delta\ph\cdot\l(-\ddot\ph(t)\frac {ml^2}3 + l\Re\l(-i\exp(-i\ph(t))F_l(t)\r)\r) \end{align} This equation must be valid for all virtual displacements $\delta\ph$, e.g. $\delta\ph=1$ which gives us the equation of motion $$ \ddot\ph(t)\frac {ml^2}3 = l\Re\l(-i\exp(-i\ph(t))F_l(t)\r) $$ Let us consider a force $F_l(t)$ that always acts orthogonal to the bar and has constant absolute value $\hF$. We choose the orientation of the force such that it drives the bar in mathematically positive direction. It can be represented as $$ F_l(t) = i\hF\exp(i\ph(t)) $$ Therewith we get $$ \ddot\ph(t)\frac {ml^2}3 = l\Re\l(\hF\r) = l\hF. $$ The bar has been modeled as a line segment to simplify the calculation. The restriction of the motion to a plane is a further simplification.

The general rigid body model is a domain $B\subset\nR^3$ embeded through a rigid-body motion \begin{align} \vr(\vr^\rmL,t) &= \vr_0(t) + R(t)\cdot \vr^\rmL \end{align} with $\vr^\rmL\in B$. Thereby, $\vr^\rmL$ are point coordinates in the rigid body reference frame and (for simplicity) $R$ is a rotation matrix (with $R R^T = \m1$ and $\det(R)=1$). Let $\vF_k$ be external forces imprinted to the body at points $\vr_k^\rmL$ $(k=1,\ldots,n)$. The corresponding points in space are $\vr_k = \vr_0 + R\vr_k^\rmL$.

The variation of $\vr_0$ is $\delta\vr_0$.

We look at the variation of $R$ somewhat more closely. The derivative of $R R^T = \m1$ gives $$\m0=\delta(\m1)=\delta(R\cdot R^T) = \delta R \cdot R^T + R \cdot \delta R^T.$$ That means the matrix $$ \delta\Phi:=\delta R \cdot R^T = -R \cdot \delta R^T = -(\delta R\cdot R^T)^T $$ is skew-symmetric and therefore only has 3 relevant components $\delta\ph_1:=\delta\Phi_{32},\delta\ph_2:=\delta\Phi_{13},\delta\ph_3:=\delta\Phi_{21}$. With the vector $\delta\vph:=(\delta\ph_1,\delta\ph_2,\delta\ph_3)$ of these three relevant components the product of $\delta\Phi$ with any vector $\vec a$ can be represented as cross product $$ \delta\Phi\cdot \vec a = \delta\vph \times \vec a. $$ Now, we are ready to calculate the virtual displacement of the rigid body motion $$ \delta\vr(\vr^\rmL)=\delta\vr_0 + \delta R \vr^\rmL $$ Augmenting with the factor $\m{1}=R^TR$ gives \begin{align} \delta\vr(\vr^\rmL)&=\delta\vr_0 + \delta R R^T R \vr^\rmL\\ &=\delta\vr_0 + \delta\vph\times R\vr^\rmL \end{align} In the same way we can compute the velocity of the points of the body \begin{align} \vv(\vr^\rmL,t):=\dot\vr(\vr^\rmL,t) &= \dot\vr_0 + \dot R \vr^\rmL\\ &= \vv_0 + \vom \times R\vr^\rmL \end{align} with $\vom\times \l(R\vr^\rmL\r) := \dot R R^T \l(R\vr^\rmL\r)$. Alembert's principle gives \begin{align} 0 &= \ddt\int_{\vr^\rmL\in B} \delta\vr(\vr^\rmL)^T \cdot (-\rho \vv(\vr^\rmL,t)) \cdot d V + \sum_{k=1}^n \delta\vr_k \cdot \vF_k \end{align} Putting in all the bits gives: \begin{multline} \ddt\int_{\vr^\rmL\in B} \l(\delta\vr_0 + \delta \vph\times R\vr^\rmL\r)^T \cdot \l(\vv_0 +\vom \times R\vr^\rmL\r) \cdot \rho d V \\ - \sum_{k=1}^n \l(\delta\vr_0 + \delta \vph\times R\vr^\rmL_k\r)^T \cdot \vF_k = 0 \end{multline} Considering that in general $(\vec a\times \vec b) \cdot \vec c = \vec a\cdot \l(\vec b \times \vec c\r)$ one obtains \begin{multline} \delta\vr_0^T\cdot\l(\ddt\int_{\vr\in B} \l(\vv_0 + \vom \times R\vr^\rmL\r) \cdot \rho d V -\sum_k \vF_k \r)\\ +\delta\vph^T \cdot\l(\ddt\int_{\vr\in B} \l(R\vr^\rmL\r)\times\l(\vv_0 - \l(R\vr^\rmL\r)\times \vom\r)\rho d V\\ -\sum_k\l(R\vr^\rmL_k\r)\times\vF_k \r) = 0 \end{multline} and with $J(t):=\int_{\vr\in B} -(R\vr^\rmL)\times(R\vr^\rmL)\rho d V$ and $m:=\int_{\vr\in B} \rho d V$ one obtains $$ \delta\vr_0^T\cdot\l(\ddt \l(\vv_0 + \vom \times R\vr^\rmL\r)m -\sum_k \vF_k \r)\\ +\delta\vph^T \cdot\l( \ddt\l(m\l(R\vr^\rmL\r)\times\vv_0 + J(t)\times \vom\r) -\sum_k\l(R\vr^\rmL_k\r)\times\vF_k \r) = 0 $$ Since the variations of $\delta\vr$ and $\delta\vph$ are mutual independent their factors must vanish separately and one obtains the well known balance equations \begin{align} \ddt \l(\vv_0 + \vom \times R\vr^\rmL\r)m &= \sum_k \vF_k \\ \ddt \l(m\l(R\vr^\rmL\r)\times\vv_0 + J(t)\times \vom\r) &= \sum_k\l(R\vr^\rmL_k\r)\times\vF_k \end{align}

  • $\begingroup$ Question: I've been thinking about this, and does the restriction on the use of Newton's laws have anything to do with the rod being one-dimensional object in a 2d plane? My intro mechanics textbook proves conservation of angular momentum with the assumption that the internal rigid body forces are central. But is that only possible here if the beam were to bend? $\endgroup$ – user47050 May 24 '14 at 13:22
  • $\begingroup$ @user47050 The 1d-object in 2d plane simplifies matters but it is not essential for the method. Now, I present the general case of a 3d-rigid body in 3d-space in the answer. Could you expand a bit more what you mean by the second part of the question? I do not fully understand yet what you mean by "But is that only possible here if the beam were to bend?" $\endgroup$ – Tobias May 25 '14 at 19:06
  • $\begingroup$ @user47050 I had to correct some bits here. I had only limited time to write the stuff. I think the text makes at least clear how to extend the method from the simple 1d rod in 2d to the full 3d rigid body motion. Just leave a comment if there is something wrong in the text. $\endgroup$ – Tobias May 26 '14 at 5:15

I can't quite follow your approach (for example, I don't understand what $w(t)$ is) but here are some comments.

In your approach, each $\mathrm{d}m$ has to apply a force on its adjacent $\mathrm{d}m$. Normally you would take the rod to be an continuous elastic material with suitable elastic parameters, so the force of one $\mathrm{d}m$ on another would be an elastic force. I take it you want to get more microscopic than that.

One possible model: take your rod to be a ball-and-massless-stick system with a stiff restoring bending force between adjacent pairs of balls to model the bending modulus (if that's the right word) in a real rod. You'd have to be careful to get that bending force right as the rod rotates. Fix the length of the rods for simplicity. Apply your external force to the first ball, fix the position of the last, but remove the restoring bending force to allow it to pivot. The problem then becomes a large set of coupled equations. It sounds messy and complicated but do-able to me.

Later you can generalize by removing the condition that the length of the rods is fixed. Add an additional restoring force to model Young's modulus.

This problem would yield more easily to simulation than solution. In fact, here is a simulation of exactly your problem, except the "far end" of the rod is free rather than fixed to a pivot. (Does not run on my MacBook using Chrome. It does run with Firefox.) It's written in GlowScript, based on VPython, a wonderful tool for physics simulations.

  • $\begingroup$ Sorry, I that as angular velocity, but as we can see I'm an idiot who doesn't know how to type an omega. $\endgroup$ – user47050 May 22 '14 at 17:19

Take a coordinate system with origin at the pinned end, where the rod is along the horizontal axis, and pretend there a force $F$ applied on the other end $x=\ell$.

Each section of the rod has a balance of forces $${\rm d}S=\ddot y\;{\rm d}m \\ {\rm d}M = S\; {\rm d}x$$ considering only linear mass effects. $S$ is the internal shear force and $M$ the internal moment, $\ddot y = x \ddot{\theta} $ is the linear acceleration of the rod segment and $m$ is the total mass of the rod. Note that ${\rm d}m = \frac{m}{\ell} {\rm d}x$

As a differential equation the above is:

$$ \frac{{\rm d}S}{{\rm d}x} = \frac{m}{\ell} x \ddot{\theta}$$ $$\frac{{\rm d}M}{{\rm d}x} = S$$

The above is solved with the boundary conditions $S(x=\ell)=F$ and $M(x=\ell)=0$ as

$$ S = F + m \left(\frac{\ell}{2} - \frac{x^2}{2 \ell}\right) \ddot\theta $$ $$ M = F (\ell-x) + m \left( \frac{\ell^2}{3}-\frac{\ell x}{2} + \frac{x^3}{6 \ell}\right) \ddot\theta $$

The reaction force that the pin is found with $S(x=0) = F + m \frac{\ell}{2} \ddot\theta$ and the force acceleration relationship (what you are looking for) is found by noting there is no reaction torque on the pin:

$$\left. M(x=0)=0 \right\} F = m \frac{\ell}{3} \ddot\theta $$

The equation is usually expressed in terms of the applied torque of $\tau = F \ell$ as

$$ \tau = I_{pin} \ddot\theta \\ \boxed{I_{pin} = m \frac{\ell^2}{3}} $$

So we derived the mass moment of inertia of a rod, pinned on one end using only linear mass, in a ${\rm d}S={\rm d}m \; \ddot y$ force balance relationship.

  • $\begingroup$ Did you read the following part of the question? "By that I mean only using the most basic conception of Newton's laws and the system's constraints, without the ideas of torque and moment of inertia, energy and momentum, and even without the idea that the net force on the rod gives the acceleration of the center of mass--so only using Newton's laws for point particles, or in this case infinitesimal dm sections of the rod." $\endgroup$ – Tobias May 26 '14 at 6:01
  • $\begingroup$ And then he goes and breaks down the rod into $dm$ section and proceeds from there. So I think the question is inconsistent. $\endgroup$ – ja72 May 26 '14 at 7:02
  • $\begingroup$ Yes, the balance of momentum belongs aximatically to continuum mechanics. It allone implies the symmetry of the stress tensor. That is the reason why I took to one of the principles of mechanics where one can avoid the explicit calculation of constraint force quantities (such as the stress tensor) in my answer. Maybe it is a foul compromise... Nevertheless, I like the question because it forces one to think about the stuff. $\endgroup$ – Tobias May 27 '14 at 13:40
  • $\begingroup$ Hey, I realize that this is months late, but I've possibly realized the problem with my question and thinking. I don't know if I'm right, but I think my problem was assuming that all the rigid body assumptions (like how different parts of the body interact by force that runs along the displacement that connect them) hold when you have a 1D object in 2D space. Hence part of the reason why using Newton's Laws makes no sense. $\endgroup$ – user47050 Jul 8 '14 at 13:26
  • $\begingroup$ Newton's Laws only apply to point masses. You need Euler's laws of rotation rigid body motion. If the space is 2D or 3D it does not matter really. $\endgroup$ – ja72 Jul 8 '14 at 13:34

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