In context of rotational dynamics my textbooks interminably states that there always exists a rotational analog for every linear variable and they are related to one another through similar equation. (And for the fact I know about most of these analogs and the related equations)

Though till now I haven't yet seen/learnt of rotational analog of Newton's Laws of Motion. So

  • Does there exist a rotational analog for Newton's laws of Motion?

  • If possible, then are they generalizable to non-rigid body?


To further clarify the question I am going to write down the rotational analog of Newton's laws of motion. Assuming that there exists a rotational analog of Newton's laws of motion, I suppose they would look like the following:

  1. A body continues it's state of rest and uniform rotation unless acted upon by an external torque.

  2. The torque acting on a body is directly proportional to the rate of change of angular momentum. i.e.,

$$ \boldsymbol {\tau} = \frac {d\mathbf L}{dt}$$

  1. For every torque there is an equal and opposite torque.

    • So are these statements (that I made as an analog of Newton's laws of Motion) true for all kind of rotational motion?

    • Is there any exception to them?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/244655/50583, physics.stackexchange.com/q/92957/50583 $\endgroup$
    – ACuriousMind
    Jan 26, 2020 at 19:12
  • $\begingroup$ What do you mean by rotational analog of Newton law of motion, there are three Newton's law, law of inertia which replaces inertia as rotational inertia, and rate of change of momentum causes force which replace as torque, and third law is same, if I am getting correct, analog mean you want to find the relationship between them, there are many answer suggested by A curious mind. $\endgroup$ Jan 27, 2020 at 8:51
  • $\begingroup$ Does this answer your question? Force and Torque $\endgroup$ Jan 27, 2020 at 8:53
  • $\begingroup$ Statement 1, "A body continues it's state of rest and uniform rotation unless acted upon by an external torque." is incorrect. It is correct to say that a body maintains its angular momentum unless acted upon by an external torque. That angular momentum remains constant typically does not mean that angular velocity remains constant. $\endgroup$ Jan 27, 2020 at 11:13

4 Answers 4


The following are the most common rotational analogues of linear motion terms:

1) Distance $(x)$ - Angle $(\theta)$

2) Velocity $(v)$ - Angular Velocity $(\omega)$

3) Acceleration $(a)$ - Angular Acceleration $(\alpha)$

4) Mass $(m)$ - Moment of Inertial $(I)$

5) Force $(F)$ - Torque $(\tau)$

All differential formulae still apply such as $\frac {dx}{dt}=v$ and $\frac {d\theta}{dt}=\omega$. A particularly important relation is $\tau=I\alpha.$ You can substitute all the equation of motion with their rotational counterparts and get perfectly valid formulae. For example $v=u+at$ becomes $\omega_f=\omega_i+\alpha t$. Non-rigid body dynamics can also be generalized using these terms although that becomes quite complicated.


Yes, your statements are correct. They always hold just like in linear motion. However, you must be careful with the frame of reference. Any rotational frame of reference is non-inertial and hence these will not apply in that case.

  • 1
    $\begingroup$ This is only valid for motion along or about a fixed axis. In general, with vector quantities, the story isn't as simple. $\endgroup$ Jan 26, 2020 at 17:05
  • 3
    $\begingroup$ @ja72 Considering that the OP has just started learning rotational motion, I thought this would be sufficient. $\endgroup$
    – Sam
    Jan 26, 2020 at 17:07
  • $\begingroup$ Why is $\theta$ labeled "arc length"? $\endgroup$ Jan 26, 2020 at 20:53
  • $\begingroup$ This doesn't answer the question. I know of rotational analog and the related formula. $\endgroup$
    – user249968
    Jan 27, 2020 at 3:15
  • $\begingroup$ I have clarified the question. You can have a look. $\endgroup$
    – user249968
    Jan 27, 2020 at 3:42

This kind of conceptualization may be useful at some level, but it is not accurate. You learn about kinematics and dynamics in 1D, and it is natural to want about the real 3D picture of those things. I was never comfortable with the statement of analogue between translational and rotational motion. The underlying assumption is that these two are separate mechanics working independently of each other. Just as the analogue between circuit (voltage, current) and dynamics (force, velocity) implies that similar behavior exists between these two systems.

But in real life (in three dimensions) the story is rather more convoluted.

We are afforded a certain level of separation between translational and rotational motion when things are resolved around the center of mass of a body. Then you can state the translational and rotational equations of motion as

$$ \begin{aligned} \boldsymbol{p} & = m \, \boldsymbol{v}_C \\ \boldsymbol{F} & = \frac{\rm d}{{\rm d}t} \boldsymbol{p} \\ \pmatrix{ F_x \\ F_y \\ F_z } & = m \pmatrix{a{x}_C \\ a{y}_C \\ a{z}_C} \end{aligned} \tag{1}$$

$$\begin{aligned} \boldsymbol{L}_C & = \mathbf{I}_C \boldsymbol{\omega} \\ \boldsymbol{\tau}_C & = \frac{\rm d}{{\rm d}t} \boldsymbol{L}_C \\ \pmatrix{\tau x_C \\ \tau y_C \\ \tau z_C} & = \left[ \matrix{ I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} } \right] \pmatrix{\alpha_x \\ \alpha_y \\ \alpha_z} + \pmatrix{Lz_C\, \omega_y - Ly_C\, \omega_z \\ Lx_C\, \omega_z - Lz_C\, \omega_x \\ Ly_C\, \omega_x - Lx_C\, \omega_y} \end{aligned} \tag{2}$$

Above $p$ is momentum and $L$ angular momentum, and $a$ acceleration $\alpha$ rotational acceleration, $F$ is force and $\tau$ is torque.

Equation (1) is Newton's second law, and (2) is the Euler equations of rotational motion. Point C is the center of mass, and each vector quantity is indicated with bold symbols, and their components resolved into x,y and z directions.

Note that the two sets of equations are de-coupled (independent of each other), but this only happens at the center of mass. The general equations as seen in this post are far more complex to write down in component form, so you need to familiarize your self with vector calculus before going that way. So your teacher understandably at this point reverts to the equivalency simplification.

On your last point, rigid body mechanics cannot be generalized on non-rigid bodies, because the notion of angular velocity doesn't exist. Look at Jupiter, which is spinning, but each part is spinning at a different rate. The dynamics exist, but the kinematics are all wrong. You can no longer calculate the velocity of a point, based on the rotation and the velocity of another point like you can do with rigid bodies.

  • $\begingroup$ I have noticed that you have use \mathbf to boldify $I$ ( you write $\boldsymbol{L}_C = \mathbf{I}_C \boldsymbol{\omega}$) which I use for vectors. So does that mean moment of inertia is a vector? $\endgroup$
    – user249968
    Jan 27, 2020 at 3:45
  • 1
    $\begingroup$ @JohanLiebert - Moment of inertia is an order 2 tensor typically represented as a 3x3 symmetric matrix. $\endgroup$ Jan 27, 2020 at 11:01
  • $\begingroup$ I use \boldsymbol for vectors, and \mathbf for tensors (matrices). The first is bold slanted while the second is bold upright. This is my convention as I have seen the following for mass moment of inertia tensors in literature $$\matrix{ I_C & \mathcal{I}_C & \mathrm{I}_C & \mathbf{I}_C & \boldsymbol{I}_C }$$ $\endgroup$ Jan 27, 2020 at 13:21

Re Sam's answer above - any beginners might be interested in the very easy to read online source at Georgia State Uni

enter image description here

  • 1
    $\begingroup$ Sam's answer is incorrect. It is ja72's answer that is correct. $\endgroup$ Jan 27, 2020 at 11:04
  • $\begingroup$ @David Hammen, agree. ja72's important qualification is also nicely explained in hyperphysics hyperphysics.phy-astr.gsu.edu/hbase/n2r.html#n2r aka Newtons 2nd law (rotation) is limited to rotation about a single principal axis. $\endgroup$ Jan 27, 2020 at 11:34

The existing laws of motion are sufficient.

One way to conceptualize that is that implicit in the laws of motion is that any vector (position vector, velocity vector, acceleration vector) can be composed and decomposed as vectors in Euclidean space.

Specific example:
Take the case of circular motion. Uniform circular motion can be decomposed in two components, each a harmonic oscillation and for each 1-dimensional motion component the laws of motion apply.

One can regard that as a form of Superposition Principle. Motion in more than one spatial dimension can be decomposed in motion components. Motion in more than one spatial dimension can be thought of a superposion of motions along each of the available spatial dimensions.

Other than that:
the analogy between expressions for linear motion and rotational motion goes only so far. There are overviews that present a side-by-side view of 10 or so expression. I expect that for subsequent expressions the analogy will not hold up.

Of course, in many cases the expressions for angular dynamics are more efficient. Whenever it is more practical to use the expressions for moment of inertia, angular momentum, and so on you should of course do so.

The expresssions for angular dynamics can all be derived from the laws of motion. For example, in the Principia the very first theorem that Newton states (Book I, proposition I) is a derivation of Keplers law of areas generalized for any centripetal force.

This generalized law of areas is also known as conservation of angular momentum.