When does torque equal to moment of inertia times the angular acceleration ?

Particularly, for example, if a disc is rotating and translating on a horizontal surface with friction by the help of some force.If I'm a inertial observer and I choose my rotation axis as the axis passing through the centre of mass, which is accelerating, would the equality that is $\vec{\tau} = I_{cm}*\vec{\alpha}$ still hold ?

  • $\begingroup$ Related: physics.stackexchange.com/a/183650/392 $\endgroup$ Jan 2, 2017 at 6:52
  • $\begingroup$ @ja72 I didn't see any relation $\endgroup$
    – Our
    Jan 2, 2017 at 6:57
  • $\begingroup$ See my answer. Understanding momentum equations is what derives the force/torque relationships. $\endgroup$ Jan 2, 2017 at 7:13

1 Answer 1


You have to understand how linear and angular momentum are defined first before you can derive the equations of motion.

In general (3D) the following are true:

  1. Linear momentum is the product of mass and the velocity of the center of mass. Since mass is a scalar, linear momentum and velocity are co-linear $$\mathbf{p} = m \mathbf{v}_{cm}$$

  2. Angular momentum about the center of mass is the product of inertia and rotational velocity. Inertia is a 3×3 tensor (6 independent components) and hence angular momentum is not co-linear with rotational velocity $$\mathbf{L}_{cm} = \mathtt{I}_{cm} \,\boldsymbol{\omega} $$

  3. The total force acting on a body equals rate of change of linear momentum $$ \mathbf{F} = \frac{{\rm d} \mathbf{p}}{{\rm d}t} = m\,\frac{{\rm d} \mathbf{v}_{cm}}{{\rm d}t} = m \, \mathbf{a}_{cm} $$

  4. The total torque about the center of mass equals the rate of change of angular momentum $$\boldsymbol{\tau}_{cm} = \frac{{\rm d} \mathbf{L}_{cm}}{{\rm d}t} = \mathtt{I}_{cm} \, \frac{{\rm d}\boldsymbol{\omega}}{{\rm d}t} + \frac{{\rm d} \mathtt{I}_{cm}}{{\rm d}t} \boldsymbol{\omega} = \mathtt{I}_{cm} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathtt{I}_{cm} \boldsymbol{\omega}$$

Because momentum is not co-linear with rotational velocity the components of the inertia tensor change over time as viewed in an inertial frame and hence the second part of the equation above describes the change in angular momentum direction.

In your case the direction of rotation is fixed (normal to the plane) and so angular momentum is fixed making the above equation (still in vector form) $$ \boldsymbol{\tau}_{cm} = \mathtt{I}_{cm} \boldsymbol{\alpha}$$ furthermore only the out of plane components of angular acceleration and torque are considered so the above equation reduces down to a scalar equation $$\tau_{cm} = I_{cm} \alpha$$

The motion of the center of mass is still described by $\mathbf{F} = m\,\mathbf{a}_{cm}$ or in scalar form $$\begin{align} F_x & = m \ddot{x}_{cm} \\ F_y & = m \ddot{y}_{cm} \end{align} $$ The two motions are (linear and angular) are independent of each other for a free body.

  • $\begingroup$ To understand step #4 google search "Derivative on a rotating body" $\endgroup$ Jan 2, 2017 at 7:14
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    $\begingroup$ You are not giving enough information about the conditions that are necessary for those equations to be satisfied, for example are we talking about a rigid body or a point object, or where is the rotation axis ? are they still valid if the rotation axis is moving or accelerating ? $\endgroup$
    – Our
    Jan 2, 2017 at 7:31
  • $\begingroup$ These describe the motion of a rigid body, and are valid instantaneously (at every instant). If the rotation axis is moving is irrelevant because it is covered under the linear equations. The angular equations only describe the motion about the center of mass. $\endgroup$ Jan 2, 2017 at 7:45
  • $\begingroup$ To justify, you are saying that even if the centre of mass is accelerating, these equations would be still valid, right ? $\endgroup$
    – Our
    Jan 2, 2017 at 9:07
  • $\begingroup$ @Leth -- That is correct, assuming the motion is described in terms of the translation of the center of mass and rotation about the center of mass. That's why ja72 put center of mass in italics, three times, in this answer. $\endgroup$ Jan 2, 2017 at 12:51

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