# Can the reference point in an inertial frame be accelerated when applying torque equation?

Torque about a point = rate of change of angular momentum about that point.

Say we're in an inertial frame and see a body accelerating and rotating and another point(either a part of this body or an external point) accelerating.

Can we apply T = dL/dt for that body about that accelerating point if we are in an inertial frame?

My guess is we can since angular momentum depends on the reference point. Is there any requirement for this point to be non-accelerating?

• Oh, that was a typo. I'll correct it. – xasthor Mar 22 '18 at 18:02

The law you state is only valid for the center of mass, or for a fixed point in space.

Euler' law of rotation states:

Net torque of an object about the center of mass equals the rate of change of angular momentum measured at the center of mass.

$$\boldsymbol{T}_C = \frac{{\rm d}}{{\rm d}t} (\boldsymbol{L}_C) = \frac{{\rm d}}{{\rm d}t} (\mathrm{I}_C \boldsymbol{\omega}) = \mathrm{I}_C \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times \boldsymbol{L}_C$$

where point C denotes the center of mass. This a direct equivalent to the fact that the net force on a body describes the motion of the center of mass only. The remaining motion (about the center of mass) is described by Euler's law.

The center of mass can be accelerating (and usually is) since usually both torque and force are considered at the same time

$$\boldsymbol{F} = \frac{{\rm d}}{{\rm d}t}( \boldsymbol{p} ) = \frac{{\rm d}}{{\rm d}t}(m\, \boldsymbol{v}_C ) = m\, \dot{ \boldsymbol{v}}_C$$

So now what happens at a different location A? Consider the location vector $\boldsymbol{c}$ of the center of mass, relative to A

Angular momentum at A is

$$\boldsymbol{L}_A = \boldsymbol{L}_C + \boldsymbol{c} \times \boldsymbol{p}$$

Net torque at A is

$$\boldsymbol{T}_A = \boldsymbol{T}_C + \boldsymbol{c} \times \boldsymbol{F}$$

The total derivative of angular momentum at A is

\require{cancel} \begin{aligned} \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) & = \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_C + \boldsymbol{c} \times \boldsymbol{p} ) = \boldsymbol{T}_C + \frac{{\rm d}\boldsymbol{c}}{{\rm d}t} \times \boldsymbol{p} + \boldsymbol{c} \times \underbrace{ \frac{{\rm d}\boldsymbol{p}}{{\rm d}t}}_{\boldsymbol{F}} \\ & = \boldsymbol{T}_A + \underbrace{( \boldsymbol{v}_C - \boldsymbol{v}_A ) \times \boldsymbol{p} }_{(\boldsymbol{v}_C-\boldsymbol{v}_A) \times (m\,\boldsymbol{v}_C) = -\boldsymbol{v}_A \times m\,\boldsymbol{v}_C = \boldsymbol{p} \times \boldsymbol{v}_A} \end{aligned}

I produce the following law (if no one else claims it, call it the ja72 law).

The rate of change of angular momentum at a non-fixed arbitrary point A equals the net torque at A, plus cross product of linear momentum with the speed of A.

$$\boxed{ \frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) = \boldsymbol{T}_A + \boldsymbol{p} \times \boldsymbol{v}_A }$$

The conditions where the derivate of angular momentum is exactly the net torque at a point _A_on a rigid body are as follows:

1. Body undergoes pure rotation with zero linear momentum, $\boldsymbol{p}=0$
2. Point A is fixed in space, or instaneneously fixed, $\boldsymbol{v}_A=0$
3. Point A is on the center of mass, making its motion parallel to momentum, $\boldsymbol{v}_A \parallel \boldsymbol{p}$
4. Point A lies on a line parallel to the rotation axis, but through the center of mass, $\boldsymbol{v}_A \parallel \boldsymbol{v}_C$
• In the original derivation of Torque = dL/dt, nowhere did we use that assumption. phys.libretexts.org/TextMaps/Classical_Mechanics_TextMaps/… – xasthor Mar 23 '18 at 2:43
• @xasthor - Isn't the correct form of the time derivative $$\dot{\bf L} = \sum_i ( \dot{\bf r}_{i}\times {\bf p}_{i}+{\bf r}_{i}\times \dot{\bf p}_{i})$$ The cross product is linear operator and the regular product rule applies for derivatives. – John Alexiou Mar 23 '18 at 14:23
• @xasthor - also what is ${\bf F}_{i}$ and ${\bf F}_{ij}$ that their sum equals velocity? – John Alexiou Mar 23 '18 at 14:27
• @xasthor - Euler's laws of rotation motion apply at the center of mass only. You cannot pick an arbitrary location for the net torque since a body rotates only if the line of action of a force is not at the center of mass. Also, angular momentum not at the center of mass includes the moment of momentum component which is a function of linear motion. See The derivation of Equations of Motion not at the center of mass to see why $\frac{{\rm d}}{{\rm d}t} ( \boldsymbol{L}_A ) \neq \boldsymbol{T}_A$ at any point A. – John Alexiou Mar 23 '18 at 14:54
• derivative of $r_i$ is $v_i$ which when crossed with $p_i$ =0 since they are in the same direction. hence the sum simplifies into the summation of torques. nothing about that sum specifies that L or T is about the center of mass. it seems to be about any arbitrarily point – xasthor Mar 23 '18 at 15:15