# Torque about an accelerating point

The total force acting on the pulley is zero so: $$F = mg + T_1 + T_2 \tag{A1}$$ Analyzing the torque and angular acceleration about the actual axis of rotation, the axle of the pulley, gives: $$τ_{net} = T_1 R − T_2 R = I α \tag{A2}$$ If we analyze about point P, the right edge of the pulley where T1 is applied, we get: $$τ_{net} = (F − mg) R − T_2 (2 R) = (I + m R^2 ) α \\ {}_\text{WRONG} \tag{A3}$$ Using Eqn. (A1) to eliminate $$F - mg$$ from Eqn. (A3) gives: $$τ_{net} = T_1 R − T_2 R = (I + m R^2 ) α \\ {}_\text{WRONG} \tag{A4}$$ The net torque in Eqns. (A2) and (A4) is the same, but the moments of inertia are different so the angular accelerations are also different. Note that if we think of point P as attached to the right-hand string if $$T_1 ≠ T_2$$ then it is accelerating.

My question is if we treat $$P$$ as fixed in space instead of a point attached to the right-hand string then what will be the equation if we take torque about $$P$$?

• Given your current picture, if the dimentions of the string arer negligible, then torque is the same, as the point in space is essentially the same.
– Sgg8
Commented May 6, 2023 at 15:53
• In the case of instantaneous axis of rotation, we say that the center of rotation is a point in space and does not undergo radial acceleration. So we think of it as an inertial frame of reference. So can't we analyze the torque about point P thinking that it not subject to any kind of acceleration.In that case we should get the actual torque about point P.@Sgg8
– xkcd
Commented May 6, 2023 at 19:06

In (A2) and (A3) you are forgetting the torque due to the acceleration of the center of mass. If P is fixed, then the center of mass is moving.

For the general case, the vector equations of motion that are summed at an arbitrary point A, not at the center of mass C in 3D are:

\begin{aligned}\boldsymbol{F}_{{\rm net}} & =\underbrace{m\boldsymbol{a}^{A}}_{\text{linear}}-\underbrace{m\boldsymbol{c}\times\boldsymbol{\alpha}}_{\text{angular}}+\underbrace{\boldsymbol{\omega}\times m\left(\boldsymbol{\omega}\times\boldsymbol{c}\right)}_{\text{centripetal}}\\ \boldsymbol{\tau}_{{\rm net}}^{A} & =\underbrace{{\rm I}^{A}\boldsymbol{\alpha}}_{\text{angular}}+\underbrace{\boldsymbol{c}\times m\boldsymbol{a}^{A}}_{\text{linear}}+\underbrace{\boldsymbol{\omega}\times{\rm I}^{A}\boldsymbol{\omega}}_{\text{gyroscopic}} \end{aligned} \tag{1}

NOTE: The $$\times$$ operation signifies vector cross product. Also bold quantities are vectors while italic quantities are scalar.

with the following descriptions $$\begin{array}{c|l} \text{Symbol} & \text{Description}\\ \hline \boldsymbol{F}_{{\rm net}} & \text{net force vector on body}\\ \boldsymbol{\tau}_{{\rm net}}^{A} & \text{net torque vector on body about point A}\\ m & \text{mass of body}\\ {\rm I}^{A} & \text{mass moment of inertia tensor about A}^{[1]}\\ \boldsymbol{c} & \text{relative position of COM from point A}\\ \boldsymbol{\omega} & \text{angular velocity vector of body}\\ \boldsymbol{a}^{A} & \text{linear acceleration of point A}\\ \boldsymbol{\alpha} & \text{angular acceleration of body} \end{array}$$

[1] The MMOI tensor at A is a 3×3 matrix that derives from the MMOI tensor at the center of mass as found in references, rotated to be aligned with the inertial coordinate frame, and translated from the COM to the reference point using the parallel axis theorem.

Projecting into 2D the gyroscopic terms vanish, but all other terms must be included. In your question, when the reference point is point P, you forgot to include the linear term in the net torque equation. This corresponds to the torque due to the acceleration of the center of mass.

In 2D the equations are a vector equation for forces, and a scalar equation for torques

\begin{aligned}\boldsymbol{F}_{{\rm net}} & =m\boldsymbol{a}^{A}-m\boldsymbol{c}\,\alpha-m\,\boldsymbol{c}\,\omega^{2}\\ \tau_{{\rm net}}^{A} & =I^{A}\alpha+m\left(c_{\perp}a^{A}\right) \end{aligned}

The term $$m\left(c_{\perp}a^{A}\right)$$ is missing from your equations. here $$c_{\perp}$$ represents the perpendicular distance from the center of mass to a line through point A and along the acceleration vector $$\boldsymbol{a}^A$$.

I would advise against doing dynamics in 2D, or by component, because things you might assume cancel out in 2D, might not cancel out actually. It is better to view the equations of motion in 3D primarily and in the end, do the necessary projection to 2D. Problems in dynamics are difficult because there are a lot of details to consider.

Also, do the dynamics at the center of mass because the resulting equations are much simpler when $$\boldsymbol{c}=0$$

\begin{aligned}\boldsymbol{F}_{{\rm net}} & =\underbrace{m\boldsymbol{a}^{C}}_{\text{linear}}\\ \boldsymbol{\tau}_{{\rm net}}^{C} & =\underbrace{{\rm I}^{C}\boldsymbol{\alpha}}_{\text{angular}}+\underbrace{\boldsymbol{\omega}\times{\rm I}^{C}\boldsymbol{\omega}}_{\text{gyroscopic}} \end{aligned} \tag{2}

The 2D version of the above is what (A1) equations look like. Here $${\rm I}^{C}$$ is the mass moment of inertia about the center of mass.

I would only recommend using (1) instead of (2) for cases where there are significant simplifications. Some examples are, when the body is non-rotating then

\small \begin{aligned}\boldsymbol{F}_{{\rm net}} & =\underbrace{m\boldsymbol{a}^{A}}_{\text{linear}}\\ \boldsymbol{\tau}_{{\rm net}}^{A} & =\underbrace{\boldsymbol{c}\times m\boldsymbol{a}^{A}}_{\text{linear}} \end{aligned}

or when the reference point is fixed (non-moving)

\small \begin{aligned}\boldsymbol{F}_{{\rm net}} & =-\underbrace{m\boldsymbol{c}\times\boldsymbol{\alpha}}_{\text{angular}}+\underbrace{\boldsymbol{\omega}\times m\left(\boldsymbol{\omega}\times\boldsymbol{c}\right)}_{\text{centripetal}}\\ \boldsymbol{\tau}_{{\rm net}}^{A} & =\underbrace{{\rm I}^{A}\boldsymbol{\alpha}}_{\text{angular}}+\underbrace{\boldsymbol{\omega}\times{\rm I}^{A}\boldsymbol{\omega}}_{\text{gyroscopic}} \end{aligned}

You will find (1) in the Newton-Euler equations of motion Wikipedia article in matrix form

$$\begin{Bmatrix}\boldsymbol{F}_{{\rm net}}\\ \tau_{{\rm net}}^{A} \end{Bmatrix}=\begin{bmatrix}m & -m[\boldsymbol{c}\times]\\ m[\boldsymbol{c}\times] & {\rm I}^{A} \end{bmatrix}\begin{Bmatrix}\boldsymbol{a}^{A}\\ \boldsymbol{\alpha} \end{Bmatrix}+\begin{Bmatrix}\boldsymbol{\omega}\times m\left(\boldsymbol{\omega}\times\boldsymbol{c}\right)\\ \boldsymbol{\omega}\times{\rm I}^{A}\boldsymbol{\omega} \end{Bmatrix}$$

In addition to the vector form of the parallel axis theorem which is $${\rm I}^{A}={\rm I}^{C}-m[\boldsymbol{c}\times][\boldsymbol{c}\times]$$