# How would I calculate the torque of the motor shown in the diagram? Torque due to force of gravity:
-Payload: 0.2kg x 0.3m x 9.81 = 0.5886
-Hand: 0.1kg x 0.25m x 9.81 = 0.24525
-Forearm: 0.3kg x 0.1m x 9.81 = 0.2943

Torque due to force of gravity = 1.13Nm

What about torque due to angular acceleration?
I know it is Torque = Moment of inertia x angular acceleration, However, the motor will be rotating the hand and the forearm around the x-axis.

So, how would I calculate the torque for this motion?

I assume you are trying to measure the torque needed to turn the arm horizontally to right or left.

If that's the case we calculate the angular moment, I as:

$$I= \Sigma m_i r^2_i \ = \\ 0.2kg*0.3^2m+0.1kg*0.25^2m+0.3kg*0.1^2m \\=0.18+0.00625+0.003=0.027Nm^2$$

Not that we don't have g factor, and torque is:

$$T=\alpha*I$$

• thank you for your reply!! So do I ignore the torque due to the force of gravity? So I'm talking about this movement: makeagif.com/gif/pronation-and-supination-of-the-forearm-pKemMR Feb 17, 2022 at 18:30
• no, if you needto lift you need the torque in X direction, if you need to turn you use the torque in z or y directio. if you dneed both you are dealing with 90 degre right hand moment and gyro precession. and it bocomes substantially more invonlved. Feb 17, 2022 at 18:40
• Ah ok thank you!! So the above calculations from your answer will be all I need. So I will only need the torque due to angular acceleration if I'm understanding correctly? Feb 17, 2022 at 18:47
• yes, as long as you measure the arm length from its hing and keep the motions separate you're fine. Feb 17, 2022 at 20:05
• So, T=a*I, where I has been calculated to be 0.027Nm^2 from your calculations. To find a, I will need to specify how fast I want the arm to turn to a specific angle? Is that correct? Feb 17, 2022 at 20:39