# Torque and angular acceleration , moments , counterweight

As mentioned before i am from a mechanical engineering background trying to design a product using counterweight. I was thrown into confusion regarding torque and moment even after doing my own research. The way mechanical engineer termed moment and torque is quite different from how it is defined in physics which confuses me in terms of understanding the concept. I have a few questions so i will try my best to list it down in a more organized way hopefully it can be understood as i'm not very good at phrasing. sorry in advanced.

$$Torque = force \times distance \tag{1}$$

Torque = moment of inertia $$\times$$ angular acceleration $$\tag{2}$$

The torque found in equation 1 can it be subbed into equation 2 and use that to find either moment of inertia or angular acceleration ? or it's a two entire different thing ?

1. I understand that moment of inertia is a calculation for the whole entire system

But how do i translate that understanding to torque = moment of inertia x acceleration . Can it still be done in this way to understand if $$T_1 > T_2$$ , it will turn anti-clockwise?

1. If i were to shift my mass 1 (counterweight) to the right , it will turn clockwise ? In this case i will have to use $$Torque = force \times moment$$ ?

I have done my research and did some experiment and it is seen that the heavier side of the beam ( in this case the right side of the beam ) will start to rotate and will eventually come to a stop at $$90 ^\circ$$ form - heavier side will be directly below the pivot and lighter side will be above the pivot . Why is this so? Is it due to the acceleration ? or am i missing out something?

Sorry i have been thinking about this for days and have tried to sought out some help from my engineering peers. However, we are not taught in-depth regarding torque and most of the time we used the term $$moment = force \times distance$$.

## 1 Answer

The torque you get by using force*distance will be equal to Ia where I is the moment of inertia of system about axis of rotation and a in the angular acceleration of system about axis of rotation. In your case it would be |m1gr1-m2gr2|=Ia(I haven't taken sign conventions into account but magnitude vice they are equal)

Where a is angular acceleration of the entire system about the pivot.

Your second question is why it would stop when the system becomes 90 degree with pivot. this is because the torque about pivot is 0 hence the angular acceleration would be (0. (Strictly speaking it would oscillate a bit before stopping though)

• I'm not sure I fully understood your question, Please let me know. Feb 25, 2019 at 5:49
• wow thanks a lot ! so it doesn't matter how much my counterweight should shift as long as it it not in equilibrium , it will start rotating and oscillates a bit before coming to a stop at 90 degree? But why is the torque at pivot = 0 ? Is there a term for it? Feb 25, 2019 at 5:57
• Torque isn't actually the product of force and distance , it is the Vector product (or cross product) of force and distance which means we need to take the product of the component of force which is perpendicular to the distance. Imagine the plank to be at 90 degree. now the force acts vertically downwards, if we draw a line from pivot to the force we find that the component of force perpendicular to the line is 0, hence the torque is 0. The thing oscillates because, imagine the plank exceeds 90 degree, the torque applied by weight would oppose the motion. Feb 25, 2019 at 6:33
• In Your case the force was already perpendicular to the distance so we can simply take the product of force and distance, this isnt true in all cases, (like the case where plank becomes 90 degree or any other angle) Feb 25, 2019 at 6:35
• Wow thanks a lot ! You are a lifesaver ! Can i clarify with you on another question. Feb 25, 2019 at 7:19