# Help understanding torque - practical application

Dear PF,
This is my first post, so please pardon any faux pas.

I've had some difficulty wrapping my head around the concept of torque. I have taken multiple classes on dynamics, statics, etc. I understand it theoretically, but not practically.

A friend presented me with this problem (non-commercial) that requires an application of torque. He's moved on from the project, but I'd still like to understand it. Any help is very much appreciated, thank you!

// Problem

What motor is needed to turn a camera dolly supporting both a man and a camera, as shown in the attached picture? I've attached a drawing, along with the work I've down on the problem.

Here is a link to a selection of electric motors, rated by horsepower, rpm, and AC/DC. How do I make a selection? How do I know a motor will provide enough torque to start them spinning and get them to at least 10 rpm?

http://www.mcmaster.com/#electric-motors/=t4fns3

#### Attachments

• 140.4 KB Views: 477
• 145.8 KB Views: 458

Related Mechanical Engineering News on Phys.org
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

jack action
Gold Member
In a free rotating component (no load), at constant speed, the motor torque is fighting the friction coming from the plain or roller bearings. If the speed are high enough, aerodynamic forces might also become important (or hydrodynamic forces if the component is in a liquid).

In addition, while getting to that constant speed, the acceleration comes also into play. The motor torque fights then friction and the inertia.

But most importantly, you must consider the power required by the motor more than its torque. The power is always constant from the motor to the rotating component, but the torque and rpm can be set to anything with the proper gear ratio.

1 person
jim hardy
Gold Member
2019 Award
Dearly Missed
You'll want a gearmotor, and maybe a variable speed one.

I see you've calculated moment of inertia
How much torque will you need in order to accelerate your inertia to your 10 RPM in whatever time you want it to accelerate?

I see you did a T=I$\alpha$
but i think you need to work in radians
http://hyperphysics.phy-astr.gsu.edu/hbase/n2r.html

And are there rolling supports under your man and camera? If so, take a fish scale and see how much force you need to drag them along . That times six feet is the torque you'll need to keep 'em turning once they've been accelerated.

1 person
Eureka! You guys are awesome.

I think I've got a handle on it now.

1. F=ma to T=I(alpha): If there is no load, to keep constant w, input T has to equal resistive f.
2. To accelerate w to a higher w, just use T=I(alpha), considering any friction or aerodynamic forces.
3. Same for if there is a resistive load, then dealing with a net T.

In addition to that, the equation Power = (rotational speed) x (Torque) is key. If a motor has a set rpm under no load, can translate that to a load through a gear ratio. Also, knowing the motor's power will allow you to calculate the Torque.

In that light, as long as it can overcome friction forces, a motor can turn nearly any load as long as it spins at a low enough rpm. It was just accelerate really slowly.

I guess the same goes for linear. On a frictionless plane, an ant can move a boulder. It will just accelerate really, really slowly.

Is that the case?

I think everything I posted above is right, but I've been lead to another question while returning to the original problem.

The loads (man and camera) are perpendicular to the rotation of the motor. Will these loads only translate into in-plane friction that the motor has to overcome and contribute to the system's moment of inertia? Is that all there is to it?

Also, I'm having trouble calculating the friction the motor has to overcome.
assume: friction coefficient ~ .2

But that would only conclude to a friction in N, not N*m. Is there a different way to calculate rotary friction?

Last edited:
jack action
Gold Member
I think everything I posted above is right, but I've been lead to another question while returning to the original problem.

The loads (man and camera) are perpendicular to the rotation of the motor. Will these loads only translate into in-plane friction that the motor has to overcome and contribute to the system's moment of inertia? Is that all there is to it?

Also, I'm having trouble calculating the friction the motor has to overcome.
assume: friction coefficient ~ .2

But that would only conclude to a friction in N, not N*m. Is there a different way to calculate rotary friction?
For the friction, all you have to do is multiply the friction force with the distance between the friction force and the center of rotation.
--> Rolling Bearing Friction

And, yes, for the camera, the load has to be supported somewhere, which will translate into a friction force in the plane of rotation. From your drawing, it would be at the rolling support under the camera.

1 person
A-ha! That makes sense. Thank you.

However, I see that since the rollers are directly below the loads (person, camera) it makes sense that would mark the distance from the axis of rotation for calculating rotational friction.

However, what if the rollers were not right below them? What if they were halfway towards the axis of rotation? Would the distance be to the rollers, or to the load for calculating rotational friction?

jack action
Gold Member
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque.
In your case the force that creates the torque is the friction force (and its is perfectly perpendicular to the lever arm), so it is the distance from that force to the center of rotation that counts as the lever arm's length.

Have I got it?

I think I've got a handle now on how to choose a motor. Is this the correct mindset? Please let me know if I've got it or not.

Also, if anyone can point me towards a similar problem with a known solution, I'd love to practice.

Thanks again for everyone who has helped me understand this. I hope this thread can also help those that were in a similar situation.

//

Using SI units (radians, seconds, meters, Newtons, kg):

[1]
Choose desired angular velocity (ω) for system
Choose desired angular acceleration (a) for system
Know load in-plane with rotation (F-ip)
Know load perpendicular to plane of rotation (F-perp)
Know distances loads are to axis of rotation (r)
Known moment of inertia of system (I)
Know coefficient of friction of ball bearings (f-coeff)
[1.1] (R-torque) = (F-ip) * r​
[1.2] (f-rotary) = (F-perp) * (f-coeff) * r​

[2]
Find motor with known (ω) setting and power (P)
Calculate applied torque (T) at that (ω) setting
[2.1] P = T * ω​
Take these variables of ω and T as input variables (ω-in and T-in)

[3]
Implement a gear ratio (R) to translate the motor's (ω-in) setting to desired (ω-out)
[3.1] R = (ω-in)/(ω-out)​
Use this (R) to calculate new output torque (T-out)
[3.2] R = (T-out)/(T-in)​
Calculate any addition rotary friction that gears produce (f-gears) using eq. [1.2]

[4]
Is (T-out) > (R-torque + f-rotary + f-gears)?
If not, the system will not rotate. Choose a more powerful motor or variable motor with higher torque setting.​
If so, the difference (T-net) will go towards angular acceleration (a)​
[4.1] (T-net) = I * a​
Is target (a) met?
If not, choose a more powerful motor or variable motor with higher torque setting.​
If so, this motor is satisfactory.​

[5]
Take note that while these conditions may be met, the motor may be much more powerful than is required. A less powerful motor may be satisfactory.
Also consider any factor of safety required for the resistive load.
This does not consider what type of motor is being used, only its characteristics.

jack action