Torque Required - Physics Help

Timeismoney08 · Oct 22, 2018

I have a small application where I need to spec out an actuator.

The load is 7kg and needs to move 90 degrees around a 1cm diameter in 1 second. What is the torque required to achieve this?

Can someone help me work out the physics here?

Thanks for the help!

Ken Roach · Oct 22, 2018

F = m * A

Force equals mass times acceleration; it really is that simple.

You know the mass; 7 kilograms.

When you say "move 90 degrees in 1 second" do you mean that it needs to start, move 90 degrees, and stop in 1 second ?

A one centimeter torque arm seems unusually short, unless this is a long cylindrical load. Can you describe it in more detail ?

An arc of 90 degrees with a radius of 1 centimeter is (pi * 2 cm / 4) = 1.57 cm.

So you have to accelerate to get to 7.85 mm in 500 ms.

Distance traveled under constant acceleration is straightforward: s = 1/2 a * t^2.

.00785 meters = 1/2 * acceleration * (0.5 seconds)^2

Acceleration = 0.0628 meters/sec2

Because we know the mass and acceleration, we can calculate the linear force:

F = m * a = 7 kg * 0.0628 meters/sec2 = 0.4396 newtons.

With a torque arm of 0.01 meters, the torque required is 0.004396 newton-meters.

Timeismoney08 · Oct 22, 2018

Ken Roach said:
Force equals mass times acceleration; it really is that simple.

You know the mass; 7 kilograms.

When you say "move 90 degrees in 1 second" do you mean that it needs to start, move 90 degrees, and stop in 1 second ?

A one centimeter torque arm seems unusually short, unless this is a long cylindrical load. Can you describe it in more detail ?

An arc of 90 degrees with a radius of 1 centimeter is (pi * 2 cm / 4) = 1.57 cm.

So you have to accelerate to get to 7.85 mm in 500 ms.

Distance traveled under constant acceleration is straightforward: s = 1/2 a * t^2.

.00785 meters = 1/2 * acceleration * (0.5 seconds)^2

Acceleration = 0.0628 meters/sec2

Because we know the mass and acceleration, we can calculate the linear force:

F = m * a = 7 kg * 0.0628 meters/sec2 = 0.4396 newtons.

With a torque arm of 0.01 meters, the torque required is 0.004396 newton-meters.

Thank you for the help.

The part is very close to on center of the rotary motion. It's rotating on a bearing and connected on the outside of the OD and driven by a rod.

I'm comfortable selecting actuators for linear applications, but just needed some help converting that to rotational motion.

Ken Roach · Oct 22, 2018

It's rotating on a bearing and connected on the outside of the OD and driven by a rod.

Are you driving this load from a shaft at the center of rotation, or are you pushing on a rod that's connected to an arm ?

I may have been thinking about the load incorrectly. Explain more about the mechanism.

Timeismoney08 · Oct 23, 2018

Ken Roach said:
Are you driving this load from a shaft at the center of rotation, or are you pushing on a rod that's connected to an arm ?

I may have been thinking about the load incorrectly. Explain more about the mechanism.

Pushing on an arm that rotates the bearing and part in the middle of the bearing.

L D[AR2P#0.0] · Oct 23, 2018

Kens figures use a radius of 1cm but you refer to a diameter of 1cm.....

OkiePC · Oct 23, 2018

This is one of those many cases where a picture would be worth a thousand words.

Torque Required - Physics Help

Timeismoney08

Member

Ken Roach

Lifetime Supporting Member + Moderator

Timeismoney08

Member

Ken Roach

Lifetime Supporting Member + Moderator

Timeismoney08

Member

L D[AR2P#0.0]

Lifetime Supporting Member

OkiePC

Lifetime Supporting Member

Similar Topics