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Old October 22nd, 2018, 09:01 PM   #1
Timeismoney08
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Torque Required - Physics Help

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!
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Old October 22nd, 2018, 10:14 PM   #2
Ken Roach
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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.
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Old October 22nd, 2018, 10:28 PM   #3
Timeismoney08
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Quote:
Originally Posted by Ken Roach View Post
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.
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Old October 22nd, 2018, 10:58 PM   #4
Ken Roach
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Quote:
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.
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Old October 23rd, 2018, 10:39 AM   #5
Timeismoney08
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Quote:
Originally Posted by Ken Roach View Post
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.
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Old October 23rd, 2018, 02:15 PM   #6
L D[AR2,P#0.0]
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Kens figures use a radius of 1cm but you refer to a diameter of 1cm.....
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Old October 23rd, 2018, 02:50 PM   #7
OkiePC
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This is one of those many cases where a picture would be worth a thousand words.
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It's not all the variables I am most concerned with, it's the undiscovered constants.
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