How do I determine θ'' and tension in a speed-governing mechanism?

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To determine θ'' and tension in a speed-governing mechanism, the initial parameters include mass (m = 0.2 kg), length (L = 10 cm), angular velocity (w = 500 rpm), and angle (θ = 40°). The angular velocity is converted to radians per second, yielding ω = 52.36 rad/s. The tension is calculated using the vertical component of forces, resulting in T = 2.56 N. The discussion highlights confusion regarding the use of polar coordinates and the distinction between net force and centripetal acceleration. Understanding the frame of reference is crucial for accurately analyzing the forces involved.
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Homework Statement


In the speed-governing mechanism m = 0.2 kg, L = 10cm, (rods AB have negligible mass), w = 500 rpm, θ = 40°,θ' = 0 . Determine θ'' at that instant and the tension in the rod.
upload_2017-1-7_0-31-35.png

Homework Equations


F=[ω][/2]r

The Attempt at a Solution


First, I calculated ω = 500 rpm * 2π /60 = 52.36 rad/s.
From the vertical components, I determined T*cosθ = mg
So T = 2.56 N
With this, I can calculate the horizontal component of T, but then I have a different ω for the balls and I'm confused. Should I use polar coordinates? If so, how should I start?
 
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EserRose said:
F=[ω][/2]r
What is F here? If it's a force, the dimensions don't match.
EserRose said:
then I have a different ω for the balls
How do you mean? Remember, this is not equilibrium.
 
haruspex said:
What is F here? If it's a force, the dimensions don't match.

Ooos I meant m*ω^2 *r

haruspex said:
How do you mean? Remember, this is not equilibrium.

I see that, so I should find the net force then?
 
EserRose said:
so I should find the net force then?
Sort of.. depends whether you want to use the inertial frame + centripetal view or object frame + centrifugal view.
In the latter case it would be net force; in the former you might describe it as the excess acceleration.
 
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