Rotations from angular acceleration and final angular velocity

AI Thread Summary
Hank and Finn are rotating a 5.4 m bar about its center, applying forces of 32 N and 18 N, resulting in a constant angular acceleration of 0.35 rad/s². They achieve an angular speed of 3.1 rad/s, leading to a calculated angular displacement of 13.7 radians. This displacement translates to approximately 2.18 rotations around the pole. The discussion clarifies that torque calculations are not necessary for this specific problem, which focuses on angular motion rather than mass determination. The participants confirm the method used is correct, despite initial uncertainties about the calculations.
NY152
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Homework Statement


At a fair, Hank and Finn play with a horizontal 5.4 m long bar able to rotate about a pole going through its exact center. Hank pushes with 32 N at one end of the bar and Finn pushes with 18 N in the opposite direction at the other end. (Assume both forces are always perpendicular to the bar.) The bar rotates from rest with a constant angular acceleration of 0.35 rad/s^2.

Hank and Finn were able to spin the bar up to an angular speed of 3.1 rad/s. How many turns around the pole did they make to push the bar?

Homework Equations


w^2=wi^2+2(alpha)theta

The Attempt at a Solution


(3.1)^2=0+(2(.35)theta
theta =13.7
theta/2pi=#of rotations
=2.18 rotations

I don't have the correct answer for this so I have no idea if I'm doing this correct or not.
 
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Your method and result are fine.

Please don't make pleas for help in your thread titles. The question didn't really involve torque calculations, did it? I'll change the title to something appropriate.
 
gneill said:
Your method and result are fine.

Please don't make pleas for help in your thread titles. The question didn't really involve torque calculations, did it? I'll change the title to something appropriate.
Sorry about that, and there are two forces acting on it, but you find the mass later by doing the sum of the torques=I*alpha
That wouldn't apply to this would it?
 
NY152 said:
Sorry about that, and there are two forces acting on it, but you find the mass later by doing the sum of the torques=I*alpha
That wouldn't apply to this would it?
It would apply to a question where it asks you to find the mass, but that wasn't the question asked this time.
 
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