How Fast Does a Pivoting Meter Stick Swing?

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The discussion centers on calculating the speed of the center of mass of a pivoting meter stick, weighing 0.200 kg, at the bottom of its swing. The torque acting on the stick is calculated to be 0.981 N*m, derived from the force and the radius of rotation. Participants suggest that while torque can be used to analyze the problem, applying the conservation of energy principle may provide a simpler solution. The conversation highlights the relationship between torque, moment of inertia, and angular acceleration, questioning how to derive angular velocity from these concepts. Ultimately, the focus is on finding an efficient method to determine the speed of the meter stick's center of mass during its swing.
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1. A .200 kg stick pivots about one end. The meter stick starts balanced above this pivot. What would be the speed of the center of mass of the meter stick at the bottom of the swing?



2. Torque=force sin (theta)*radius


3. The torque acting on the object is .981 N*m (9.81*.2*sin 90*.5) (sin 90 because it is 90 degrees above the horizontal axis.)
But how do you get angular velocity from torque? I also know that torque=moment of intertia*angular acceleration. But how do you get angular velocity from this?
Is this even a torque problem?
 
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You can indeed solve this problem by considering the torques acting on the ruler. However, it may be somewhat simpler to consider conservation of energy.
 

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