Angular momentum of a uniform bar w/ two forces applied

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The discussion focuses on calculating the angular momentum of a uniform bar rotating about its left end after 6 seconds. The moment of inertia (I) is determined to be (1/3)mL^2, leading to the expression for angular momentum (L) as L = (16m/75)*6sec*alpha. Participants discuss incorporating forces into the calculation, emphasizing the role of torque in changing angular momentum. The net torque is calculated as 2.4, which is equated to the derivative of angular momentum. The thread highlights the importance of understanding the relationship between torque, angular momentum, and the forces acting on the rotating body.
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1. The uniform bar shown in the diagram has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum of the bar 6.0 s after the motion has begun? The forces acting on the bar are shown.

170580A.jpg




2. L = Iw, L = I(alpha)t



3. I know that I = (1/3)mL^2 = 16m/75

So, L = (16m/75)*6sec*alpha

Am I going about this correctly so far? How do I bring the forces into the equation?
 
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What do you apply to a rigid, rotating body to increase/decrease it's angular momentum?
 
... torque. I still don't see what I'm doing. net torque = 4*.6*sin(90) = 2.4 = the derivative of L
 
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