Angular momentum of a uniform bar w/ two forces applied

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SUMMARY

The discussion focuses on calculating the angular momentum of a uniform bar with a length of 0.80 m, rotating about an axis at its left end. The moment of inertia (I) is determined using the formula I = (1/3)mL^2, resulting in I = 16m/75. The angular momentum (L) is calculated using L = Iw and L = I(alpha)t, where the net torque is found to be 2.4. The participant seeks clarification on incorporating forces into the angular momentum equation and the role of torque in modifying angular momentum.

PREREQUISITES
  • Understanding of angular momentum and its formulas (L = Iw, L = I(alpha)t)
  • Knowledge of moment of inertia calculations for rigid bodies
  • Familiarity with torque and its effect on rotational motion
  • Basic principles of rotational dynamics
NEXT STEPS
  • Study the relationship between torque and angular momentum in rigid body dynamics
  • Learn about the effects of applied forces on the angular momentum of rotating bodies
  • Explore advanced applications of moment of inertia in different geometries
  • Investigate the role of angular acceleration in rotational motion analysis
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of rotating systems will benefit from this discussion.

Linus Pauling
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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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