Rotational angular mommentum?

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A barbell consists of two small balls, each with mass 400 grams (0.4 kg), at the ends of a very low mass rod of length d = 35 cm (0.35 m). The center of the barbell is mounted on the end of a low mass rigid rod of length b = 0.525 m (see Figure). The apparatus is started in such a way that although the rod rotates clockwise with angular speed 80 rad/s, the barbell maintains its vertical orientation.

calculate Lrot
Lrot= I[tex]\omega[/tex]
I=mr^2

i tried
(.8kg)*.35^2*80rads/s
and
(.8kg)*.525^2*80rads/s
both are wrong can someone help me?

and i also need help finding the the translational angular momentum too
 
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  • #2
Doc Al
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Can you post the figure? (Post the entire problem if you can.)
 
  • #3
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yes here it is
 

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  • #4
Doc Al
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i tried
(.8kg)*.35^2*80rads/s
I'd say that is correct. There's no angular momentum of the barbell about its center of mass, since it doesn't rotate.

and i also need help finding the the translational [STRIKE]angular [/STRIKE]momentum too
What have you tried? The direction of the translational momentum (not angular) depends on where it is in its motion.
 
  • #5
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(.8kg)*.35^2*80rads/s is wrong :( and im just so lost right now. web assign keeps telling me im wrong and you say im right so im so confused right now
 
  • #6
Doc Al
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(.8kg)*.35^2*80rads/s is wrong :( and im just so lost right now. web assign keeps telling me im wrong and you say im right so im so confused right now
Oops... my bad. That should be 0.525 m, not 0.35. (I got the distances mixed up.) So your other choice was correct:
(.8kg)*.525^2*80rads/s
You need both magnitude and direction. What did you put for the direction?
 
  • #7
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zero magnitude and no direction
 
  • #8
Doc Al
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zero magnitude and no direction
:confused: You just gave the formula for calculating the magnitude, so how can it be zero magnitude?
 
  • #9
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the manitude of Ltrans = r*psin theta right? so it would be .8kg*80rads/s*.525m?
 
  • #10
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for Lrot is 0 magnitude and no direction. i got that part right
 
  • #11
Doc Al
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the manitude of Ltrans = r*psin theta right? so it would be .8kg*80rads/s*.525m?
Yes.

for Lrot is 0 magnitude and no direction. i got that part right
So they wanted Lrot about the center of mass, not the axis of rotation (point B in the diagram)? In which case the attempts in your first posts were irrelevant?
 
  • #12
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i have no idea :(
 
  • #13
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the manitude of Ltrans = r*psin theta right? so it would be .8kg*80rads/s*.525m?
its wrong i tried -33.6 and positive 33.6 both are wrong :(
 
  • #14
same figure in the book
-------------------------------------
11.X.6
The barbell in the previous exercise is mounted on the end of a low-mass rigid rod of length b = 0.9m (Figure 11.22). The apparatus is started in such a way that although the rod rotates clockwise with angular speed w1=15 rad/s, the barbell maintains its vertical orientation.

same figure in the book
Figure 11.22 A barbell pivoted on a low-mass rotating rod. The barbell does not rotate.


(a) Calculate Lrot (both direction and magnitude).
(b) Calculate Ltrans,B (both direction and magnitude).
(c) Calculate Ltot,B (both direction and magnitude).

Answer:
(a)Lrot=0 ; (b) Ltrans = 9.72 kg · m2/s into page; (c)Ltot = 9.72 kg · m2/s into page
 

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