Rotational angular mommentum?

[ohheytai]

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$$\omega$$
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

 

[Doc Al]

Can you post the figure? (Post the entire problem if you can.)

 

[ohheytai]

yes here it is

 

[Doc Al]

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.

 

[ohheytai]

(.8kg)*.35^2*80rads/s is wrong :( and I am just so lost right now. web assign keeps telling me I am wrong and you say I am right so I am so confused right now

 

[Doc Al]

(.8kg)*.35^2*80rads/s is wrong :( and I am just so lost right now. web assign keeps telling me I am wrong and you say I am right so I am 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?

 

[ohheytai]

zero magnitude and no direction

 

[Doc Al]

zero magnitude and no direction

You just gave the formula for calculating the magnitude, so how can it be zero magnitude?

 

[ohheytai]

the manitude of Ltrans = r*psin theta right? so it would be .8kg*80rads/s*.525m?

 

[ohheytai]

for Lrot is 0 magnitude and no direction. i got that part right

 

[Doc Al]

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?

 

[ohheytai]

i have no idea :(

 

[ohheytai]

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 :(

 

[vietsubject]

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