Finding angular velocity using conservation of angular momentum

AI Thread Summary
The discussion revolves around a physics problem involving a rotating rod and a bug crawling to its end. The user attempts to apply the conservation of angular momentum but initially miscalculates by assuming the bug's initial angular velocity is zero. A respondent points out that the bug's initial angular velocity should match that of the rod, which clarifies the user's misunderstanding. The user acknowledges this correction and expresses gratitude for the guidance. The conversation highlights the importance of correctly applying initial conditions in angular momentum problems.
cosurfr
Messages
5
Reaction score
0
Hello everyone, I hope that someone can help me solve this rather than solving it for me, here's the problem and what I've done so far...oh and thanks in advance!

Homework Statement



A thin rod has a length of 0.25 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.32 rad/s and a moment of inertia of (1.1 x 10^-3 kg x m^2). A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass = 4.2 x 10^-3kg) gets where it’s going, what is the angular velocity of the rod?


Homework Equations


*Lf=angular momentum final
*Lo=angular momentum initial


Conservation of angular momentum: L=Iw -> Lf=Lo -> Iwf=Iwo, wo=0
Iwf=0

The Attempt at a Solution


Okay so I thought this problem seemed fairly straight forward, so hopefully I'm just doing a little algebra incorrectly or something, but here's my whirl at the answer
Iwf=0
(Moment of Inertia of rod)(ang vel of rod)+(mass of bug)(length of rod)^2(w)=0

(1.1 x 10^-3 kg x m^2)(0.32 rad/s) + (4.2 x 10^-3kg)(0.25m)^2(w)=0

(4.2 x 10^-3kg)(0.25m)^2(w)= -(1.1 x 10^-3 kg x m^2)(0.32 rad/s)

*****Now just solve for w? Or is this completely wrong? I did do the calculations but didnt want to put all that up if this wasnt even the right way to set up the problem. If someone could let me know if I'm on the right track and lend me a hint of how to proceed i'd really appreciate it. Thanks

Cosurfr
 
Physics news on Phys.org
cosurfr said:
A thin rod has a length of 0.25 m and rotates ...has an angular velocity of 0.32 rad/s ...

Why did you use w0 = 0?

Go back towards the top of your proof, and use this given value for w0 instead ... see what happens.
 
Thanks so much. Thats exactly where i went wrong. The initial ang vel of the bug would have been the same as the initial ang vel of the rod (constant). That cleared it all up. Thanks a ton for not just doing it for me but rather helping me find the way. This website is the best FYI

cosurfr
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top