Question

finding translational and rotational speeds...
Pls help!

Answer 1

whats you poblem??

Answer 2

pls help~

Answer 3

@gerryliyana ?

Answer 4

yes.., but i'm not sure..,

Answer 5

any ideas @furnessj ???

Answer 6

I've had a look at this, but never ever heard of translational speed... (i'm in the UK, it may be a colloqial physics term?). Rotational speed implies the rod spins on a point (you would think half way along) but I can't see how to include the f in any calculation without (it must be included, why else is it mentioned). So, no, no ideas.. Are there any formulas from yuor course @Cecily ?

Answer 7

wait @furnessj so it MUST be rotating about midpoint M... err... or centre of mass ?

Answer 8

Ok @Cecily

first of all.., (for inelastic, e = 0)

e = - (v1' - v2')/(v1 - v2)
0 = - (v1' - v2')/(v1 - v2)
v1' = v2'

Base on the Momentum Law

m1 v1 + m2 v2 = m1 v1' + m2 v2'
m v + M 0 = m v' + M v' (because v' = v1' = v2')

m v + 0 = (m+M) v'
m v = (m+M) v'

v' = mv/(m + M)

Answer 9

that's for translational...,

Answer 10

I got the same answer for what I thought translational speed must be. Just using v = rw for rotational velocity, and the fact that the velocity is this v' we get this answer
\[\omega = \frac{ m v }{ (M+m) } f L\]

Answer 11

but i'm not sure for what I though translational speed.., wait a sec,

Answer 12

@furnessj do you mean r=fL ?

Answer 13

yes, fL is the radius of rotation, so you are just converting a translated horizontal velocity into an angular (rotational) one.

Answer 14

I don't quite understand... v=rw for sure, then?
why w=v'r

Answer 15

sorry, the r shuold be on the bottom! \[ω = \frac{mv}{ (M+m) fL}\]

Answer 16

yes this makes more sense... but doesn't seem like the final answer.

Answer 17

I think the centre of mass is changed and this should be mentioned, am I correct?
(centre of mass shifting up after collision)

Answer 18

well thx @gerryliyana for the translational speed. I am OK with that now, but I am still digesting for rotational...

Answer 19

My friend used conservation of angular momentum and she gets \[\omega=\frac{ 3mvf }{ L((m+M)+mf) }\]

Answer 20

I don't have time to check this but there was no angular momentum before, not sure how that can work..

Answer 21

I have got the method to solve from this youtuve video with numerical inputs. posting here for your reference. Thanks again both!! **gratitude**
http://www.youtube.com/watch?v=K-KftkOGuHE&feature=plcp

Answer 22

congrats Cecily..., :).., btw how old are you?? where ur school??

Answer 23

thank you for the video.., :)

Answer 24

Great video to find! And I learned something :)