Can someone explain how they got this?

[gvcalamike]

The question was:

A solid cylinder or radius 10cm (.1m) and mass 12kg starts from rest and rolls without slipping a distance L=6m down a roof that is inclined at the angle 30 degrees. What is the angular speed of the cylinder about its center as it leaves the roof?


Step 1: I got the vertical height of the roof: (6.0m)sin 30 = 3m

.: I really don't know when to use the conservation of energy or conservation of mechanical energy, but it seems like most of the problems I've been working use them. Can someone explain when to use it? Nonetheless, I guessed here and here's what I did. Btw, I got the answer right after I got a hint.

Step 2: Going with my guess I did this:
Ki = 0
Ui = mgh = (12)(9.8)(3.0) = 352.8 J
Uf = 0

Okay, here's where I get confused. I can get Kf = 1/2 mv² + 1/2 Iw² (w = omega). Then they used:
1/2 mr²w² +1/2(1/2mr²)w²

So how does 1/2 mv²become 1/2mr² w²

and 1/2 Iw² become 1/2 (1/2mr²)w²
I have an idea, utilizing what you have at hand to come up with something, but I don't really understand how they came about.

So Kf would end up being (3/4)mr²w² and I can solve it from here, but can anyone explain how Kf came about? Thanks

0+352.8 = 3/4mr²w²

 

[tiny-tim]

hi gvcalamike!

(have an omega: ω )

… I really don't know when to use the conservation of energy or conservation of mechanical energy, but it seems like most of the problems I've been working use them. Can someone explain when to use it?

generally, we can use conservation of (mechanical) energy if there's no collision (or jerk) …

so if a body rolls or slides down a hill that curves gently to become horizontal, energy is conserved

but if the hill is straight, so that it meets the flat ground at a "corner", then the body collides with the ground, with a jerk, and energy is lost

So how does 1/2 mv²become 1/2mr² w²

and 1/2 Iw² become 1/2 (1/2mr²)w²

because v = ωr …

this is the rolling constraint, ie the relation between v and ω …

the instantaneous point of contact is stationary, but the c.o.m. is moving with speed v, so as a matter of geometry, the angular speed must be v/r