Rotational Kinetic Energy Test Review Help

[km41]

Homework Statement

There is a system consisting of two masses, m1=20 kg, m2=30kg on a pulley and m2 is 2 meters above the ground, while m1 is on the ground.
This is the question: the system released from rest a 30 kg block that is 2m above a ledge. The pulley is a disk with radius of 10 cm and a mass of 5 kg.
a) Find the speed of the 30 kg block before it hits the ledge.
b) Find the angular speed of the pulley at that time.
c) Find the tensions in the strings.

Homework Equations

I'm going to assume: Initial=Final
1. KE(rotational)+KE (linear)+ PE= KE(rotational)+KE (linear)+ PE
2. KE= (1/2)Inertia x (Omega Squared)
3. Inertia= Mass x (Radius Squared)
4. Velocity (linear)= radius x omega

I can't think of any other, but there probably is

The Attempt at a Solution

a. I think I have to use the first equation, The KER cancels out and the KE (linear) is out leaving only the PE. I can't really figure it out from there?

 

[tiny-tim]

welcome to pf!

hi km41! welcome to pf!

I'm going to assume: Initial=Final
1. KE(rotational)+KE (linear)+ PE= KE(rotational)+KE (linear)+ PE
…
a. I think I have to use the first equation, The KER cancels out and the KE (linear) is out leaving only the PE. I can't really figure it out from there?

(why are you assuming Initial=Final? )

yes, use the first equation

but the KER does not cancel out, it starts 0, and it increases until just before mass 2 hits the ledge

show us that equation with your figures in ​

 

[km41]

What I meant to say was KER was 0. Sorry about that, I'm just flustered.
0 + 0 + m2gh= m1gh+ 1/2(m1 x (v squared) + m2 x (v squared) + I x (Omega squared))
??

 

[tiny-tim]

hi km41!

(have an omega: ω and try using the X² and X₂ icons just above the Reply box )

yes, that's correct …

now find a formula expressing ω in terms of v, and solve ​

 

[km41]

ω = v/r

So..
0 + 0 + m2gh= m1gh+ 1/2(m1 x ( v squared) + m2 x ( v squared) + I x (ω squared)

30 x 9.8 x 2 = 0 + 1/2 ( 20 x v squared + 30 x v squared + I (v/r)squared

 

[tiny-tim]

ω = v/r

yes

0 + 0 + m2gh= m1gh+ 1/2(m1 x ( v) + m2 x ( v) + I x (ω)

30 x 9.8 x 2 = 0 + 1/2 ( 20 x v + 30 x v + I (v/r)

your gh terms are wrong, and you've written v instead of v²

try again ​

 

[km41]

How is the gh wrong?

Isn't it 30 x 9.8 x 2
(Mass 2 x gravity x height from the ledge)

 

[tiny-tim]

yes but you need some m₁gh also

 

[km41]

In the Initial or Final or Both?
I'm not even sure how to do that??

 

[km41]

Ok nevermind I got it.

m2gh= m1gh + .5 (m1(vsq) + m2 (vsq) + .5m(vsq))

 

[km41]

Tiny- tim, Thank you so much for your help

I solved the rest of it and I got it. Thanks!