Uniform disk on incline problem,solving w/out height or length while only given angle

nopistons93

1. The problem statement, all variables and given/known data
A solid uniform disk of radius 0.250 m and mass 55.0 kg rolls down an incline that makes an angle of 15 degrees with the horizontal. The disk starts from rest from the top of the ramp. Find the speed of the disk's center of mass when it reaches the bottom of the ramp(linear speed), its angular speed omega at that point and its KE rotational and KE translational energies.


2. Relevant equations
V(linear) = [tex]omega[/tex]*R
[tex]\alpha[/tex]=R*[tex]\omega[/tex]²
v=R*[tex]\omega[/tex]
a=R*[tex]\alpha[/tex]

PE = KE_rotational + KE_{translational}

I = 1/2mr²

3. The attempt at a solution

I solved for I and got 1.72 kg * m^2 however without height or length I can't solve for the PE side inorder to get the KE.

I've seen the equation [tex]\alpha[/tex]=2gsin[tex]\Theta[/tex]/3R and thats the only thing I've noticed that would allow me to solve for either V or [tex]\omega[/tex] in this case. Am I missing something here, some equation which relates another way to this problem?

nopistons93

ok so deriving the formula a = g*sin(theta)/1.5 , I found out that a = 1.69 m/s^2

since alpha = a/r , I found alpha to be 6.76 m/s^2.

since alpha=r*omega^s , i found omega to be 5.20 m/s again.

Solving for v from (v=r*omega), I found v equal to 1.30 m/s

i then plugged these numbers into the KE formulas to solve for them. Does that sound right?

cryptoguy

This equations is actually easily derived. Just start working out the problem assuming you do have h (height), use PE = KE. You will end up with [tex]Vcm^2 = \frac{4gh}{3}[/tex]. You then need to remember the equation [tex]V^2 = Vi^2 + 2ad[/tex]. Since Vi = 0, you will be able to plug in that equation for V^2 and get the equation you mentioned (noticed that h (height) cancels out, so it is not needed)

nopistons93

im not understanding whre the Vcm^2 = 4gh/3 is coming from though..

cryptoguy

all right well you start out with mgh = 1/2 I(omega)^2 + 1/2 1/2 mv^2. I = 1/2mr^2 as you said. omega = v/r. After a little algebra you should arrive at Vcm^2 = 4gh/3

nopistons93

ok exactly what I was looking for there man. thank you very much\

nopistons93

hey crypto,

im at the point..

2*a*d*sin(theta)=4gh/3

that reduces to a = 2g/3sin(theta) for some reason..

cryptoguy

that's right, it's pretty much the same equation as you wrote, divide both sides by r (radius), and you will get [tex]\alpha = \frac{2gsin(\theta)}{3R}[/tex]

nopistons93

well my equation yields 25.42 while the correct one yields 1.69

cryptoguy

you mean for a (acceleration)? didn't you get 1.69 a couple posts back?

nopistons93

Yes, just putting radius in the formula gives you alpha. I do understand I think I'm just writing down my algebra wrong.