
#1
Jul608, 01:07 PM

P: 11

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 = 1/2mr^{2} 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? 



#2
Jul608, 01:40 PM

P: 11

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? 



#3
Jul608, 02:03 PM

P: 134





#4
Jul608, 02:25 PM

P: 11

Uniform disk on incline problem,solving w/out height or length while only given angle
im not understanding whre the Vcm^2 = 4gh/3 is coming from though..




#5
Jul608, 03:22 PM

P: 134

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




#6
Jul608, 03:26 PM

P: 11

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




#7
Jul608, 03:44 PM

P: 11

hey crypto,
im at the point.. 2*a*d*sin(theta)=4gh/3 that reduces to a = 2g/3sin(theta) for some reason.. 



#8
Jul608, 03:50 PM

P: 134

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]




#9
Jul608, 03:57 PM

P: 11

well my equation yields 25.42 while the correct one yields 1.69




#10
Jul608, 04:08 PM

P: 134

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




#11
Jul608, 04:10 PM

P: 11

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



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