
#1
Apr908, 08:22 PM

P: 360

1. The problem statement, all variables and given/known data
A cylinder with mass m1 = 4.00kg, and radius 30cm, rotates about a vertical, frictionless axle with angular velocity of 8.00 rev/s. A second cylinder, this one have a mass of m2 = 3.00kg, and radius 20cm, initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed. A) calculate the final angular speed B)Find the energy lost in the system due to the interaction of the two cylinders. 3. The attempt at a solution Just wondering if this is right: I converted the 8 rev/s to 50.265 rad/s. I wasn't sure if i needed to but I did it anyway. [tex] L_{i} = L_{f}[/tex] [tex]I_{1}\omega_{1i}^2 = (I_{1} + I_{2})\omega_{f}^2[/tex] [tex]\omega_{f} = \sqrt{\frac{I_{1}\omega_{1i}^2}{(I_{1} + I_{2})}} [/tex] [tex] = \sqrt{\frac{(.5)(4)(0.3^2)(50.265^2)}{(0.5)(4)(0.3^2) + (0.5)(3)(0.2^2)}[/tex] = 43.53 rad/s Then for part B, I found the loss of energy due to friction using the change in kinetic energy. [tex]0.5(I_{1} + I_{2})\omega_{f}^2  0.5(I_{1})\omega_{f}^2^2 [/tex] [tex] 0.5[(0.5)(4)(0.3^2) + (0.5)(3)(0.2^2)](43.53)^2  0.5((.5)(4)(0.3^2)(50.265)^2 [/tex] = 51.17 J Did I solve this problem correctly? 



#2
Apr908, 08:48 PM

HW Helper
P: 1,542

Angular momentum is [tex]L = I\omega [/tex]. [tex]\omega [/tex] is NOT squared.




#3
Apr908, 08:49 PM

P: 360

crap. Other then that is that the right process (don't mind the numbers now)




#4
Apr908, 08:58 PM

HW Helper
P: 1,542

[SOLVED] Angular speed problem
Yes, I think your approach is OK.



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