
#1
Nov511, 10:45 PM

P: 59

Hi guys so this is my first time posting a question on this amazing forum, i hope i am doing this right since i read that no one will help me if i am missing something. Anyways, i am dealing with a physics problem that it driving me crazy (mainly because my teacher does not explain anything!) Here is the problem.
1. The problem statement, all variables and given/known data An 7.50cmdiameter, 320 g sphere is released from rest at the top of a 1.90mlong, 20.0 incline. It rolls, without slipping, to the bottom. b. What fraction of its kinetic energy is rotational? So the known data is: mass= 0.0320 kg radius= 0.0375 m delta x= 1.90m wi = 0 vf = 3.02 m/s wf = 80.5 rad/s Find ratio between rotational and kinectic energies. 2. Relevant equations Kroll = 1/2 (I* w^2) + 1/2(M)(vf)^2 = Krot + Kcm 3. The attempt at a solution ok so from part A of the problem i found the velocity at the bottom of the incline to be 3.02 m/s, then i found the final angular velocity. With that all i thought i had to do was plug the data into the previous equation and get an answer: Since moment of inertia of a cylinder is 2/5MR^2 Kroll = 1/2 (2/5 ) (0.032kg)(0.0375m)^2 ( 80.5 rad/s)^2 + 1/2 ( 0.032 kg)(3.02 m/s)^2 I am getting 0.204 for the Kroll and for the Krot i am getting 0.06... when i devide 0.06/0.204 i get 0.3 which is not the answer :( i have no idea what am i doing wrong! please help me, this homework was due 2 days ago, i am just doing it to understand the concept no for the points, so i would really appreciate your help, wonderful people, thanks! 



#2
Nov511, 11:18 PM

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The mass is 0.32kg, not 0.032kg, but that shouldn't affect the ratio.
Maybe you have too much roundoff error. .3 is about 5% different than the correct answer. 



#3
Nov511, 11:23 PM

P: 59

ok but am i doing this correctly? i mean, am i supposed to divide Krot by Krolling to get a ratio? or is it the other way (Kroll/krot) ?
I don't get it, no matter how i do it, i am still getting around 0.29... and when round i get 0.3 but it's wrong :/ 



#4
Nov511, 11:44 PM

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Kinetic and Rotational energy of a sphere
The fraction of the kinetic energy that is rotational is what you are calling Krot / Kroll .
Why are you rounding to 1 significant figure? 



#5
Nov611, 12:33 AM

P: 59

i am rounding to two sig figures :( 0.30, cause i get 0.298




#6
Nov611, 12:40 PM

P: 59

please i really need to understand this, i am not getting the answer.. :/




#7
Nov611, 12:57 PM

P: 294

The question is What fraction of its kinetic energy is rotational?, your answer is the relationship between the rotational and translational kinetic energies.




#8
Nov611, 01:12 PM

P: 59

ok, i dont know what that means then, i thought you had to get the ratio between them,,, if so then i am getting 0.30, which is supposed to be wrong...




#9
Nov611, 01:51 PM

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So, Krot/Kroll is the fraction he needs. BTW: For a solid sphere rolling without slipping, it can be shown that this fraction is exactly 2/7 . 



#10
Nov611, 04:53 PM

P: 59

hey bro, thanks for helping me, the answer was indeed 2/7 .. but can you explain me how you got that answer?



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