Kinetic and Rotational energy of a sphere


by hyddro
Tags: energy, kinetic, rotational, sphere
hyddro
hyddro is offline
#1
Nov5-11, 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 7.50-cm-diameter, 320 g sphere is released from rest at the top of a 1.90-m-long, 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!
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SammyS
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#2
Nov5-11, 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 round-off error.

.3 is about 5% different than the correct answer.
hyddro
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#3
Nov5-11, 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 :/

SammyS
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#4
Nov5-11, 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?
hyddro
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#5
Nov6-11, 12:33 AM
P: 59
i am rounding to two sig figures :( 0.30, cause i get 0.298
hyddro
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#6
Nov6-11, 12:40 PM
P: 59
please i really need to understand this, i am not getting the answer.. :/
Villyer
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#7
Nov6-11, 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.
hyddro
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#8
Nov6-11, 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...
SammyS
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#9
Nov6-11, 01:51 PM
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Quote Quote by Villyer View Post
The question is What fraction of its kinetic energy is rotational?, your answer is the relationship between the rotational and translational kinetic energies.
It's clear that what hyddro calls Kroll is the sum of the rotational KE, and the KE of the center of mass, the latter being the translational KE.

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 .
hyddro
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#10
Nov6-11, 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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