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posted by  bigtexmexbear on 11/10/2007 12:03:28 PM  |  status: Live  

linear speed of cylinder when it reaches bottom of ramp

Course Textbook Chapter Problem
Algebra Based Physics N/A N/A N/A
Question Details:
A solid cylinder of mass 3.0Kg and radius 0.2 m starts from rest at the top of a ramp, inclined 15o, and it rolls to the bottom with out slipping. ( for a cylinder I = 0.5MR2) The upper end of the ramp is 1.2m higher than the lower end . Find the linear speed of the cylinder when it reaches the bottom of the ramp. (g = 9.8 m/s2).
Choices:
1.  4.7 m/s
2.  4.3 m/s
3.  4.0 m/s
4.  2.4 m/s
5.  2.2 m/s
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posted by ktqu on 11/10/2007 3:13:23 PM  |  status: Live
Asker's Rating: Lifesaver   
bigtexmexbear's comment:
"thank you very much"
Response Details:
Use conservation of energy.
KEi +PEi = KEf +PEf
Since it is rolling KE must be split into translational and rotational components.
KEr = 1/2 Iω2. (ω = v/r) KEr = 1/2 (1/2 MR2)(v/R)2 = 1/4Mv2
KEt = 1/2 mv2
 
(1/4Mvi2 + 1/2Mvi2) + (mghi) = (1/4Mvf2 + 1/2Mvf2) + (mghf)
Since mass is in every term it can be cancelled out.
The final height (hf) is zero because it is at the bottom of the ramp.
The initial height (hi) is 1.2 m as stated in the problem.
The initial velcity (vi) is 0 because it starts from rest.
Therefore, ghi = 3/4vf2
9.8(1.2) = 3/4vf2
vf2 = 15.68
vf = 3.96 m/s
 
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