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Projected cylinder!

  • Thread starter yasar1967
  • Start date
73
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1. A solid uniform cylinder of mass M and Radius is projected up an incline of angle Ө. It rolls without slipping from an initial speed V๐ of the center of mass. What distance s does the center of cylinder travel before it starts to fall back?



2. I=½MR², V²=V๐²+2as, F=ma, a=R²α ...



3. MgsinӨ+f=ma=mR²α , f=Iα/R

0²=V๐²-2as , I=I=½MR² , a=2gsinӨ
s=V๐²/4gsinӨ
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But the answer page of the book says s= (3V๐²/4g) sinӨ [exactly as it's written---Prob. 9/67 PHYSICS for.... 3rd Edition; Fishbane-Gasiorowich-Thornton]
---------
I've found another solution from another book which goes with the conservation of the energy:
½MV๐²+½Iω๐²=MgH , H=SsinӨ
For a solid uniform cylinder, rotational kinetic energy is 1/3th of total energy and therefore:
3/2(½MV๐²)=MgSsinӨ, s=3V๐²/4gsinӨ

Which of us is correct????
 

Answers and Replies

73
0
I'm surprised that one one yet has an answer.
Maybe it's a good idea to get in touch with the authors 'cos I believe they made a mistake either in the book's second or third edition.
 
19
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0²=V๐²-2as , I=I=½MR² , a=2gsinӨ
s=V๐²/4gsinӨ
You should not use this because this neglects the affect of the rolling of the cylinder. Also, check your acceleration with a force-diagram (unless you meant SsinӨ).

However, as you have already stated
½MV๐²+½Iω๐²=MgH , H=SsinӨ
, the conservation of energy includes the rolling. Make sure to check your calculations.

---
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Last edited:

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