Solve Projected Cylinder: Find Distance s w/ 3 Solutions

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SUMMARY

The discussion focuses on calculating the distance \( s \) that a solid uniform cylinder travels up an incline before rolling back, given its initial speed \( V_0 \) and incline angle \( \theta \). Two solutions are presented: one using kinematic equations yielding \( s = \frac{V_0^2}{4g \sin \theta} \) and another employing conservation of energy resulting in \( s = \frac{3V_0^2}{4g \sin \theta} \). The latter solution is confirmed to be correct, as it accounts for the rotational kinetic energy of the cylinder. Participants suggest contacting the book authors regarding potential errors in the provided solutions.

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  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia (I = ½MR²)
  • Knowledge of energy conservation principles in physics
  • Basic trigonometry, particularly involving angles and sine functions
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  • Review the derivation of kinematic equations for rolling objects
  • Study the conservation of energy in systems involving rotational motion
  • Explore the implications of friction in rolling without slipping
  • Investigate the effects of incline angles on motion dynamics
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Students and educators in physics, particularly those focusing on mechanics, as well as anyone involved in solving problems related to rolling motion and energy conservation.

yasar1967
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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]
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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?
 
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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.
 
yasar1967 said:
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
yasar1967 said:
½MV๐²+½Iω๐²=MgH , H=SsinӨ
, the conservation of energy includes the rolling. Make sure to check your calculations.

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