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I have 2 questions where I have some difficulty in the solution method shown in the books. enter image description here

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In case of the question number 5 I DONOT understand why there is an $mga cos \theta$ in the equation of potential energy.

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Here in the above question I think there should one more term $ mgl sin\alpha$ in the equation of V.

Please let me know if I an erring in any of the questions.

Sharmi C
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1 Answers1

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To the first doubt, the potential energy with reference to the horizontal plane, is given by the height of the cylinder center regarding the base plane. This height is composed of two terms: first the height of the contact point which is $(s-x)\sin\theta$ and the other $a \cos\theta$ which is the cylinder center elevation regarding the point of contact. Then

$$ V = \left((s-x)\sin\theta + a\cos\theta\right)mg $$

Cesareo
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  • I understood. I just want to know that should I follow the above approach for all such questions dealing with other types of bodies say a sphere or a hoop. Thank you – Sharmi C Mar 25 '20 at 14:15
  • For the second question should we proceed similarly to get $ V= \left((s-x)\sin\theta + r\cos\theta\right)mg $ where I assume s is length of inclined plane and r is radius of sphere – Sharmi C Mar 25 '20 at 14:29