maheshmeghwal9
  • maheshmeghwal9
In the arrangement shown in Fig. 1.24 the masses m of the bar and M of the wedge, as well as the wedge angle alpha , are known.The masses of the pulley and the thread are negligible. The friction is absent. Find the acceleration of the wedge M.
MIT 8.01 Physics I Classical Mechanics, Fall 1999
schrodinger
  • schrodinger
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maheshmeghwal9
  • maheshmeghwal9
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maheshmeghwal9
  • maheshmeghwal9
Answer is \[\LARGE{\color{blue}{a=\frac{mg \sin \alpha }{M+2m(1-\cos \alpha)}.}}\]\[\color{red}{\text{But how to get this answer?}}\]
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Let x, v and a be displacement, velocity and acceleration of wedge. All these values are positive (to the right). Velocity of m wrt wedge is (first line x-component, second line y-component) -v cosα -v sinα Velocity of wedge wrt ground is: v 0 Adding them up gives velocity of m wrt ground: v (1 – cosα) -v sinα KE of wedge is KE1 = ½ M v² KE of mass is KE2 = ½ m v² [(1 - cosα)² + sin²α] = m v² (1- cosα) Altitude of m above ground is -x sinα , as it goes x down along the inclined when wedge moves of x. PE of mass is PE = -m g x sinα The system is conservative. So total energy is conserved. Adding up all derivatives of above energies must give zero: 0 = Mva +2mva(1 - cosα) - mg v sinα Simplifying by v and solving for a leads to the solution.

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Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
* altitude above ground is actually -x sinα + H where H is height of the pulley. H disappears in the derivation.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
I also have a solution based on forces and N's laws, although I prefer the more global energy approach.
yrelhan4
  • yrelhan4
also sir, what would be the displacement of the wedge? this is a common question, but i always end up getting stuck. @Vincent-Lyon.Fr

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