# 113&122 - ( 29 ( 29 ( 29 ( 29 j sin 2 / 2 / 2 2 2 , 1/2...

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Spring, 2004 2.113 Given: A particle of mass m slides down a frictionless chute and enters a circular loop of diameter d as shown. It is released from rest. Find: The minimum height h so that the particle will make a complete circuit of the loop without having lost contact. Solution: When a particle loses contact with the loop, the normal force disappears. Hence we want the minimum height for the normal force to never disappear (in other words, for it to barely disappear when it’s most likely to do so. If you draw a f.b.d. at several points around the loop, and f is dependent on the position as shown, j r j sin sin 2 2 2 mg R v m N R v m s m N mg ma F n n - = = = + = . Note this shows N is the smallest where j sin is maximized and v 2 is minimized. To find where v 2 , note that the only force that is doing work on the particle is gravity.
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Unformatted text preview: ( 29 ( 29 ( 29 ( 29 j sin 2 / 2 / 2 2 2 , 1/2 so , 2 2 2 d d h g v h g h g v and h mg mv W T down down down--= ⇒ ∆ = ∆ = ∆ = = ∆ So v 2 is minimized when j sin is maximized. j sin is maximized at the topmost point on the loop, where j sin = 1 ( 29 ( 29 ( 29 ( 29 d h g d d h g v-=--= ⇒ 2 2 / 2 / 2 2 ( 29 ( 29 ( 29 --=--= ⇒ = = + ⇒ = ∑ 1 4 2 / 2 sin 2 2 d d h mg mg d d h g m N R v m s m N mg ma F n n r j & -= ⇒ d d h mg N 5 4 If the height is just enough for the particle to never lose contact, N = 0 here 4 / 5 5 4 5 4 d h d h d d h mg = ⇒ = ⇒ -= ⇒ 2.122 Done in the notes...
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## This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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