Ph1a-2007-HW3-soln

Ph1a-2007-HW3-soln - Ph1a Solutions 3 Che-Fung Chan...

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Ph1a Solutions 3 Che-Fung Chan (cchan@caltech.edu), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 3.1 QP3 (5 points) 3.1.a (0.5 points) In terms of the lengths in the diagram, we have: l 1 = ( p 1 - x 1 ) + πR + ( p 1 - p 2 ) = 2 p 1 - x 1 - p 2 + πR l 2 = p 2 + πR + ( p 2 - x 2 ) = 2 p 2 - x 2 + πR 3.1.b (1 point) The free body diagrams are in Fig. 1. M M F F T T g g 2 1 1 2 Figure 1: QP3.b In the vertical direction, we have Newton’s Laws: m 1 a 1 = T 1 - m 1 g m 2 a 2 = T 2 - m 2 g 3.1.c (1 point) Taking two time derivatives of the equations in part a gives: d 2 l 1 dt 2 = 0 = - a 1 - d 2 p 2 dt 2 d 2 l 2 dt 2 = 0 = 2 d 2 p 2 dt 2 - a 2 From this, we get a 2 = - 2 a 1 , or a 1 = ka 2 where k = - 1 2 . 3.1.d (1 point) We can get the relationship between the tensions by considering Newton’s Laws applied to the (massless) 2 nd pulley. For massless objects, the condition is that net force must be zero to avoid the unphysical result of in±nite acceleration. The force diagram is Fig. 2 m P 2 a P 2 = T 1 - 2 T 2 - m P 2 g 1
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2 T T 1 T 2 Figure 2: QP3.d T 1 - 2 T 2 = 0 So we get T 1 = 2 T 2 , or T 1 = k 2 T 2 with k 2 = 2. 3.1.e (1 point) Solving for a 1 and T 2 is straight-forward: T 2 = m 2 p a 1 k P + m 2 g = m 2 p a 1 k + g P m 1 a 1 = ( k 2 T 2 ) - m 1 g m 1 a 1 = m 2 k 2 ( a 1 k + g ) - m 1 g a 1 = ( m 2 k 2 - m 1 ) kg km 1 - k 2 m 2 T 2 = m 1 m 2 g ( k - 1) km 1 - k 2 m 2 Using the correct values of k = - 1 / 2 and k 2 = 2, we get a 1 = 2 m 2 - m 1 m 1 + 4 m 2 g T 2 = 3 m 1 m 2 g m 1 + 4 m 2 Using the incorrect assumption of k = - 1 and k 2 = 1 / 3, we get a 1 = m 2 - 3 m 1 3 m 1 + m 2 g T 2 = 6 m 1 m 2 g 3 m 1 + m 2 3.1.f (0.5 point) If m 1 = m 2 the formula for a 2 reduces to: a 2 = a 1 k = ( k 2
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Ph1a-2007-HW3-soln - Ph1a Solutions 3 Che-Fung Chan...

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