Ph1a-2007-HW4-soln

Ph1a-2007-HW4-soln - Ph1a Solutions 4 Chefung...

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Ph1a Solutions 4 Chefung Chan ([email protected]), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 4.1 QP6 (5 points) 4.1.a (1 point) Writing down Newton’s laws in the vertical direction at the top of the loop (in the lab frame) gives: F net = ma c = F N F g F N = m v 2 top R L g ! Conserving energy allows us to ±nd v top : E i = 1 2 kx 2 = 1 2 mv 2 top + mg (2 R L ) v 2 top = 2 m 4 gR L F N = m kx 2 m 4 L R L g ! F N = 2 R L 5 mg 4.1.b (1 point) At P , the velocity v p is found from conservation of energy, noting that h p = R P (1 cos θ p ). E i = 1 2 2 = 1 2 mv 2 p + mgh p v p = r 2 m 2 p (1 cos θ p ) The x and y components of the velocity are given by: v x = v p cos θ p v y = v p sin θ p To ±nd h max , realize that v x doesn’t change after the car ramps o² from point P .T h u s ,w eg e t h max when v y =0. E i = 1 2 2 = 1 2 mv 2 x + mgh max h max = 1 2 mg ( 2 mv 2 p cos 2 θ p ) h max = 2 2 mg cos 2 θ p 2 g ± 2 m 2 p (1 cos θ p ) ² h max = 2 sin 2 θ p 2 mg + R p (1 cos θ p )cos 2 θ p 1
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4.1.c (0.5 points) The energy lost to the friction is just W f = F f d = µmgd .
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Ph1a-2007-HW4-soln - Ph1a Solutions 4 Chefung...

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