HW_04_S

# HW_04_S - Ph1a Solutions 4 Chefung Chan([email protected]

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Unformatted text preview: Ph1a Solutions 4 Chefung Chan ([email protected]), Fall 2004 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 parenleftBigg v 2 top R L- g parenrightBigg Conserving energy allows us to find v top : E i = 1 2 kx 2 = 1 2 mv 2 top + mg (2 R L ) v 2 top = kx 2 m- 4 gR L F N = m parenleftBigg kx 2 m- 4 gR L R L- g parenrightBigg F N = kx 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 kx 2 = 1 2 mv 2 p + mgh p v p = radicalbigg kx 2 m- 2 gR 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 find h max , realize that v x doesn’t change after the car ramps off from point P . Thus, we get h max when v y = 0. E i = 1 2 kx 2 = 1 2 mv 2 x + mgh max h max = 1 2 mg ( kx 2- mv 2 p cos 2 θ p ) h max = kx 2 2 mg- cos 2 θ p 2 g parenleftbigg kx 2 m- 2 gR p (1- cos θ p ) parenrightbigg h max = kx 2 sin 2 θ p 2 mg +...
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HW_04_S - Ph1a Solutions 4 Chefung Chan([email protected]

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