08homework07 - Husain, Zeena Homework 7 Due: Oct 14 2003,...

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Unformatted text preview: Husain, Zeena Homework 7 Due: Oct 14 2003, 4:00 am Inst: H L Berk 1 This print-outshouldhave 21questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A pendulum made of a string and a sphere is able to swing in a vertical plane. The pendulum is released from a position of 63 from vertical as shown in the figure below. The string hits a peg located a distance d below the point of suspension and rotates about the peg. 9 . 8 m / s 2 9 kg 1 1 m 63 d Find the smallest value of d (highest peg position) in order for the sphere to swing in a full circle centered on the peg. Correct answer: 8 . 59756 m. Explanation: Let : m = 9 kg , = 11 m , = 63 , and g = 9 . 8 m / s 2 . g m r 8 . 6m The radius of the sphere about the peg r = - d. (1) Relative to the the point of suspension, the initial potential energy is U i =- mg cos (2) and at the top of the circle, the final potential energy is U f =- mg ( d- r ) =- mg ( - 2 r ) . (3) From this, K f = mv 2 2 , and conservation of energy, we find- mg ( - 2 r ) + mv 2 2 =- mg cos . (4) T mg From a free-body diagram taken near the top of the circle centered on the peg, we have- T- mg =- mv 2 r . (5) The sphere will be at its maximum height at the top of the circle of radius r . For the highest possible peg position, when the sphere is at the top of the circle, the tension in the string is zero, T = 0 . This condition just barely lets the sphere complete a full circle. Thus, Eq. 5 can be rewritten mg r 2 = mv 2 2 . (6) Using Eq. 6 to eliminate mv 2 2 in the Eq. 4, we obtain- mg ( - 2 r ) + mg r 2 =- mg cos - 2 + 4 r + r =- 2 cos 5 r = 2 (1- cos ) . (7) Therefore r = 2 5 1- cos (8) = 2(11 m) 5 1- cos(63 ) = 2 . 40244 m , so d = - r = 11 m- 2 . 40244 m = 8 . 59756 m . Husain, Zeena Homework 7 Due: Oct 14 2003, 4:00 am Inst: H L Berk 2 002 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A golf ball ( m = 19 . 9 g ) is struck a blow that makes an angle of 46 . 2 with the horizontal. The drive lands 159 m away on a flat fairway. If the golf club and ball are in contact for 9 . 54 ms, what is the average force of impact? (Neglect air resistance.) Correct answer: 82 . 3772 N. Explanation: Note that the range of the golf ball is given by: L = v 2 sin2 g Then the initial velocity of the ball is: v = p Lg/ sin2 = 39 . 4914 m / s The average force exerted is the change in its momentum over the time of contact: F = mv t = (19 . 9 g)(0 . 001 kg / g)(39 . 4914 m / s) (9 . 54 ms)(0 . 001 s / ms) = 82 . 3772 N 003 (part 1 of 1) 10 points The linear impulse delivered by the hit of a boxer is 140 Ns during the 0 . 196 s of contact....
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08homework07 - Husain, Zeena Homework 7 Due: Oct 14 2003,...

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