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Unformatted text preview: Bautista, Aldo Homework 9 Due: Nov 1 2005, 4:00 am Inst: Maxim Tsoi 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Consider two masses, 4 . 4 kg and 7 . 2 kg, con nected by a string passing over a pulley having a moment of inertia 9 g m 2 about its axis of rotation, as in the figure below. The string does not slip on the pulley, and the system is released from rest. The radius of the pulley is . 26 m. The acceleration of gravity is 9 . 8 m / s 2 . 26 cm . 26 m 9 g m 2 7 . 2 kg 4 . 4 kg Find the linear speed of the masses after m 2 descends through a distance 26 cm. As sume mechanical energy is conserved during the motion. Correct answer: 1 . 10277 m / s. Explanation: Let : I = 9 g m 2 , R = 0 . 26 m , m 1 = 4 . 4 kg , m 2 = 7 . 2 kg , h = 26 cm , v = R , I = 1 2 M R 2 , and K disk = 1 2 I 2 = 1 4 M v 2 . Consider the free body diagrams 7 . 2 kg 4 . 4 kg T 2 T 1 m 2 g m 1 g a a If we neglect friction in the system, the mechanical energy is constant and we can state that the increase in kinetic energy equals the decrease in potential energy. Since K i = 0 (the system is initially at rest), we have K = K f K i = 1 2 m 1 v 2 + 1 2 m 2 v 2 + 1 2 I 2 , where m 1 = 4 . 4 kg and m 2 = 7 . 2 kg have a common speed. Because v = R , this expression becomes K = 1 2 m 1 + m 2 + I R 2 v 2 . From the figure, we see that m 2 loses potential energy while m 1 gains potential energy. That is, U 2 = m 2 gh and U 1 = m 1 gh . Apply ing the principle of conservation of energy in the form K + U 1 + U 2 = 0 gives v = v u u t 2( m 2 m 1 ) g h m 1 + m 2 + I R 2 = 1 . 10277 m / s . Comment: One could work this problem in a systematic way. Begin with the workenergy theorem: W ext i >f = K f K i + U f U i + W dissip i >f The assumption that the mechanical energy is conserved and the system starts from rest implies that K f = U i U f Bautista, Aldo Homework 9 Due: Nov 1 2005, 4:00 am Inst: Maxim Tsoi 2 K f = K 1 + K 2 + K pulley = 1 2 m 1 v 2 + 1 2 m 2 v 2 + 1 2 I w 2 = 1 2 m 1 v 2 + 1 2 m 2 v 2 + 1 2 I v R 2 U i U f = U i 1 U f 1 + U i 2 U f 2 = m 1 g h + m 2 g h , this leads to the sam expression for the veloc ity v as given above. 002 (part 1 of 2) 10 points A uniform solid disk of radius 1 . 6 m and mass 60 . 2 kg is free to rotate on a frictionless pivot through a point on its rim. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 6 m Pivot 9 . 8 m / s 2 If the disk is released from rest in the po sition shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle?...
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