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Unformatted text preview: Bautista, Aldo Midterm 3 Due: Nov 16 2005, 10:00 pm Inst: Maxim Tsoi 1 This printout should have 17 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 Assume: The buoyant force of the air is neg ligible. Given: The density of the oil is 822 kg / m 3 . The density of the water salt mixture is 1030 kg / m 3 . A rectangular parallelepiped block of uni form density floats in a container which con tains oil and salt water as shown. The block sticks up above the oil by a dis tance d 1 = 0 . 43 m. The oil thickness is d 2 = 0 . 38 m. The blocks depth in the salt water is d 3 = 0 . 17 m. The horizontal area of the block is 0 . 034 m 2 . M d 1 d 3 d 2 Calculate the mass of the block. Correct answer: 16 . 5736 kg. Explanation: The mass of the salt water displaced is m 3 = w d 3 a = (1030 kg / m 3 )(0 . 17 m)(0 . 034 m 2 ) = 5 . 9534 kg . The mass of the oil displaced is m 2 = o d 2 a = (822 kg / m 3 )(0 . 38 m)(0 . 034 m 2 ) = 10 . 6202 kg . Since the mass of the block m b is equal to the liquid displace, we have m b = m 2 + m 3 = (10 . 6202 kg) + (5 . 9534 kg) = 16 . 5736 kg . 002 (part 1 of 1) 10 points A 24 kg mass and a 16 kg mass are suspended by a pulley that has a radius of 4 . 7 cm and a mass of 7 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 1 . 8 m apart. Treat the pulley as a uniform disk. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 8 m 4 . 7 cm 7 kg 24 kg 16 kg Determine the speeds of the two masses as they pass each other. Correct answer: 1 . 80115 m / s. Explanation: Let : M = 7 kg , R = 4 . 7 cm , m 1 = 24 kg , m 2 = 16 kg , h = 1 . 8 m , v = R , I = 1 2 M R 2 , and K disk = 1 2 I 2 = 1 4 M v 2 . From conservation of energy K 1 + K 2 + K disk = U or m 1 v 2 2 + m 2 v 2 2 + M v 2 4 = ( m 1 m 2 ) g h 2 m 1 + m 2 + M 2 v 2 = ( m 1 m 2 ) g h , Bautista, Aldo Midterm 3 Due: Nov 16 2005, 10:00 pm Inst: Maxim Tsoi 2 where h 2 is the height. Taking no slipping into account, we can solve for v v = s 2( m 1 m 2 ) g h 2( m 1 + m 2 ) + M (1) = s 141 . 12 J 43 . 5 kg = 1 . 80115 m / s . Alternative Solution: The forces in the vertical directions for m 1 and m 2 , and the torque on the pulley, give us m 1 g T 1 = + m 1 a m 2 g T 2 = m 2 a , so T 1 = m 1 ( g a ) T 2 = m 2 ( g + a ) , and I = [ T 1 T 2 ] R 1 2 M R 2 a R = h m 1 ( g a )] R [ m 2 ( g + a ) i R 1 2 M a = m 1 g m 1 a m 2 a m 2 g a = 2( m 1 m 2 ) M + 2( m 1 + m 2 ) . The velocity is v 2 f = v 2 i + s a y v f = r 2 a h 2 = s 2( m 1 m 2 ) g h 2( m 1 + m 2 ) + M , which is the same as Eq. 1....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner

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