PHYS18Fall2009MT1solns

PHYS18Fall2009MT1solns - Physics 18 Fall 2009 Midterm 1...

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Physics 18 Fall 2009 Midterm 1 Solutions For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat ! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck ! 1. A typical laboratory centrifuge rotates at 4000 RPM. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. (a) What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? (b) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0 ms (10 - 3 seconds) encounter with a hard floor? ———————————————————————————————————— Solution (a) Since it’s spinning, the test tube is subjected to a centripetal acceleration a cent = v 2 /r , where v is the rotational velocity and r is the rotational distance. Now, we know the rotational frequency, f = 4000 rotations/min ÷ 60 sec/min 67 rotations per second. This means that the period is T = 1 /f = 1 / 67 = 0 . 015 seconds. Now, the period is the time it takes to go around one orbit in the centrifuge, and is the total distance around (2 πr ) divided by how fast it’s going around ( v ). So, T = 2 πr v , which means that v = 2 πr/T . Thus, the acceleration is a cent = v 2 r = (2 π ) 2 T 2 r. Plugging in the numbers gives a cent = (2 π ) 2 . 015 2 × 0 . 1 = 1 . 75 × 10 4 m/s 2 , which is 1790 g ! (b) If we drop a test tube, it experiences a big acceleration when it hits the ground. Since it goes from some velocity v drop to zero over a time Δ t , then the acceleration is a = v drop / Δ t . What’s the initial speed? Since it’s dropped from a height - h , under the constant acceleration due to gravity then v 2 f = v 2 i - 2 g ( - h ) = 0 + 2 gh , since the tube is assumed to have dropped from rest. The final velocity it gets is just v f = v drop = 2 gh . Thus, a = 2 gh Δ t = 2 × 9 . 8 × 1 10 - 3 = 4400 m/s 2 , which is about 450 g . So, the centrifuge subjects the tube to accelerations even greater than hitting the ground! 1
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2. An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a 6.0 kg mass from the rope. The boot is suspended by the ropes and does not touch the bed.
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PHYS18Fall2009MT1solns - Physics 18 Fall 2009 Midterm 1...

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