Exam 3A (2016).pdf

# 8 pts k i 1 2 m c v 2 c 1 2 m h v 2 h 21 53 50 2 13

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(8 pts) K i = 1 2 m C v 2 C + 1 2 m H v 2 H = (21 . 5)(3 . 50) 2 + (13 . 6)(7 . 4) 2 2 = 504 . 056 J K f = 1 2 ( m C + m H ) v 2 f = (21 . 5 + 13 . 6)(4 . 63793) 2 2 = 377 . 508 J K i - K f K i 100% ' 25 . 1% Copyright c 2016 University of Georgia. 5

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Physics 1111 Exam #3A-12:30 Solutions 17 November 2016 Last Name: V: Potter’s Wheel (24 points) One type of potter’s wheel uses a large, heavy disk as a flywheel to keep the small disk that holds the pot turning smoothly for longer. The potter uses their foot to spin the big disk up to speed. One such disk has a diameter of 0 . 950 m and a mass of 32 . 0 kg. Assume the potter can give the wheel a constant angular acceleration of 0 . 380 rev/s 2 . (a) How many revolutions will the wheel make in reaching a speed of 1 . 20 rev/s, starting from rest? (8 pts) For this part, since the given values are in revolutions and the question asks for revolutions, we can keep working in revolutions. ! 2 f = ! 2 i + 2 ( Δ ) = ) Δ = ! 2 f - ! 2 i 2 Plugging in numbers: Δ = (1 . 20 rev/s) 2 - (0) 2(0 . 380 rev/s 2 ) ' 1 . 89 rev . It is also okay to convert all values to radians first, so long as you convert back to revolutions at the end. (b) To give this acceleration, the potter’s foot provides a tangential frictional force at 3 / 4 of the way from the center to the edge. Assuming this is the only force causing rotation, what is the magnitude of this force? (8 pts) For this part, we need Newton’s Second Law for rotation: X = I For a disk, the moment of inertia is I = 1 2 mR 2 Also, remember that we must convert to rad/s 2 . The force is applied at a distance (3 / 4) R from the axis. Putting these things together, we have 3 4 R F sin 90 = 1 2 mR 2 F = 2 3 mR = 2 3 (32 . 0 kg)(0 . 475 m)(2 . 38761 rad/s 2 ) ' 24 . 2 N . (c) When rotating at the speed given above, how much kinetic energy is stored in the wheel? (8 pts) Again, we must work with the angular quantities in units of radians. K = 1 2 I ! 2 = 1 2 1 2 mR 2 = 1 4 mR 2 ! 2 = 1 4 (32 . 0 kg)(0 . 475 m) 2 (7 . 53982 rad/s) 2 ' 102 . 6 J . Copyright c 2016 University of Georgia. 6
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