Exam3-solution-Sp03

Exam3-solution-Sp03 - Physics 221. Exam III Spring 2003 The...

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Physics 221. Exam III Spring 2003 The situation below refers to the next three questions: A solid cylinder of radius R and mass M with initial velocity v 0 rolls without slipping up the inclined plane. mg N f S R M 102. While the cylinder is rolling up, the frictional force is _________________ and the cylinder is rotating ___________________ a. directed uphill , clockwise. b. directed downhill, with constant angular speed *c. directed uphill , counterclockwise d. zero, clockwise e. directed downhill, with constant angular acceleration. It is rolling counterclockwise because it’s moving to the left. Static friction needs to point up because we need a net torque into the page (to produce a net angular acceleration into the page) since the cylinder is slowing down. Weight and the normal cannot produce this torque, so friction has to be the force responsible for it. Also, the “tendency to move” of the bottom of the cylinder is down the ramp. Friction is opposing that (and not the motion of the CM). 103. If the cylinder moves up a vertical height H before it starts rolling back down the inclined plane, its initial speed v 0 must be: a. 0 = vg H b. 0 10 7 = H c. 0 10 3 = H d. v 0 g H = *e. 0 4 3 = H Page 1 of 21
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Physics 221. Exam III Spring 2003 Only conservative forces (weight) do work here, so mechanical energy is conserved, 22 2 2 2 11 Rolling without slipping: 1 Moment of inertia of a cylinder: 2 1 2 3 4 4 3 ω =+ = = =  +=   = = iC M C M f CM CM CM CM CM CM EM v I g h vR I MR v Mv MR Mgh R vg h gh v 104. If the mass and the radius of the cylinder moving uphill are doubled, the magnitude of the acceleration of the center of mass changes by a factor of: a. 1/2 *b. 1 (does not change) c. 2 d. 4 e. 8 To begin with, the calculation in the previous part ended up being independent of M and R. That is an indication of the acceleration being the same. And: We saw a demo in class with two completely different cylinders going down a ramp and reaching the bottom at the same time. Or you can do the calculation. From the free-body diagram above, 2 2 sin 1 2 sin sin 23 sC M CM ss CM CM CM CM mg f ma ma mR fR mR f R aR ma mg ma a g θ α 2 −= == = = = Page 2 of 21
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Physics 221. Exam III Spring 2003 105. A solid wood door a = 1 m wide and b = 2 m high is hinged on one of the long sides (as a regular door) and has a total mass M = 40 kg. Initially the door is at rest. It is struck with a handful of sticky mud of mass m = 0.5 kg traveling perpendicular to the door at 12 m/s before impact. The mud hits the door at its geometrical center. What is ω f , the final angular speed of the door? ( I door with respect to hinges = Ma 2 /3) a. ω f = 0.10 rad/s *b. ω f = 0.22 rad/s c. ω f = 0.43 rad/s d. ω f = 0.56 rad/s e. ω f = 0.80 rad/s There is no external torque on the system door-mud ball, so angular momentum is conserved. The mud hits the door at distance a/2 from the hinges. 2 2 2 1 32 23 4 6 0.22 rad/s 43 i f a Lmv a LM am mv M m a mv M ma ωω ω =  =+   == + Page 3 of 21
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Physics 221. Exam III Spring 2003 106. Four small identical balls are arranged in a square using four massless rods. The system can rotate about three different axis, as shown in the figures below. Rank the moments of inertia of the system in each case.
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This note was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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Exam3-solution-Sp03 - Physics 221. Exam III Spring 2003 The...

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