HW08-solutions - baum(lmb2768 – HW08 – distler...

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Unformatted text preview: baum (lmb2768) – HW08 – distler – (56295) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod of mass m and length L is hinged with a frictionless hinge at one end. The moment of inertia about the center of mass is 1 12 mL 2 . Attached to the end of the rod opposite to the hinge there is a mass of magnitude 2 m . The rod is released from rest in the horizontal position. L m 2 m What is the speed of the mass 2 m when the rod passes through the vertical position? Consider the mass at the end of the rod to be a point particle. 1. v 2 m = 3 g L 2 2. v 2 m = 2 g L 3. v 2 m = 2 g L 3 4. v 2 m = 15 g L 7 correct 5. v 2 m = 2 g L 5 6. v 2 m = g L 7. v 2 m = 7 g L 15 8. v 2 m = g L 9. v 2 m = g L 2 10. v 2 m = 5 g L 2 Explanation: Let us measure heights from the point at the end of the rod when it is vertical. The ini- tial energy is all potential, with a magnitude E i = U rod i + U 2 m i = mg L + (2 m ) g L = 3 mg L. Energy is conserved. When the rod is ver- tical, the energy of the mass 2 m is all kinetic: E = K 2 m mass = 1 2 (2 m ) v 2 2 m = m ( ω L ) 2 . The final potential energy of the rod is U rod f = 1 2 mg L and the kinetic energy is K rod f = K rot f + K tr f = 1 2 I ω 2 + 1 2 mv 2 rod = 1 2 1 12 m L 2 ω 2 + 1 2 m ω L 2 2 = 1 6 mL 2 ω 2 . Alternately, I rod,end = 1 3 L 2 , so K rod f = 1 2 I rod,end ω 2 = 1 2 1 3 mL 2 ω 2 = 1 6 mL 2 ω 2 . Since v m = ω L, U rod i + U 2 m i = U rod f + K 2 m f + K rod f 3 mg L = 1 2 mg L + m ( ω L ) 2 + 1 6 mL 2 ω 2 18 mg L = 3 mg L + 6 m ( L ω ) 2 + mL 2 ω 2 15 mg L = 7 m ( L ω ) 2 = 7 mv 2 2 m , v 2 m = 15 7 g L. baum (lmb2768) – HW08 – distler – (56295) 2 002 (part 1 of 2) 10.0 points A circular disk, a ring, and a square, have the same mass M and width 2 r . disk 2 r ring 2 r square 2 r For the moment of inertia about their cen- ter of mass about an axis perpendicular to the plane of the paper, which of the follow- ing statements concerning their moments of inertia is true? 1. I disk > I square > I ring 2. I ring > I disk > I square 3. I disk > I ring > I square 4. I ring > I square > I disk 5. I square > I ring > I disk correct 6. I square > I disk > I ring Explanation: In the ring, the same mass of the disk is concentrated at the maximum distance from the axis, so I ring > I disk In the square, the same mass of the ring lies at distances which are at least the radius of the ring, so I square > I ring and I square > I ring > I disk . Alternative Solution: Since the moment of inertia I α mr 2 , I disk = 1 2 M r 2 I ring = M r 2 For the square, using I rod cm = 1 12 m 2 , m = M 4 , = 2 r and d = r I square = 4 1 12 m 2 + md 2 = 4 1 12 M 4 (2 r ) 2 + M 4 r 2 = 4 3 M r 2 ....
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HW08-solutions - baum(lmb2768 – HW08 – distler...

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