HW08-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online