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# HW18.sol - homework 18 RAHMAN TARIQUE Due Apr 7 2008 3:00...

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homework 18 – RAHMAN, TARIQUE – Due: Apr 7 2008, 3:00 am 1 Question 1, chap 13, sect 3. part 1 of 1 10 points A solid cylinder of mass M = 29 kg, radius R = 0 . 22 m and uniform density is pivoted on a frictionless axle coaxial with its symmetry axis. A particle of mass m = 3 . 1 kg and initial velocity v 0 = 7 . 8 m / s (perpendicular to the cylinder’s axis) flies too close to the cylinder’s edge, collides with the cylinder and sticks to it. Before the collision, the cylinder was not ro- tating. What is its angular velocity after the collision? Correct answer: 6 . 24483 rad / s (tolerance ± 1 %). Explanation: Basic Concept: Conservation of Angu- lar Momentum, L particle z + L cylinder z = const . The axle allows the cylinder to rotate without friction around a fixed axis but it keeps this axis fixed. Let the z coordinate axis run along this axis of rotation; then the axle may exert arbitrary torques in x and y directions but τ z 0. Consequently, the z componenent of the angular momentum must be conserved, L z = const, hence when the particle collides with the cylinder L before z, part + L before z, cyl = L z, net = L after z, part + L after z, cyl . Before the collision, the cylinder did not rotate hence L before z, cyl = 0 while the particle had angular momentum vector L before part = vectorr × vector P 0 = vectorr × mvectorv 0 . Both the radius-vector vectorr and the velocity vectorv 0 of the particle lie in the xy plane ( to the z axis), and according to the picture, at the moment of collision the radius vector has mag- nitude | vectorr | = R equal to the cylinder’s radius and direction perpendicular to the particle’s velocity. Hence, its angular momentum is parallel to the z axis and has magnitude | vector L before part | = L before z, part = Rmv 0 . Altogether, before the collision L before z, net = Rmv 0 and therefore, after the collision we should also have L after z, net = Rmv 0 . (1) After the collision, the cylinder and the particle rotate as a single rigid body of net moment of inertia I net = I cyl + I part = 1 2 MR 2 + mR 2 . (2) relatve to cylinder’s axis. For a rigid rotation like this, the angular momentum points in z direction and its magnitude is L z, net = ωI net . (3) Combining eqs. (1), (2) and (3) together, we immeditaly obtain ω after = L after z, net I net = Rmv 0 1 2 MR 2 + mR 2 = v 0 R × m 1 2 M + m = 6 . 24483 rad / s . Question 2, chap 13, sect 3. part 1 of 2 10 points A figure skater on ice spins on one foot. She pulls in her arms and her rotational speed increases.

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HW18.sol - homework 18 RAHMAN TARIQUE Due Apr 7 2008 3:00...

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