A2.0kg disk traveling at5.0m/s strikes a1.2kg stick of length 4.0 m that is lyin

Question

A2.0kg disk traveling at5.0m/s strikes a1.2kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is1.60kg ? m2.

Suppose the disk is traveling with the same velocity as in the example, but along a line that is a distancer< 2.00 m from thexaxis as shown in the figure. The disk collides elastically with the stick and the final velocity of the disk is found to be3.41m/s along its original line of motion. Determine the distancer.

Explanation / Answer

Conservation of Momentum:
2 x 5 = 2vdf + 1.6

Conservation of Energy:
2(52)/2 = 2v2df/2 + 0.82


vt = (4/2) = 2
5 = vdf + 0.8
25 = v2df + 0.82

w = 10 rad/s, vt = 20 m/s, vdf = -3 m/s

2 x 5 = 2vdf + 1.2vs + 1.6

2(52)/2 = 2v2df/2 + 0.6vs2 +0.82
vs = rw
1.59 = (0.6r+0.8)
6.686 = (0.3r2 + 0.4)2

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