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posted by  payara on 7/23/2008 1:32:18 PM  |  status: Closed  

ch17 prb129

Course Textbook Chapter Problem
Classical Mechanics Vector Mechanics for Engineers: Dynamics (8th) by Johnston, Beer, Clausen 17 129
Question Details:
A 40-g bullet is fired with a horizontal velocity of 600 m/s in the lower end of a sleder 7-kg bar of L=600mm. Knowing that the h=240mm and that the bar is initially at rest, determine (a) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impusive reation at C, assuming that the bullet becomes embedded in .001 s.
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posted by zsm28 on 7/23/2008 4:02:43 PM  |  status: Live
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Response Details:
m = 40-g = 0.040 kg, v = 600 m/s, M = 7-kg, L = 600 mm = 0.600 m, h = 240 mm = 0.240 m
initial angular momentum = mv(L - h)
final angular momentum = mω(L - h)2 + Iω
moment of inertia about C is I = (moment of inertia of the part above C) + (moment of inertia of the part below C) = (1/3)*(mass of of the part above C) * h2 + (1/3)*(mass of of the part below C) * h2 = (1/3)*(M/L)*h * h2 + (1/3)*(M/L)*(L - h) * h2 = (M/3L)*[h3 + (L - h)3] = M(L2 - 3Lh + 3h2)/3 = 7*(0.6002 - 3*0.600*0.240 + 3*0.2402)/3 = 0.2352 kg-m2
angular momentum conservation:
mv(L - h) = mω(L - h)2 + Iω
∴ω = mv(L - h)/[m(L - h)2 + I}
ω = 35.9 rad/s
impulse reaction F
F*(L - h)*t = Iω
∴F = Iω/[(L - h)*t] = 2.35*104 N
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