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posted by  chiefdog on 7/17/2008 5:04:39 PM  |  status: Live  

Plane Motion work/energy ( I have most of the work )

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Weight of wheel is 15lbs. Point A should be at the center of the wheel (r = 6”). Weight of bar is 10 lbs and length of bar (from O to A) is 12”. Radius of gyration of wheel about center is 4.5”. Starts at rest and rotates to horizontal position due to external applied Moment of 20 lb-ft. No slipping of wheel. Find angular velocity of rod OA when it reaches horizontal position. Answer: 5.24 rad/s (Give direction)
 
HERES what I have so far!
 
T1 + V1 +V' = T2 + V2
T1 = 0
V1 = mgh = WwH + WbH = 15*.5 + 10*.5 = -12.5
V' = 20  
T2 = (1/2) *(m) *(ω)^2 + (1/2)*(m)*(ω)^2 + (1/2) *(m) *(ω)^2 + (1/2)*(m)*(ω)^2
V2 = 0
 
therefore I end up with  -12.5+20 = .2598ω^2
solving for ω gives me 5.37 rad/s    I am somewhat close but I dont know what my error
 
 
 
 
 
 
 
 
 
 
 
 
 
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posted by zsm28 on 7/17/2008 8:07:45 PM  |  status: Live
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Response Details:
W = 15 lb, r = 6" = 0.5', R = 4.5" = 0.375', moment of inertia of the wheel about its center Iw = MR2 = 15/32 * 0.3752 = 0.0659
w = 10 lb, L = 12" = 1', moment of inertia about O is Ir = ML2/3 = (10/32)*12/3 = 1/9.6
M = 20 lb-ft, angular displacement θ = π/2 (from rest to horizontal position)
Work done by M is M*θ = 20*π/2 = 10π
initial energy = 0
final energy:
potential energy of the rod = w*(L/2) = 10*(1/2) = 5
potential energy of the wheel = W*L = 15*1 = 15
rotational kinetic energy of the rod = (1/2)Ir * ω2 = 1/19.2 ω2 = 0.0521 ω2 (we want to find ω)
rotational kinetic energy of the wheel = (1/2)Iw * ω'2 = (1/2)Iw * (2ω)2 = 2Iw * ω2 = 0.1318 ω2
(since "no slipping of wheel, vA = ωL = ω'r, so ω' = ωL/r = ω*1/0.5 = 2ω)
translational kinetic energy of wheel = MvA2/2 = (W/g)(ωL)2/2 = WL2/(2g) ω2 = 15*12/(2*32) ω2 = 0.234375 ω2
Work one = total final energy
10π = 5 + 15 + (0.0521 + 0.1318 + 0.234) ω2
10π = 20 + 0.4179 ω2
ω = 5.23 rad/s
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