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rigid body translation and rotation (easy 9 karma pt)

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Scholar
Karma Points: 207
Respect (92%):
Date Posted: 7/20/2008 5:55:33 PM  Status: Live
rigid body translation and rotation (easy 9 karma pt)
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
 Gear A (φ = 12”) drives Gear B (φ = 36”), which is attached to the hoisting drum
(φ = 48”),. The load, L, is lifted from its rest position and acquires an upward velocity of 3 ft/s in a vertical rise of 4 ft with constant acceleration. As the load passes this position, calculate
    1. The acceleration of point C on the cable in contact with the hoisting drum.

Answer: 4.64 ft/s2 (you figure out CW or CCW)

    1. The angular velocity and angular acceleration of Gear A.
Answer: 4.5 rad/s, 1.69 rad/s2 (you figure out the direction CW or CCW
 
 
 
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Answers:

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Oracle
Karma Points: 14,581
Date Posted: 7/21/2008 12:41:17 AM  Status: Live
Asker's Rating: Helpful   
Response:
the acceleration of the load = a = v2/(2d) = (3 ft)2/(2*4 ft) = 9/8 = 1.125 ft/s2
it is also the tangential acceleration of C, so at = 1.125 ft/s2
the linear speed of C is v = 3 ft/s, the radius of C = rc = 24" = 2 ft
so the angular speed of C = ωc = v/rc = 3/2 = 1.5 rad/s  (CCW)
the radial acceleration of C = ar = ωc2 * rc = 4.5 ft/s
a) accelration of C = a = √(at2 + an2) = 4.64 ft/s2
b) ωB = ωc = 1.5 rad/s   (CCW)
ωB rB = ωA rA
∴ωA = ωB*(rB/rA) = 1.5*(18/6) = 4.5 rad/s (CW)
angular acceleration of C = αc = at/rc = 1.125/2 = 0.5625 rad/s2   (CCW)
angular acceleration of B = αB = αc = 0.5625 rad/s2   (CCW)
αB rB = αA rA
∴αA = αB*(rB/rA) = 0.5625*(18/6) = 1.69 rad/s2 (CW)



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