The circular concrete culvert rolls with an angular velocity of ? = 0.54 rad/s w

Question

The circular concrete culvert rolls with an angular velocity of ? = 0.54 rad/s when the man is at the position shown. At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kG = 3.4 ft

Part A)

Determine the angular acceleration of the culvert. The combined weight of the culvert and the man is 500 lb . Assume that the culvert rolls without slipping, and the man does not move within the culvert

It is NOT 0.639!

Thank you in advanced!!

Explanation / Answer

Moment of Inertia of System about it's Mass = m* Kg^2 = 500/32.2 * 3.4^2 = 179.5 ft^2
Let the angular acceleration of Culvert = a

Writing the moment equation=
-500 * 0.5 =  -500/32.2 * agx * 4 -500/32.2* agy * 0.5 - 179.5* a

Where ,
agx = 4a - 0.125
agy = 0.5 a

Substituing Values -
-500 * 0.5 =  -500/32.2 * (4a - 0..125) * 4 -500/32.2* 0.5a * 0.5 - 179.5* a
Solving for a

Angular acceleration of the culvert, a = 0.596 rad/s^2

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