Rolling motion of an unbalanced disk

AI Thread Summary
The problem involves a 5kg wheel with a radius of 300 mm and a mass center offset by 100 mm from its geometric center. The wheel rolls without sliding, with an angular velocity of 8 rad/s. Using the moment of inertia formula, I = mk², the calculated moment of inertia is 1125 kg m². The angular acceleration is derived from the relationship a = rα, resulting in an angular acceleration of 12000 rad/s². This calculation incorporates both tangential and normal components of acceleration.
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Homework Statement


The mass center G of a 5kg wheel of radius R = 300 mm is located at a distance r = 100 mm from its geometric center C. The centroidal radius of gyration is k = 150 mm. As the wheel rolls without sliding, its angular velocity varies and it is observed that it is = 8 rad/s in position shown. Determine the corresponding angular acceleration of the wheel.
200911222253246339452720406837507452.jpg



Homework Equations


Ac=Ag=Ac + Ag/c = Ac + (Ag/c)tangent + (Ag/c) Normal
I= mk^2
a=r\alpha


The Attempt at a Solution


200911222316396339452859963087501839.jpg

 
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I= 5 kg(150mm)^2 = 1125 kg m^2a=r\alpha = 100 mm (8 rad/s) = 800 rad/s^2Ac = Ag = I \alpha = (1125 kg m^2)(8 rad/s) = 9000 rad/s^2Ag = Ac + (Ag/c) tangent + (Ag/c) Normal = 9000 rad/s^2 + (Ag/300mm)Tangent + (Ag/300mm)NormalAg = 9000 rad/s^2 + (Ag/300mm)(cos30) + (Ag/300mm)(sin30) = 9000 rad/s^2 + (Ag/3)Ag = 9000 rad/s^2 + 3000 rad/s^2 = 12000 rad/s^2 Therefore, the angular acceleration of the wheel is 12000 rad/s^2
 
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