Rolling motion of an unbalanced disk

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SUMMARY

The discussion focuses on calculating the angular acceleration of a 5 kg wheel with a radius of 300 mm and an off-center mass center located 100 mm from its geometric center. Using the moment of inertia formula I = mk², the moment of inertia is determined to be 1125 kg m². The angular acceleration is calculated using the relationship a = rα, resulting in an angular acceleration of 12000 rad/s² when the angular velocity is 8 rad/s.

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  • Understanding of rotational dynamics and angular motion
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  • Knowledge of kinematics involving rolling motion
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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[tex]\alpha[/tex]


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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