Angular acceleration of off balance wheel starting from rest

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SUMMARY

The discussion focuses on calculating the angular acceleration of a 30 kg wheel with a radius of 0.25 m and a radius of gyration (KG) of 0.15 m, starting from rest. The moment of inertia (I) is calculated using the formula I = mk², resulting in 0.675 kg·m². The net force acting on the wheel is determined to be 294.3 N, leading to the conclusion that the angular acceleration is 10.3 rad/s². The calculations involve the application of Newton's second law and the relationship between linear and angular motion.

PREREQUISITES
  • Understanding of rotational dynamics and angular acceleration
  • Familiarity with the moment of inertia formula (I = mk²)
  • Knowledge of Newton's second law for rotation (M = Iα)
  • Basic principles of force analysis in vertical motion
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about torque and its relationship to angular acceleration
  • Explore the implications of radius of gyration in rotational motion
  • Investigate the effects of off-center mass on angular dynamics
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Students in physics or engineering courses, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to angular motion and dynamics.

Arin
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Homework Statement


A 30 kg wheel has a center of mass 0.1 m left from the center of the wheel and radius of gyration KG = 0.15 m. Find the angular acceleration if the wheel is originally at rest. The radius of the wheel is 0.25m.

Homework Equations


I=mk^2
T=f*d
M=I*a
Fn acting bottom in Y direction = m*g

The Attempt at a Solution


I=mk^2=(30)*(15^2)=0.675kg*m^2
Fx=0

Fn=(30*9.81)=294.3N
Sum of force in Y: Fn-mg=mass*angular acceleration*radius to center
294.3-294.3=30*a*0.1

Moment about center =ig*a*radius to center of mass
(294.3*.1)=(.675)*a*.1

... lost from here?

Apparently answer is 10.3 rad/s^2

Help please! Been working on this for hours
 
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