Solve Rotational Questions for Translational Speeds & Angular Speed

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The discussion focuses on solving a physics problem involving two objects connected by a string over a frictionless pulley with moment of inertia. Using the conservation of energy principle, the equation 1/2m1v1^2 + 1/2m2v2^2 + 1/2Iω^2 = 9mg is established to relate the translational speeds of the objects and the angular speed of the pulley. By rearranging this equation, expressions for v1 and v2 are derived, allowing for the calculation of the angular speed ω. The final results indicate that the translational speeds of the objects as they pass each other are v1 and v2, while the angular speed of the pulley is ω. This approach effectively integrates concepts of rotational dynamics and energy conservation.
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Homework Statement



Consider two objects with m1>m2 connected by a light string that passes over a pulley having a moment of inertia of I about the axis of rotation. The string does not slip on the pulley or stretch. The pulley turns without friction. The two object are released from rest by a vertical distance of 9h.

Use the principle of conservation of energy to find the translational speeds of the objects as they pass each other?

Find the angular speed of the pulley at this time?

Homework Equations





The Attempt at a Solution

 
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Conservation of energy: 1/2m1v1^2 + 1/2m2v2^2 + 1/2Iω^2 = mgh Using this equation and substituting the values given, we have 1/2m1v1^2 + 1/2m2v2^2 + 1/2Iω^2 = 9mg We can solve for v1 and v2 by rearranging the equationv1^2 = 2(9mg - 1/2m2v2^2 - 1/2Iω^2)/m1 v2^2 = 2(9mg - 1/2m1v1^2 - 1/2Iω^2)/m2 We can then solve for ω using either equationω^2 = 2(9mg - 1/2m1v1^2 - 1/2m2v2^2)/I Therefore, the translational speeds of the objects as they pass each other is v1 and v2, and the angular speed of the pulley is ω.
 
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