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A solid uniform disk with a mass M and a radius R is pivoted around a horizontal axis through its center, and a small object with mass M is attached to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular velocity of the small object at the bottom.

Respuesta :

lcmendozaf

Answer:

[tex]\omega = \sqrt{\frac{4g}{3R}}[/tex]

Explanation:

By conservation of energy:

The initial energy of the system is: [tex]M*g*R[/tex]

The final energy of the system is: [tex]1/2*M*R^2*\omega^2+1/2*(1/2*M*R^2)*\omega^2[/tex]

[tex]M*g*R=1/2*M*R^2*\omega^2+1/2*(1/2*M*R^2)*\omega^2[/tex]

Solving for ω:

[tex]\omega=\sqrt{\frac{4g}{3R}}[/tex]

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