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A circular turntable rotates at constant angular velocity about a vertical axis. There is no friction and no driving torque. A circular pan rests on the turntable and rotates with it: see the sketch at right. The bottom of the pan is covered with a layer of ice of uniform thickness, which is, of course, also rotating with the pan. The ice melts but none of the water escapes from the pan. Is the angular velocity now greater than, the same as, or less than the original velocity

Respuesta :

Answer:

the moment of inertia of the water increases, therefore the angular velocity must decrease

Explanation:

To analyze this exercise we see that the system is isolated therefore the angular momentum is conserved

initial instant. The plate with the ice cap (solid)

          L₀ = I₀ w₀

where Io is the moment of inertia of the plate plus the ice disk

          I₀ = ½ M r² + ½ m r²

where M is the mass of the plate and m is the mass of the ice

final instant.  When the ice has melted, therefore we have water that is a liquid and as the system is rotating it accumulates towards the periphery of the system,

           L_f = I w

in this case the moment of inertia is

            I = ½ M r² + I_water

the moment of inertia of water if it is concentrated in a thin ring is

          I_{water} = mr²

we can see that the moment of inertia increases

how angular momentum is conserved

          L₀ = L_f

          (½ M r² + ½ m r²) w₀ = (½ M r2 + I_{water})  w

          w = [tex]\frac{ \frac{1}{2} Mr^2 + \frac{1}{2} m r^2 }{\frac{1}{2} Mr^2 + I_{water} } \ w_o[/tex] w

we can see that the moment of inertia of the water increases, therefore the angular velocity must decrease

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