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A uniform cylindrical turntable of radius 1.60 m and mass 27.7 kg rotates counterclockwise in a horizontal plane with an initial angular speed of 4π rad/s. The fixed turntable bearing is frictionless. A lump of clay of mass 2.47 kg and negligible size is dropped onto the turntable from a small distance above it and immediately sticks to the turntable at a point 1.50 m to the east of the axis. (a) Find the final angular speed of the clayand turntable.

Respuesta :

Answer:

[tex]\omega_f = 3.46\pi[/tex]

Explanation:

As we know that there is no external torque on the system of disc and the sand

so the angular momentum of the system will remains conserved

so here we can say

[tex]L_i = L_f[/tex]

[tex]I_o\omega = (I_o + mr^2)\omega_f[/tex]

here we know that

[tex]I_o = \frac{1}{2}mR^2[/tex]

[tex]I_o = \frac{1}{2}(27.7)(1.60)^2 = 35.46 kg m^2[/tex]

Now from above equation

[tex]35.46(4\pi) = (35.46 + 2.47(1.5^2))\omega_f[/tex]

[tex]35.46(4\pi) = 41.02\omega_f[/tex]

[tex]\omega_f = 3.46\pi[/tex]

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