A small turntable with I = 10 kg m 2 is rotating at 5 rad/s. A small piece of cl

Question

A small turntable with I = 10 kg m2 is rotating at 5 rad/s. A small piece of clay (assumed to be a point mass) of mass m = 2 kg is dropped onto the turntable a distance D = 1.5 m from the center (above right). The clay sticks to the turntable and both ultimately rotate at the same angular velocity. Determine that final angular velocity and determine the change in kinetic energy due to this process. Additionally, explain why the kinetic energy increased or decreased.

A small turntable with I = 10 kg m^2 is rotating at 5 rad/s. A small piece of clay (assumed to be a point mass) of mass m = 2 kg is dropped onto the turntable a distance D = 1.5 m from the center (above right). The clay sticks to the turntable and both ultimately rotate at the same angular velocity. Determine that final angular velocity and determine the change in kinetic energy due to this process. Additionally, explain why the kinetic energy increased or decreased.

Explanation / Answer

Moment of inertia of clay ,I'= m*D^2 = 2* (1.5)^2 = 4.5 Kg.m^2
Use conservation of angular momentum,
I*wi = (I+I')*wf
10*5 = (10+4.5)*wf
wf= 3.45 rad/s
Answer: Final angular velocity = 3.45 rad/s^2

Initial kinetic energy = 0.5*I*wi^2
                                       = 0.5*10*(5)^2
                                       = 125 J

Final kinetic energy = 0.5*(I+I')*wf^2
                = 0.5*(10+4.5)* (3.45)^2
                = 86.3 J
Kinetic energy has decreased.
Decraese in kinetic energy = 125 - 86.3 = 38.7 J
Kinetic energy has decreased because energy has lost as heat

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