Question #7476

A circular disc rotates on a thin air film with a period of 0.3s. Its moment of inertia about its
axis of rotation is 0.06 kg m2. A small mass is dropped onto the disc and rotates with it. The
moment of inertia of the mass about the axis of rotation is 0.04 kg m2. Determine the final
period of the rotating disc and mass. (10

Expert's answer

I - moment of inertia.

I=I1+I2

T1=2pi*sqrt(I1/m1r)=>m1gr=I1*(2pi/T1)^2=26.3

T2=2pi*sqrt((I1+I2)/m1gr)={m2->0}=0.38

I=I1+I2

T1=2pi*sqrt(I1/m1r)=>m1gr=I1*(2pi/T1)^2=26.3

T2=2pi*sqrt((I1+I2)/m1gr)={m2->0}=0.38

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