# Answer to Question #7844 in Mechanics | Relativity for bhawana

Question #7844

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.

Expert's answer

We have the following superposition just before contact:

Tm + Td = 0.04w + 0.06w = 0.1w

Tthe mass could be any distance a from the axis

0.04 = ma²,

but since it is a small enough mass these may remain indeterminate.

The kinetic energy before is

0.5*0.06*(0.3/(2πR)² rad/sec),

equal to

KE2=0.5*1*(x/(2πR))².

The denominators cancel, leaving

x = sqrt(0.06*0.3)=0.13 S.

Tm + Td = 0.04w + 0.06w = 0.1w

Tthe mass could be any distance a from the axis

0.04 = ma²,

but since it is a small enough mass these may remain indeterminate.

The kinetic energy before is

0.5*0.06*(0.3/(2πR)² rad/sec),

equal to

KE2=0.5*1*(x/(2πR))².

The denominators cancel, leaving

x = sqrt(0.06*0.3)=0.13 S.

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