Question

potter's wheel, with rotational inertia 48 kg m, is spinning freely at 40 rpm. The potter drops a lump of clay onto the wheel, where it sticks a distance 1.2m from the rotational as if the subsequent anplar speed of the wheel and clay is 32 rpm what is the mass of the clay?

Answer 1

Since there is no external force applied, So angular momentum will remain conserved. So

Li = Lf

Ii*wi = If*wf

Ii = Initial moment of inertia of potter's wheel = 48 kg.m^2

wi = Initial angular velocity of wheel = 40 rpm = 40*2*pi/60 = 4.19 rad/sec

If = final moment of inertia of potter's wheel + lump of clay = Ii + m*r^2

m = mass of lump = ?

r = distance of lump of clay from center of wheel = 1.2 m

wf = final angular velocity of wheel = 32 rpm = 32*2*pi/60 = 3.35 rad/sec

So,

Ii*wi = If*wf

48*4.19 = (48 + m*1.2^2)*3.35

m = (48*4.19/3.35 - 48)/1.2^2

m = 8.358 kg = 8.4 kg = mass of the lump of clay

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