Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.7 rev/s . The wheel can be considered a uniform disk of mass 4.7 kg and diameter 0.34 m . The potter then throws a 2.8-kg chunk of clay, approximately shaped as a flat disk of radius 8.0 cm , onto the center of the rotating wheel.

What is the frequency of the wheel after the clay sticks to it? Ignore friction.

Express your answer using two significant figures.

Answer #1

using conservation of angular momentum

I
_{initial} = I
_{final}

and moment of inertia for wheel is

I_{wheel} = M R^{2}/2

= 4.7
x 0.17^{2}/2

I_{wheel} = 0.067915 kg-m^{2}

now

the moment of inertia for clay is,

I_{clay} = M R^{2}/2

= 2.8
x 0.08^{2}/2

=
0.00896 kg-m^{2}

I_{wheel}_{initial}
= ( I_{wheel} + I_{clay} )
_{final}

0.067915 x 1.7 = ( 0.067915 + 0.00896 )
_{final}

_{final}
= 0.067915 x 1.7 / ( 0.067915 + 0.00896 )

_{final}
= 1.50 rev/sec

**the frequency of the wheel after the clay sticks to it
is
_{final} = 1.50 rev/sec**

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