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A potter's wheel having a moment of inertia of 46 kg⋅m2 is spinning freely at 44 rpm . The potter drops a lump of clay onto the wheel, which sticks to the wheel at a distance 1.5 m from the rotational axis. Consequently, the wheel is spinning at 35 rpm . What is the mass of the lump of clay?

1 Answer

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Answer:m=5.25 kg

Step-by-step explanation:

Given

Initial angular momentum
I_1=46 kg-m^2


N_1=44 rpm


\omega _1=(2\pi N_1)/(60)=4.60 rad/s

Potter drop a lump of clay such that speed decreases to 35 rpm


N_2=35 rpm


\omega _2=(2\pi \cdot 35)/(60)=3.66 rad/s

Moment of inertia after dropping lump of let say mass m


I_2=46+m* 1.5^2


I_1\omega _1=I_2\omega _2


46* 4.6=I_2* 3.66


I_2=57.81 kg-m^2


46+m* 1.5^2=57.81


m=5.25 kg

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