A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 kg directly o

Question

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 kg directly onto the middle of the wheel. The clay is in the shape of a pancake and has a radius of 0.19 m. Assume both the wheel and the chunk of clay can be modeled as solid cylinders (I = ½ MR2 ). What is the new tangential velocity of the wheel and the clay?

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King 1 week 2021-07-22T11:25:19+00:00 1 Answers 2 views 0

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    2021-07-22T11:27:14+00:00

    Answer:

    v_{f,w} = 1.791\,\frac{m}{s}, v_{f,c} = 0.972\,\frac{m}{s}

    Explanation:

    The situation can be modelled by applying the Principle of Angular Momentum Conservation:

    I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}

    The final angular speed is:

    \omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}

    \omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev}  \right)

    \omega_{f} \approx 5.116\,\frac{rad}{s}

    The tangential velocities of the wheel and the clay are, respectively:

    v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )

    v_{f,w} = 1.791\,\frac{m}{s}

    v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )

    v_{f,c} = 0.972\,\frac{m}{s}

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