A 60 cm diameter potter’s wheel with a mass of 30 kg is spinning at 180 rpm. Using her hands, a potter forms a 14 cm-diameter pot that is ce

Question

A 60 cm diameter potter’s wheel with a mass of 30 kg is spinning at 180 rpm. Using her hands, a potter forms a 14 cm-diameter pot that is centered on and attached to the wheel. The pot’s mass is negligible compared to that of the wheel. As the pot spins, the potter’s hands apply a net frictional force of 1.3 N to the edge of the pot. If the power goes out, so that the wheels motor no longer provides any torque, how long will it take the wheel to come to a stop? You can assume that the wheel rotates on frictionless bearings and that the potter keeps her hands on the pot as it slows.

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Sapo 2 weeks 2021-08-28T18:25:07+00:00 1 Answers 0 views 0

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    2021-08-28T18:26:20+00:00

    Answer:

    It will take the wheel 278.9 s to come to a stop

    Explanation:

    Mass of the potter’s wheel, M = 30 kg

    Diameter of the potter’s wheel, d₁ = 60 cm = 0.6 m

    Radius, r₁ = d/2 = 0.6/2

    r₁ = 0.3 m

    The moment of inertia of the wheel, I = 0.5Mr_1^{2}

    I = 0.5*30*0.3^{2}\\I = 1.35 kg.m^2

    d₂ = 14 cm = 0.14 m

    r₂ = 0.14/2 = 0.07 m

    Angular velocity, \omega = 180 rpm

    \omega = \frac{180*2\pi }{60} \\\omega = 18.85 rad/s

    Frictional Force, F = 1.3 N

    The torque generated:

    \tau = F*r_{2}\\\tau = 1.3*0.07\tau = 0.091 Nm

    Torque can also be calculated as:

    \tau = I \alpha\\\tau = I \frac{\omega }{t} \\0.091 = 1.35*\frac{18.8  }{t} \\t = (18.8*1.35)/0.091\\t = 278.9 s

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