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15 January, 17:43

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 15 seconds. How much friction force does the brake pad apply to the shaft?

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  1. 15 January, 17:58
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    frictional force = 0.52 N

    Explanation:

    diameter of turn table (D1) = 30 cm = 0.3 m

    mass of turn table (M1) = 1.2 kg

    diameter of shaft (D2) = 1.2 cm = 0.012 m

    mass of shaft (M2) = 450 g = 0.45 kg

    time (t) = 15 seconds

    acceleration due to gravity (g) = 9.8 m/s^{2}

    radius of turn table (R1) = 0.3 / 2 = 0.15 m

    radius of shaft (R2) = 0.012 / 2 = 0.006 m

    total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

    I = 0.5 (M1) (R1) ^{2} + O. 5 (M2) (R2) ^{2}

    I = 0.5 (1.2) (0.15) ^{2} + O. 5 (0.45) (0.006) ^{2}

    I = 0.0135 + 0.0000081 = 0.0135081

    ω₁ = 33 rpm = 33 x / frac{2π}{60} = 3.5 rad/s

    α = - ω₁/t = - 3.5 / 15 = - 0.23 rad/s^{2}

    torque = I x α

    torque = 0.0135081 x (-0.23) = - 0.00311 N. m

    torque = frictional force x R2

    - 0.00311 = frictional force x 0.006

    frictional force = 0.52 N
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