slader A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearin

Question

slader A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions?

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Maris 2 months 2021-08-05T10:24:29+00:00 1 Answers 2 views 0

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    2021-08-05T10:26:19+00:00

    Answer:

    The value of torque on the rod is = 3.94 N – m

    Explanation:

    mass = 50 kg

    Radius = 0.5 m

    Moment of inertia ( I ) = \frac{MR^{2} }{2}

    I = 50 \frac{0.5^{2} }{2}

    I = 6.25 kg- m^{2}

    Angular velocity \omega = 120 \frac{Rev}{min} = 12.6 \frac{rad}{sec}

    Δ\theta = No. of rev × 2\pi

    Δ\theta = 20 × 2\pi

    Δ\theta = 126 rad

    From work energy theorem  

    Work done is equal to change in kinetic energy.

    W_{AB}  = K_{B} - K_{A} ——– (1)

    Where A & B represents the initial & final states.

    Since K_{A} = 0

    W_{AB}  = K_{B}

    W_{AB}  = \frac{1}{2} I \omega^{2}

    W_{AB}  = \frac{1}{2} 6.25 (12.6)^{2}

    W_{AB} = 496 J

    We know that

    W_{AB} = T Δ\theta

    Where W_{AB} = work done

    T = torque applied on the rod

    Δ\theta = Angular displacement of the rod.

    496 = T × 126

    T = 3.94 N – m

    Therefore the value of torque on the rod is = 3.94 N – m

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