A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 11 N·m a

Question

A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 11 N·m about the central axis of the disk. About that axis and with torque applied for 19 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

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Hưng Khoa 7 months 2021-07-14T07:03:09+00:00 1 Answers 11 views 0

Answers ( )

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    2021-07-14T07:05:06+00:00

    Answer

    Given,

    Rotational inertia = 2.0 x 10-3 kg·m²

    Torque = 11 N.m

    time, t = 19 ms

    a) Angular momentum

      \tau = \dfrac{\Delta L}{\Delta t}

     L is angular momentum

      \Delta L = \tau \Delta t

      \Delta L = 11\times 19 \times 10^{-3}

      \Delta L = 0.209\ Kg m^2/s

    b) Angular velocity

      We know.

         L = I \omega

         \omega = \dfrac{L}{I}

         \omega = \dfrac{0.209}{2\times 10^{-3}}

         \omega = 104.5\ rad/s

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