A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is first rotated

Question

A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is first rotated about its midpoint. It is then rotated about one of its ends, and in both cases uniformly accelerates from 0 rad/s to 2.4 rad/s in 3.6 s. Find the torque exerted by the twirler on the baton when it is rotated about its middle. Find the torque exerted by the twirler on the baton when it is rotated about its end.

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King 4 years 2021-08-15T23:22:43+00:00 1 Answers 46 views 0

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  1. Philomena
    0
    2021-08-15T23:24:10+00:00
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    Answer:

    Part a)

    When rotated about the mid point

    \tau = 0.021Nm

    Part b)

    When rotated about its one end

    \tau = 0.042 Nm

    Explanation:

    As we know that the angular acceleration of the rod is rate of change in angular speed

    so we will have

    \alpha = \frac{\Delta \omega}{\Delta t}

    \alpha = \frac{2.4 - 0}{3.6}

    \alpha = 0.67 rad/s^2

    Part a)

    When rotated about the mid point

    I = 2mr^2

    I = 2(0.48)(0.18)^2

    I = 0.0311 kg m^2

    now torque is given as

    \tau = 0.0311 (0.67)

    \tau = 0.021Nm

    Part b)

    When rotated about its one end

    I = m(2r)^2

    I = (0.48)(0.36)^2

    I = 0.0622 kg m^2

    now torque is given as

    \tau = (0.0622)(0.67)

    \tau = 0.042 Nm

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