A student sits on a rotating stool holding two 2.6 kg masses. When his arms are extended horizontally, the masses are 0.71 m from the axis o

Question

A student sits on a rotating stool holding two 2.6 kg masses. When his arms are extended horizontally, the masses are 0.71 m from the axis of rotation, and he rotates with an angular velocity of 1.8 rad/sec. The student then pulls the weights horizontally to a shorter distance 0.23 m from the rotation axis and his angular velocity increases to ω2. For simplicity, assume the student himself plus the stool he sits on have constant combined moment of inertia Is = 3.8 kg m2 . Find the new angular velocity ω2 of the student after he has pulled in the weights. Answer in units of rad/s.

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Nguyệt Ánh 4 years 2021-09-03T01:54:52+00:00 1 Answers 8 views 0

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  1. Ben
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    2021-09-03T01:55:54+00:00
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    Answer:

    2.8 rad/s

    Explanation:

    In absence of external forces, the total angular momentum of the system must be conserved.

    The angular momentum when the arms of the student are extended horizontally is given by:

    L_1 = (I_0 + 2I)\omega_1

    where:

    I_0=3.8 kg m^2 is the moment of inertia of the student+stool

    I=mr^2 is the moment of inertia of each mass, where:

    m = 2.6 kg is one mass

    r = 0.71 m is the distance of each mass from the rotation axis

    \omega_1=1.8 rad/s is the initial angular velocity

    So we have

    L_1=(I_0+2mr^2)\omega_1

    When the student pulls the weights to a distance of r’ = 0.23 m, the angular momentum is:

    L_2=(I_0+2I')\omega_2

    where:

    I'=mr'^2 is the new moment of inertia of each mass, with

    r’ = 0.23 m

    Since the angular momentum must be constant, we have:

    L_1=L_2\\(I_0+2mr^2)\omega_1 = (I_0+2mr'^2)\omega_2

    and solving for \omega_2, we find the final angular velocity:

    \omega_2 = \frac{I_o+2mr^2}{I_0+2mr'^2}\omega_1=\frac{3.8+2(2.6)(0.71)^2}{3.8+2(2.6)(0.23)^2}(1.8)=2.8 rad/s

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