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    A turn-table on a frictionless vertical axis has a radius of 1.5 m and a moment of rotational inertia of 150 kgm^2. It is at rest until a 75 kg person who had been standing still (and who can be treated as a point), begins to walk around the rim at a speed of 2.0 m/s with respect to the turn-table.
    a) what is the speed of the person with respect to the Earth?
    b)How much work did the person do in bringing the turn-table up to speed?

    © BrainMass Inc. brainmass.com December 24, 2021, 4:45 pm ad1c9bdddf
    https://brainmass.com/physics/angular-momentum/rotational-angular-speed-turn-table-5757

    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attached file for answer
    a) what is the speed of the person with respect to the Earth?
    Moment of inertia of turntable= 150 Kg m^2 =It

    Radius of turntable= 1.5 m
    Mass of the person= 75 Kg
    "Moment of inertia of the person with respect to the vertical axis=
    M r ^2 =" 168.75 Kg m^2 =Ip

    Let the person be moving with angular velocity w p with respect to the earth
    Let the turntable be moving with angular velocity w t with respect to the earth

    w p= Vp/r where Vp is the velocity of the person with respect to the earth
    w t= Vt/r where Vt is the velocity of the rim of the turntable with respect to the earth

    Since the angular momentum is conserved and since the turntable and the person were initially
    at rest

    Ip w p + It w t=0
    Vector equation
    or 168.75 * Vp/r +150 * Vt/r =0

    or 168.75* Vp + 150 * Vt=0 (since r is common, r is not zero and RHS= 0 r can be removed from the equation)

    Thus Vp and Vt are in opposite direction.
    If we use Vp and Vt to denote the magnitude of velocities and treat the direction of Vp as positive and
    the direction of Vt as negative

    The equation becomes
    Scalar equation
    168.75* Vp - 150 * Vt=0

    or Vt=168.75/150= 1.125 Vp --------(1)

    We also know from the data in the problem that the person is moving at a speed of 2.0 m/s with respect to the turn-table.

    Thus Vp- Vt =2 Vector equation
    If we use Vp and Vt to denote the magnitude of velocities and treat the direction of Vp as positive and
    the direction of Vt as negative
    Scalar equation
    Vp+ Vt =2 --------(2)

    Solving simultaneously Equation 1 and 2

    Vp+1.125 Vp =2
    or Vp= 0.94 m/s
    Vt=1.125 Vp = 1.06 m/s
    The speed of the person with respect to the Earth= 0.94 m/s

    b)How much work did the person do in bringing the turn-table up to speed?

    w p =Vp/r= 0.63 rad/s
    w t =Vt/r= 0.71 rad/s

    Rotational Kinetic energy = 1/2 I w ^2

    Rotational Kinetic energyof person = 1/2 Ip w p^2= 33.49 Joules
    Rotational Kinetic energyof turntable = 1/2 It w t^2= 37.81 Joules
    Total= 71.30 Joules

    This is the total amount of work done by the person= 71.30 Joules
    as initially both the person and turntable
    were at rest

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:45 pm ad1c9bdddf>
    https://brainmass.com/physics/angular-momentum/rotational-angular-speed-turn-table-5757

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