A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of

Question

A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.74 rad/sec. The moment of inertia of the student plus the stool is 7 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.28 m from the rotation axis.

(a) Find the new angular speed of the student.
rad/s

(b) Find the kinetic energy of the student before and after the objects are pulled in.

before: J
after: J

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Hồng Cúc 4 years 2021-07-30T23:22:19+00:00 1 Answers 14 views 0

Answers ( )

  1. MichaelMet
    0
    2021-07-30T23:24:16+00:00
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    Answer:

    A) ω2 = 1.394 rad/s

    B) KE_i = 1.7085 J and KE_f = 3.2196J

    Explanation:

    We are given;

    Mass of objects: m1 = m2 = 2kg

    Radius: r1 = 0.9m ; r2 = 0.28m

    Angular velocity: ω1 = 0.74 rad/s

    Moment of inertia of stool and object: I = 3kg.m²

    A) Using the law of conservation of angular momentum:

    I1•ω1 = I2•ω2

    The values of the moment of inertia according to the parallel axis theorem will be:

    I1 = I + 2mr1²

    So, I1 = 3 + 2(2 x 0.9²)

    I1 = 6.24 kg.m²

    Similarly,

    I2 = I + 2mr2²

    I2 = 3 + 2(2)(0.28²)

    I2 = 3.3136 kg.m²

    Since I1•ω1 = I2•ω2

    Thus, ω2 = I1•ω1/I2

    ω2 = 6.24 x 0.74/3.3136

    ω2 = 1.394 rad/s

    B) The initial kinetic energy is:

    KE_i = (1/2)•I1•ω1²

    KE_i = (1/2) x 6.24 x 0.74²

    KE_i = 1.7085 J

    The final kinetic energy is;

    KE_f = (1/2)•I2•ω2²

    KE_f = (1/2) x 3.3136 x 1.394²

    KE_f = 3.2196J

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