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A person stands, hands at their side, on a platform that is rotating at a rate of 1.50 rev/sec. When they raise their hands to a horizontal position, the rate of rotation decreases to 0.70 rev/sec. By what factor has their moment of inertia changed going from hands at side to hands outstretched?

a. 2.14
b. 1.05
c. 0.467
d. None

Respuesta :

okpalawalter8

Answer:

The value is k =  2. 14

Explanation:

From the question we are told that

    The angular speed of the platform when the hands are on the side is [tex]w_1 = 1.50 \ rev / sec[/tex]

     The angular speed of the platform when the hands are in horizontal position is

    [tex]w_2 = 0.70 \ rev / sec[/tex]      

Generally the angular momentum of the system (the person and the platform) when the person's hand is at the side is

          [tex]L_1 = I_1 * w_1[/tex]

Generally the angular momentum of the system (the person and the platform) when the person's is stretched out horizontally  is

          [tex]L_2 = I_2 * w_2[/tex]

Generally from the law of angular momentum conservation we have that

         [tex]L_1 = L_2[/tex]

=>      [tex]I_1 * w_1 = I_2 * w_2[/tex]

=>     [tex]I_2 = \frac{w_1}{w_2} * I_1[/tex]

=>     [tex]I_2 = \frac{1.50 }{0.70} * I_1[/tex]

=>     [tex]I_2 = 2.14 I_1[/tex]

So the factor by which  moment of inertia changed is  k =  2. 14

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