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strojnjashka [21]
3 years ago
12

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

music. How does the skater use knowledge of the conservation of momentum to do this?
Physics
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

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   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

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        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

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Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

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         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

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substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

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