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Calculate the angular momentum, in kg.m^2/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.420 kg · m2. kg · m2/s (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia, in kg · m2, if his angular velocity drops to 1.95 rev/s. kg · m2 (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque, in N m, was exerted if this takes 23.0 seconds? (Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.) N.m

Answer :

davidlcastiblanco

Answer:

a. [tex]p_1=15.83 kg*m^2/s[/tex]

b. [tex]I_2= 3.07 kg*m^2[/tex]

c. [tex]T=-0.5124N*m[/tex]

Explanation:

[tex]p=I*w[/tex]

[tex]w=2\pi*v[/tex]

[tex]w_1=2\pi*6=12\pi rad/s[/tex]

a.

[tex]w_1=2\pi*6=12\pi rad/s[/tex]

[tex]p_1=I_1*w_1[/tex]

[tex]p_1=0.420kg*m^2*12\pi rad/s[/tex]

[tex]p_1=15.83 kg*m^2/s[/tex]

b.

[tex]w_2=2\pi*1.95=3.9\pi rad/s[/tex]

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

[tex]I_2=\frac{12\pi}{3.9\pi}= 3.07 kg*m^2[/tex]

c.

[tex]w_f=2\pi*3=6\pi rad/s[/tex]

[tex]a=\frac{w_f-w_i}{t}=\frac{3\pi-12\pi}{23s}=-1.22s[/tex]

[tex]T=I*a[/tex]

[tex]T=0.420kg*m^2*-1.22m/s^2[/tex]

[tex]T=-0.5124N*m[/tex]

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