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iren [92.7K]
3 years ago
8

An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

mass is r1 = 0.45 m from the pole. At that point she grabs hold of the pole.
a. What is the skater's angular speed when she first grabs the pole? rad/s
b. What Is the skater's angular speed after she now pulls her center of mass to a distance of r_2 = 0.46 m from the pole? rad/s
Physics
1 answer:
IrinaVladis [17]3 years ago
7 0

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

r_1 = Distance between center of mass of person and pole = 0.45 m

r_2 = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

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And

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= 0.4 mr²

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ωf = ωi + α t

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Order of I ( least to greatest ) :

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Order of ωf: ( least to greatest)

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