Answer:
1.78 rad/s
1.70344 rad/s
Explanation:
v = Velocity = 0.8 m/s
m = Mass of person = 60 kg
[tex]r_1[/tex] = Distance between center of mass of person and pole = 0.45 m
[tex]r_2[/tex] = New distance between center of mass of person and pole = 0.46 m
I = Moment of inertia
Angular speed is given by
[tex]\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s[/tex]
The angular speed is 1.78 rad/s
In this system the angular momentum is conserved
[tex]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[/tex]
The new angular speed is 1.70344 rad/s
The skater's angular speed when she first grabs the pole and after is mathematically given as
Question Parameter(s):
An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole.
her center of mass is r1 = 0.45 m from the pole.
Generally, the equation for the Angular speed is mathematically given as
[tex]\omega_1=\frac{v}{r_1}\\\\\omega_1=\dfrac{0.8}{0.45}[/tex]
w1=1.78 rad/s
b) In conclusion, after she now pulls her center of mass to a distance of r_2 = 0.46 m from the pole
[tex]\omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}[/tex]
w2=1.70344ra d/s
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