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The man is walking with speed v1 = 1.26 m/s to the right when he trips over a small floor discontinuity. Estimate his angular velocity just after the impact. His mass is 85 kg with center-of-mass height h = 0.79 m, and his mass moment of inertia about the ankle joint O is 63 kg·m2, where all are properties of the portion of his body above O; i.e., both the mass and moment of inertia do not include the foot. The angular velocity is positive if counterclockwise, negative if clockwise.

Answer :

boffeemadrid

Answer:

1.343 rad/s clockwise

Explanation:

m = Mass of person = 85 kg

h = Center of mass height = 0.79 m

I = Moment of inertia = [tex]36\ kgm^2[/tex]

Here the angular momentum is conserved over the joint O

[tex]-mv_1h=I\omega_2\\\Rightarrow \omega_2=\dfrac{-mv_1h}{I}\\\Rightarrow \omega_2=\dfrac{-85\times 1.26\times 0.79}{63}\\\Rightarrow \omega=−1.343\ rad/s[/tex]

The angular velocity is 1.343 rad/s clockwise after the impact.

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