Question

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 1

1.343 rad/s clockwise

Step-by-step explanation:

m = Mass of person = 85 kg

h = Center of mass height = 0.79 m

I = Moment of inertia =
`36\ kgm^2`

Here the angular momentum is conserved over the joint O

`-mv_1h=I\omega_2\\\Rightarrow \omega_2=(-mv_1h)/(I)\\\Rightarrow \omega_2=(-85* 1.26* 0.79)/(63)\\\Rightarrow \omega=−1.343\ rad/s`

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