In a judo foot-sweep move, you sweep your opponent\'s left foot out from under h

Question

In a judo foot-sweep move, you sweep your opponent's left foot out from under him while pulling on his gi (uniform) toward that side. As a result, your opponent rotates around his right foot and onto the mat. The figure below shows a simplified diagram of your opponent as you face him, with his left foot swept out. The rotational axis is through point O. The gravitational force Fg on him effectively acts at his center of mass, which is a horizontal distance of d = 28 cm from point O. His mass is 74 kg, and his rotational inertia about point O is 61 kg·m2

(a) What is the magnitude of his initial angular acceleration about point O if your pull Fa on his gi is negligible?

(b) What is it if your pull Fa on his gi is horizontal with a magnitude of 300 N and applied at height h = 1.4 m?

Explanation / Answer

a) torque = I x alpha

torque = (h x Fa ) + (d x Fg)

when h is negligible.

torque = 0.28 x 74 x 9.81 = 203.26

206.23 = 61 x alpha

alpha = 3.33 rad/s^2

b) torque = (h x Fa ) + (d x Fg)

(1.4 x 300) + (0.28 x 74 x 9.81) = 61 xalpha

alpha = 10.22 rad/s^2

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