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. Figure 10-41 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 F_g on him effectively acts at his center of mass, which is a horizontal distance of d = 28 cm from point O. His mass is 68 kg, and his rotational inertia about point O is 70 kg mid dot m^2. What is the magnitude of his initial angular acceleration about point O if your pull F_a on his gi is negligible?___________rad/s^2 What is it if your pull F_a on his gi is horizontal with a magnitude of 300 N and applied at height h = 1.4 m?______________rad/s^2

Explanation / Answer

Given: mass 'M' = 68 Kg; location of COM from point O is 'd' = 28 cm = 0.28 m; moment of inertia of the man about point O is 'I' = 70 Kg-m2

(a) Now the torque of weight of the man about O is '0' = M*g*d = 68*9.8*0.28 = 189.6 N-m

So the initial angular acceleration of the man '0' = 0/I = 189.6/70 = 2.66 rad/s2

(b) with the force Fa = 300 N acting at a height of h = 1.4 m

The new torque about o will be '' = 0  + Fa*h = 189.6 + 300*1.4 = 609.6 N-m

Hence the new angular acceleration '' = /I = 609.6/70 = 8.71 rad/s2

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