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To throw an 80 kg opponent with a basic judo hip throw, you intend to pull his uniform with a force vecF and a moment arm d_(1) = 0.30 m from a pivot point on your right hip. You wish to rotate him about the pivot point with an angular acceleration alpha" of "-6.0rad//s^(2) - that is, with an angular acceleration that is clockwise in the figure. Assume that his rotational inertia I relative to the pivot point is 15kg*m^(2). (a) What must the magnitude of vecF be if, before you throw him, you benf your opponent forward to bring his center of mass to your hip? (b) What must the magnitude of vecF be if your opponent remains upright before you throw him, so that vecF_(g) has a moment arm d_(2)=0.12m? |
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Answer» Solution :(a) We can relate your pull `vecF` on your opponent to the given angular acceleration `alpha` via Newton.s second law for rotation `(tau_("net")=Ialpha)`. Calculations: As his feet leave the floor, we can assume that only three forces act on him: your pull `vecF`, a force `vecN` on him from you at the pivot point, and the gravitational force `vecF_(G)`. To USE `tau_("net")=Ialpha`, we need the corresponding three torques, each about the pivot point. From Eq. `(tau=r_(bot)F)`, the torque due to your pull `vecF` is equal to `-d_(1)F`, where `d_(1)` is the moment arm `r_(bot)` and the sign indicates the clockwise rotation this torque TENDS to cause. The torque due to `vecN` is zero, because `vecN` acts at the pivot point and thus has moment arm `r_(bot)=0`. To evaluate the torque dur `vecF_(g)`, we can assume that `vecF_(g)` acts at your opponent.s center of mass. With the center of mass at the pivot point, `vecF_(g)` has moment arm `r_(bot)=0` and thus the torque dur to `vecF_(g)` is zero. So, the only torque on your opponent is due to your pull `vecF`, and we can write `tau_("net")=Ialpha` as `-d_(1)F=Ialpha` We then find `F=(-Ialpha)/d_(1)=(-(15kg*m^(2))(-6.0rad//s^(2)))/(0.30m)` = 300 N. (b) Because the moment arm for `vecF_(g)` is no LONGER zero, the torque due to `vecF_(g)` is now equal to `d_(2)mg` and is positive because the torque attempts counterclockwise rotation. Calculations: Now we write `tau_("net")=Ialpha` as `-d_(1)F+d_(2)mg=Ialpha` `F=-(Ialpha)/d_(1)+(d_(2)mg)/d_(1)`. From (a), we KNOW that the first term on the right is equal to 300 N. Substituting this and the given data, we have `F=300N+((0.12m)(80kg)(9.8m//s^(2)))/(0.30m)` = `613.6N~~610N`. |
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