Nastia Luikin can reduce her rotational inertia by a factor of about 3.3when changing from a straight position to a tuck position.If it takes her 0.4 s to makeone full rotation (or revolution) in a straight position,what is her angular speed when in

Nastia Luikin can reduce her rotational inertia by a factor of about 3.3when changing from a straight position to a tuck position.If it takes her 0.4 s to makeone full rotation (or revolution) in a straight position,what is her angular speed when in
Nastia Luikin can reduce her rotational inertia by a factor of about 3.3
when changing from a straight position to a tuck position.If it takes her 0.4 s to make
one full rotation (or revolution) in a straight position,what is her angular speed when in a
tuck position?Provide your answer in rev/s as well as in rad/s.
我不是要翻译。

Nastia Luikin can reduce her rotational inertia by a factor of about 3.3when changing from a straight position to a tuck position.If it takes her 0.4 s to makeone full rotation (or revolution) in a straight position,what is her angular speed when in
当Nastia Luikin从手臂伸展的姿势变为手臂抱团的姿势时,她能把她自己的转动惯量减少3.3倍.
如果她在手臂伸展的姿态旋转一周用时0.4s,那么她手臂抱团时,她的角速度是多少.（分别用：转/秒,和弧度/秒两种单位表示）
设：Nastia Luikin手臂抱团的转动惯量为：J
初始角速度：ω0=2π/0.4=5π rad/s,初始转速：n0=1/0.4=2.5 rev/s
则由角动量守恒：3.3Jω0=Jω,解得：ω=16.5π rad/s
n=ω/2π=16.5π/2π=8.25 rev/s
姿态变化时,人是有做功的,所以角动能不守恒,而合力矩为零,所以角动量守恒.

nastia luikin从伸展位变到抱团位（注：可能是体操、芭蕾或跳水，抱团后转动贯量变小，转速变快）可以将她的转动惯量减少到原来的1/3.3，如果她在伸展状态转一整圈（或一转）花0.4秒时间，她在抱团位的角速度是多少？答案要同时用转/ 秒和弧度/秒两个单位给出。
转动动能E=1/2 * J * ω *ω， J就是转动贯量，ω是角速度（rad/s)
根据能量守恒原理，姿式变化后动...

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nastia luikin从伸展位变到抱团位（注：可能是体操、芭蕾或跳水，抱团后转动贯量变小，转速变快）可以将她的转动惯量减少到原来的1/3.3，如果她在伸展状态转一整圈（或一转）花0.4秒时间，她在抱团位的角速度是多少？答案要同时用转/ 秒和弧度/秒两个单位给出。
转动动能E=1/2 * J * ω *ω， J就是转动贯量，ω是角速度（rad/s)
根据能量守恒原理，姿式变化后动能不变，则
1/2 * J0 *ω0 *ω0 =1/2 *J1* ω1 *ω1
1/2 * J0 *2π/0.4 *2π/0.4=1/2 *J0/3.3 *ω1 *ω1
则角速度ω1=sqrt(3.3)*2π/0.4 rad/s= sqrt(3.3)/0.4 rev/s

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你确定你的题目没有抄错，，nastia luikin可以减少她的转动惯量由约3.3的一个因素发生变化时，直的位置对塔克的位置。如果她需要0.4秒，使一个完整的旋转（革命）在直行位置，她是什么样的角速度在塔克的位置？证明你在转/ S以及弧度/秒

这题考查 角动量守恒 (Conservation of Angular Momentum)
使用公式 L = I w L is angular momentum, I rotational inertia, w is angular speed
L1=L2 ---> I1 w1 = I2 w2
从题目的条件可知 I2= 1...

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这题考查 角动量守恒 (Conservation of Angular Momentum)
使用公式 L = I w L is angular momentum, I rotational inertia, w is angular speed
L1=L2 ---> I1 w1 = I2 w2
从题目的条件可知 I2= 1/(3.3) I1, w1= 2.5 rev/s = 5 pi radian/s
带入 上面等式 I1*2.5 rev/s = 1/3.3 I1 *w2 --> 两边同时除以I1, 可求 w2= 8.25 rev/s =16.5PI rad/s

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