Nastia Luikin can reduce her rotational inertia by a factor of about 3.3
问题描述：
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。
答
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.