An ice dancer with her arms stretched out starts into a spinwith an angular velo

Question

An ice dancer with her arms stretched out starts into a spinwith an angular velocity of 1.0 rad/s. Her moment of inertiawith her arms stretched out is 1/2 70 kg (0.2)^2 + 1/3 4 kg (.9)^2= 1.4 + 1.08 = 2.48 kg m^2. What is her angular velocity when shepulls in her arms to make her moment of inertia 1.4 kg m^2? A. 2.67 rad/s B. 2.45 rad/s C. 2.03 rad/s D. 1.90 rad/s E. 1.77 rad/s An ice dancer with her arms stretched out starts into a spinwith an angular velocity of 1.0 rad/s. Her moment of inertiawith her arms stretched out is 1/2 70 kg (0.2)^2 + 1/3 4 kg (.9)^2= 1.4 + 1.08 = 2.48 kg m^2. What is her angular velocity when shepulls in her arms to make her moment of inertia 1.4 kg m^2? A. 2.67 rad/s B. 2.45 rad/s C. 2.03 rad/s D. 1.90 rad/s E. 1.77 rad/s

Explanation / Answer

Applying the law of conservation of momentum we have as, I11=I22 2.48*1=I*1.4 I=1.771rad/s Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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