A 56.0 kg ice skater spins about a vertical axis through her body with her arms

Question

A 56.0 kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 1.50 turns each second. The distance from one hand to the other is 1.5 m . Biometric measurements indicate that each hand typically makes up about 1.25% of body weight.

*What horizontal force must her wrist exert on her hand?

*Express the force in part (a) as a multiple of the weight of her hand.

This is a mastering physics problem ... I have tried 4 times and I am on my last chance to put in the right answer. Please HELP!

Explanation / Answer

Given that mass M=56kg number of turns made per sec is 1.5 so the angular velocity is =1.5 (2) =3 rad/s the distance from one hand to the other is d=1.5m mass of each hand is m=(0.0125)M=0.7kg then the force on each hand is F=m2 (d/2)                                                   =(0.7)(3)2(0.75)                                                  =46.63N                                                                                      

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.