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An Olympic figure skater having n moss of 40 kg glides across the ice at 2 6 met

ID: 2322329 • Letter: A

Question

An Olympic figure skater having n moss of 40 kg glides across the ice at 2 6 meters/second What is her momentum? What is her kinetic energy? Then, she leaps and lands on the toe of one skate, twirling Into a graceful pirouette, with her arms held tightly together straight up over her head As an engineer, you know she has merely converted all of her translational motion into rotational motion. It her body can be modeled as being a circular cylinder with a radius of 0.32 meters, how many rpm (revolutions per minute) Is she rotating? What is her rotational kinetic energy? What is her angular momentum? She then fully extends her arms, straight out from her shoulders. Assume her moment of inertia increases by an additional 1 kg-m^2. What is her new rotational speed in rpm?

Explanation / Answer

As all the kinetic energy of translation is converted into the kinetic energy of rotation ...Therefore

K.E tr = K.Erot

   K.Erot = o.5 Iw2

             = o.5 ( o.5 m R2) w2

    135.2 = o.52 ( 40 * 0.322 ) w2

      w = 11.48 rad / s           ( multiplying by 60 and dividing by 2pi) we will get rpm.

       w = 109.7 rev / min''

135.2 = o.52 ( 40 * 0.322 + 1 ) w2

w =10.2 rad / s

w = 98.8 rev / min

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