Please answer the question neatly with steps and show all units, for good rating

Question

Please answer the question neatly with steps and show all units, for good rating!

Page 2 An ice skater whose rotational inertia is 12.0kg m2 begins an in place spin at a rate of 1.50revs If she brings het arms and legs closer to the axis of totation for the spin resulting in a rocational inertia of 4.00kg m what will her new rate ofspin be? b) What were her rotational kinetic energies before and after the change in rotational inertia? (c How much work did she have todo to accomplish this change? (Hint: Rotational work AKe0 Two are spinning in place about of One disk whose mass is 3.0kg and radius is 15cm is tuming at 4.0rads, and the second whose mass is 6.0kg whose radius is irm turning at 2,0rads, ccw. The disks couple and turn together. How fast will they be

Explanation / Answer

a) The initial momen of inertia of the skater is I1 = 12kg-m2 (given)

Intital angular velocity is 1 = 1.5rev/s (given)

Final momen of inertia of the skater is I2 = 4kg-m2  (given)

Let the final angular velocity be 2.

Then, using the conservation of angular momentum we have

I11 = I22, putting the values we get:

(12kg-m2)(1.5rev/s) = (4kg-m2)2,

or 2 = 4.5rev/s is the new rate of spin of the skater.

b) Rotational kinetic energy before change in rotational inertia is:

KE1 = 1/2(I112) = (1/2)(12kg-m2)(1.5rev/s)2 = 13.5J

Rotational kinetic energy after change in rotational inertia is:

KE2 = 1/2(I222) = (1/2)(4kg-m2)(4.5rev/s)2 = 40.5J

c) Work done by her to accomplish the change is W = KE2 - KE1 = 40.5J - 13.5J = 27J

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---