A figure skater is spinning at a rate of 1.6 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to 65% of it's original value. What is her new rate of rotation?
____ rev/ s
____ rev/ s
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Well, you can simply use the conservation of angular momentum. (I1)(w1) = (I2)(w2) (where I1 and I2 are the moments of inertia of the skater initially and finally; w1 and w2 are her respective angular velocities)
From the conditions of the problem, the value of I2 = 65 I1/100. and w1 = 1.6 rps.
Punch the values w1 and I2 in the above written equation, I1 will get cancelled from both the sides and w2 will come out to be 2.46 or simply 2.5 rev/s (rounded to 2 significant figures).
From the conditions of the problem, the value of I2 = 65 I1/100. and w1 = 1.6 rps.
Punch the values w1 and I2 in the above written equation, I1 will get cancelled from both the sides and w2 will come out to be 2.46 or simply 2.5 rev/s (rounded to 2 significant figures).