COULD STEP BY STEPS IN DETAIL PLEASE BE ADDED IT WOULD BE AWESOME IF IT CAN BE T
ID: 1788527 • Letter: C
Question
COULD STEP BY STEPS IN DETAIL PLEASE BE ADDED
IT WOULD BE AWESOME IF IT CAN BE TYPED FOR BETTER UNDERSTANDING AS WELL,
THANK YOU!
P4: Consider a skinny skater with moment of inertia 10 kg m2, spinning on her skate tip and holding two 8 kg weights close to her body. Initially she is spinning at 12 rad/sec and holding the weights 25 cm from her axis of spin. When she unfolds her arms, the weights move out to a distance of 60 cm and her moment of inertia increases to It- . ( 11 kg m2 (Consider the weights as point masses) (a) How fast is she spinning after she unfolds her arms? (b) Does her kinetic energy change? By how much? Is this an increase or decrease? (c) After she unfolds her arms, friction slows her to a stop in 10 seconds. What is her angular acceleration (d) After unfolding her arms, how many revolutions does she complete before stopping?Explanation / Answer
(a) Ii = 10 + 2(8 x 0.25^2) = 11 kg m^2
wi = 12 rad/s
If = 11 + 2(8 x 0.60^2) = 16.76 kg m^2
Li = Lf
Ii wi = If wf
wf = 7.9 rad/s
(B) Yes, it changes.
Ki = Ii wi^2 /2 = 792 J
Kf = If wf^2 /2 = 523 J
it decreases.
change = Kf - Ki= 269 J
(C) 0 = 7.9 - alpha x 10
alpha = - 0.79 rad/s^2
(d) 0^2 - 7.9^2 = 2(0.79) (theta)
theta = 39.5 rad = 6.3 revolutions