The hoop is effected by the gravity force, the normal reaction and friction forces at the point where the hoop is currently touching the girl’s waist:
In our case, we have g =10 m/s2, k=0.5, D=0.45 m, d=0.375 m
Using (2) we get
Using (3) we get
ω =16.33 rad/s (4)
The rotations per second is equal to
We put (4) in (5)
n=2.6 rotations per second
1 minute has 60 second.
In this way, the number of rotations per minute is equal to
2.6×60=156
Answer
156
Comments
Thanks alot
Dear Tosin, please check fixed answer
Thank you so much but how did you arrive at n=1.24 For which Equation (3) where ω=16.33, inserting the value of ω in Equation (4) final answer gives 2.60. Eagerly awaiting your response.
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