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A 50-kg ice skater spins about a vertical axis through her body with outstretched arms...

A 50-kg ice skater spins about a vertical axis through her body with outstretched arms. She makes 2 turns each second. The distance from one hand to the other is 1.60 m. Each hand makes up approximately 1.25% of the body weight
 
a) What horizontal force does her wrist exert on the hand?
 
b)If she spins faster (3 turns each second), what is the percentage change of the force on the hand?
 
from what I have, I think the force of wrist exerted on hand is = 78.75 N
 
for part B, I changed the denominator in the acceleration formula to seconds for each of the three turns rather than two, plugged it in to the force equation, and got 287 N. An increase of 265%. Is this correct? Thanks!
 
 
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2 Answers

Centrifugal force is given by:
 
F=mω2R, where m is the mass of the hand, R is the radius of the circle the hand makes, ω is the angular frequency in rad/s. Mass of one hand is 1.25% of the body mass, that is 50*0.0125=0.625 kg; radius equals to 0.8 m (half the distance between two arms), angular velocity is 2*2π=4π≈12.6 rad/s. Therefore, the force on the hand is:
 
F=0.625*12.62*0.8≈79.4 N
 
If she spins with 3 turns per second, her angular velocity increases in 3/2=1.5 times. Since it goes as squared in the equation, the force changes by a factor 1.52=2.25. Therefore, the force increases to 179 N, that is 125% more.

Comments

The hands make 2 turns every second so the angular velocity is 2 turns *( 2 pi radians/turn) = 4* pi radians.
The mass of the hands (since mass is proportional to weight, we do not need to factor in gravity for this problem) is 0.0125 * 50 kg = 0.625 kg. The radius is half the distance between the hands or 0.80 m. We can solve for force exerted on the hands as it will be equal to the centripetal force.
 
F=(mass) * (radius) (angular velocity) ^2 = 0.625 kg *(0.80 m) * (4*pi radians)^2 = 78.96 N on each hand

For part B, the angular velocity changes to 6 pi radians, so the equation is:
F=(mass) * (radius) (angular velocity) ^2 = 0.625 kg *(0.80 m) * (6*pi radians)^2 = 177.65 N on each hand
 
 
The percentage change is 177.65/78.96*100%= 225%