Question

A 56.0kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each second. The distance from one hand to the other is 1.5m. Biometric measurements indicate that each hand typically makes up about 1.25 % of body weight.
Part A) horizontal force must her wrist exert on her hand F=150N

Express the force in part (a) as a multiple of the weight of her hand?

Answer 1

Answer:

Net force = 129.4, Force as multiple of weight of her hand = 18.84

Explanation:

Given Data:

Total body weight = 56.0 kg   ;

no. of turns = 2.5/second        ;

hand to hand distance = 1.5m ;

weight of hand = 1.25% of body weight ;

Solution:

mass of hand = $\frac{1.25}{100}$*56 = 0.7kg ;

radius = d/2 = 1.5/2 = 0.75m     ;

Now we need to find velocity, as we know that velocity can be calculated by dividing distance by time

v = d/t = $\frac{2.5*2*3.14*0.75}{1}$ = 11.775 m/s or 12 m/s;

a.

The formula to calculate force is given as

F = mv²/r = (0.7*11.775²)/0.75 = 129.4 N

b.

To calculate force as multiple of weight on her hand, we need to calculate the gravitational force W on her hand first.

W = gm = 9.81 * 0.7 = 6.867 N

Now the wieght on her hand can be represented by

$\frac{Force_{net} }{weight of hand}$  = 129.4 / 6.867 = 18.84