I have tried to solve this problem in multiple ways, but i keep getting it wrong please help
A 48.0 {\rm kg} ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 1.50 turns each second. The distance from one hand to the other is 1.5 {\rm m}. Biometric measurements indicate that each hand typically makes up about 1.25 \% of body weight.
What horizontal force must her wrist exert on her hand?
Express the force in part (a) as a multiple of the weight of her hand.
To solve this problem, we need to consider the forces acting on the ice skater as she spins. The given information includes the mass of the ice skater, the number of turns per second, the distance between her hands, and the percentage of body weight for each hand.
To find the horizontal force exerted by her wrist on her hand, we can follow these steps:
Step 1: Calculate the weight of each hand
To find the weight of one hand, we need to multiply the body weight of the ice skater by the percentage of body weight that corresponds to each hand.
Weight of one hand = (body weight) x (percentage of body weight for one hand)
Step 2: Calculate the centripetal force
The centripetal force acting on the ice skater is caused by the circular motion when she spins. Using the formula for centripetal force:
Centripetal force = (mass) x (angular velocity)^2 x (radius)
In this case, the mass is the mass of the ice skater, the angular velocity is given as 1.50 turns per second, and the radius of the circular motion is half the distance between her hands.
Step 3: Calculate the force exerted by her wrist
The force exerted by her wrist on her hand must be equal to the centripetal force. We need to calculate this force as a multiple of the weight of her hand.
Force exerted by the wrist = Centripetal force / Weight of one hand
Now let's calculate the answer step by step.
Step 1:
Weight of one hand = (48.0 kg) x (1.25 %) = 0.6 kg
Step 2:
Radius = 0.5 x (1.5 m) = 0.75 m
Centripetal force = (48.0 kg) x (1.50 turns/s)^2 x (0.75 m) = 102.6 N
Step 3:
Force exerted by the wrist = 102.6 N / 0.6 kg = 171 N/kg (exact answer)
So, the horizontal force exerted by her wrist on her hand is 171 times the weight of her hand.
centripetal force on the hand is mv^2/r
You have mass m; v is wr or (1.5*2PI/sec)^2 * radius r.