A 52-kg ice skater spins about a vertical axis through her body w...

Question

A 52-kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is 1.50 m. Biometric measurements indicate that each hand typically makes up about 1.25% of body weight. (b) What horizontal force must her wrist exert on her hand?

Accepted Answer

Welcome back everybody. We are looking at a ballerina and we're actually looking at this ballerina from the top down. So here's her head and the arms of this ballerina go off to the side like this? Now this ballerina is spinning and she makes this circle with the path of her spinning arms. Let me just over here a little bit and we are told a couple of things about this ballerina, we are told that she weighs kg. Were also told that her rotations per second are going to be equal to three 0.0. And we are also told that the diameter of this circle or the distance between her hands is 1.80 m. And we are asked to find what the force exerted by her wrists on her hands is horizontally outwards. So we are looking for f. W One more thing that we are also told us. We are told that the mass of one of her hands is about 1.5% or zero or . times the mass of her entire body. So according to Newton's second law, and we are told that any force is equal to the mass times acceleration. In this case, we're gonna be looking at the mass of her hands, which we already know. But what about this acceleration right here? Well, since she's spinning in a circle, we are actually going to be dealing with centripetal access And the formula for that is that the centripetal acceleration is equal to four times pi squared times the radius of our circle. All over the period. But what is the period? We have all the other variables. What is the period? The period? It's just one over rotations are second. So in order to find this, we need to find these two values. So let's go start plugging in values and see what we can do here. Our period is going to be equal one over the rotations per second, which is just three. This is equal to 0.33 seconds. Great. So now the centripetal acceleration that we're looking for is just four times pi squared times the radius, which is going to be one half of our diameter. So 0.9 divided by our period that we just found. Which is 0.33, which if you plug this into the calculator, we get that. Our centripetal acceleration is 326. m per second squared. Great. Now that we have our centripetal acceleration and we already know the mass of our hand. Is this right here? Let's plug in all these values and find what we are looking for. Which is the force of the wrist exerted on the hand. So the mass of the hand is .015 times her normal mass of her body which is 50 times our access centripetal acceleration which is 326.3. Which if you plug all of this into your calculator, you get that? The force of the wrist on the hand is 245 newtons corresponding to our final answer choice of C. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.

Watch next