A figure skater pushes off from rest and glides along a circular path, the radius of which is 3 times her height. What must be her minimum initial velocity in order to come full circle without having to push off a second time?
I have no idea. It depends on friction. Where did you get this question?
The question is a real-life question.
In that case it depends on a lot of things, mainly:
The temperature of the ice. Skates move easily over the ice because they melt a tiny film of slippery water under the blade. If the ice is very cold, the skate will have to be very, very sharp to work efficiently. On the other hand you do not want mushy ice because then you get a lot of drag. You want the skate to cut in enough to provide the necessary side force for centripetal acceleration toward the center of the turn. this is the mass of the skater m times v^2/R where v is the forward speed and R is the turn radius. If the ice is too hard or the skate too dull, there will be slipping and loss of speed. In the end the skater must have enough kinetic energy (1/2)mv^2 at the start of the turn to overcome the friction loss during the turn. which is the retding friction force times the distance which for a full turn is 2 pi R
To determine the minimum initial velocity required for the skater to come full circle without pushing off a second time, we need to consider the centripetal force and the gravitational force acting on the skater.
In circular motion, the centripetal force is provided by the net inward force toward the center of the circle, which in this case is the frictional force between the skater's blades and the ice. The gravitational force acts vertically downward.
In order to come full circle without pushing off a second time, the centripetal force must be equal to or greater than the gravitational force, to prevent the skater from falling off the circular path.
Let's break down the solution step by step:
1. We know the radius of the circular path is 3 times the skater's height. Let's denote the skater's height as h.
2. The radius of the circular path, r, can then be expressed as: r = 3h.
3. The gravitational force acting on the skater can be calculated using the skater's mass (m) and acceleration due to gravity (g). The formula for gravitational force is: F_gravity = m * g.
4. The centripetal force can be calculated using the mass of the skater (m), the minimum initial velocity (v), and the radius of the circular path (r). The formula for centripetal force is: F_centripetal = m * v^2 / r.
5. To prevent the skater from falling off the circular path, the centripetal force must be greater than or equal to the gravitational force. Therefore, we set up the following inequality: F_centripetal ≥ F_gravity.
6. Substituting the formulas, we get: m * v^2 / r ≥ m * g.
7. Simplifying, we get: v^2 ≥ r * g.
8. Substituting the value of r = 3h, we get: v^2 ≥ 3h * g.
9. Finally, taking the square root of both sides, we find: v ≥ √(3h * g).
So, the minimum initial velocity required for the skater to come full circle without pushing off a second time is v ≥ √(3h * g), where h is the height of the skater and g is the acceleration due to gravity.