(III) A skier of mass m starts from rest at the top of a solid...

Question

(III) A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface.

(b) If friction is present, does the skier fly off at a greater or lesser angle?

Accepted Answer

Everyone. Let's take a look at this practice problem dealing with conservation of energy. So in this problem, a block is placed on top of a ball with a radius of capital R and it slides down on its surface from rest in the presence of friction, does the block fly off at a greater angle than if there is no friction. And we assume that the block has a mass of capital M below the problem or given a diagram of what was described in the problem, we also give four possible choices as our answers. For choice. A in the presence of friction, the block flies off at a greater angle than if there is no friction. Choice B in the presence of friction, the block flies off at a smaller angle than if there is no friction. For choice C. In the presence of friction, the block flies off at the same angle. And for choice D not enough information is provided to come to a conclusion. So the first thing we're going to look at is actually the forces acting on the block right as it begins to slide off. So in this case, when it's getting ready to just slide off. That means we have no more normal force due to the surface of the ball acting on the block. And so we're just gonna have the weight of the block pulling straight downwards and that will have a force of magnitude capital MG. Now, if we call the angle at which the block slides off as phi, then our weight here makes an angle with respect to the radius vector and that angle is going to be phi as well. And so we can actually look at the centripetal force acting on the block. So in this case, we're just going to use the second law. So we will have the net force and some of the forces. And here we're just looking in the radio direction that's going to be equal to the mass multiplied by the acceleration. Now, the only force is gonna be the component of the weight that's acting along the radial direction. And so that is going to be mg multiplied by the cosine phi. And this is gonna be equal to M. And here for the acceleration, I'll have the cent triple acceleration. So we call our formula for centripetal acceleration that is V squared divided by R I want to solve this for cosine phi. And so the MS will cancel and I'll just divide both sides by G I have cosine phi is equal to V squared divided by the quantity of Gr. So now what I want to do is figure out a relationship between the V and the phi for this, we're gonna use conservation of energy. So what I'm going to do is draw a horizontal line connecting our vertical radius. Um That's given in the diagram that at the point where we're actually leaving the block is leaving the sphere. So that's gonna give me a right triangle. And so what I can do is use some trig to figure out basically the height at which the block has fallen. And the reason why we're gonna need that height is for our conservation of energy equation. Recall for conservation of energy. Initially, we're at rest, we have no connect energy. So we just have our gravitational potential energy for our initial energy. And this is going to be equal to our final connect energy plus our final potential energy. But we're free to choose a zero point for our gravitational potential energy will set it equal to the point where the box, the blocks um leaving the sphere. So we just have U I is equal to KF to recall your formulas for um gravitational potential energy. That is gonna be MGH, this is gonna be equal to our kinetic energy. So recall that formula that's gonna be one half MV squared. The MS will cancel out here and I can actually write H in terms of my radius and the angle phi. So in this case, we'll have g multiplying the quantity of initially have R and I need subtract from that the component of that rec triangle that we drew that is along that vertical direction. So that's just minus R cosine phi and this is gonna be equal to one half be squared. So I can solve this equation for V by multiplying both sides by two and then taking the square root. So I'll have V is equal to square root of two G. And I'll factor out an R. We'll have two gr and this is multiplying in the quantity of one minus cosine a phi so I could take this beat and plug it back into our formula and allow us to solve for the P. And so I do that real quick. So I'll have the cosign I is equal to, then I'll have or V squared. I'll just have the two gr multiplying one minus cosine phi that whole quantity is divided by Grgrs will cancel. And I can solve this for cosine phi, this is gonna be equal to two thirds. Now, at this point, I could take the arc arc cosine of both sides and find the actual angle. But we actually don't need that to answer the question. We just need to know if the angle is gonna be larger or smaller or the same if we include friction. So for this, we're going to actually look at our conservation of energy equation and include friction. So if I include friction in the conservation of energy, I'll still have my initial potential energy, then I'll have minus the absolute value of the work done by friction that's gonna be equal to KF. Now, the reason why I use the absolute values here and have a minus sign in front of it, that's because the work done by friction is always negative since the frictional force is always in the opposite direction as the displacement. So using our definition of work with the dot product in there, that dot product will always come out to be negative one for friction. And so this allows us to um indicate that the fact that the work is negative without a actually having to remember and having a plus sign there. So we can basically redo the same calculation that we did before. So for the potential energy, we will have capital MG in place of H will just have R multiplying the quantity of one minus cosine phi it will have minus the absolute value of WF which is work done by friction, that's gonna be equal to one half MV squared. And again, we can solve this for V um by multiplying both sides by two and divided by M. And then taking the square root we have V is equal to where are you at? Of two gr multiplying the quantity of one minus cosign I they'll have minus two divided by M multiplying the absolute value of WF. So at this point, I can look and compare the two speed values that I came up with and notes that they're exactly the same except inside the square root um of this one with friction, I have a second term that is negative. That means that the new speed with friction is actually going to be lower. And if I go back to my equation that relates to speed to the angle, if the beat here is going to be lower, then that means that right hand um term is actually going to be less than what it was originally without friction. And so that means the cosign P here um is actually going to be less. And as cosine um the function decreases, that means the phi is actually increasing. That's because if you remember cosine starts at one at zero degrees and goes to zero at 90 degrees. So if the cosine is decreasing, that means that the angle is actually increasing. So in this case, that means that I'm going to actually have a larger angle with friction. And so if I look at my answer choices, that means that answer choice. A here is actually correct that the block flies off at a greater angle than if there's no friction. So just kind of recap of what we've done here. We started off looking at the case where we had no friction with the blocks lying um off of the ball. And we used um our centripetal acceleration and our centripetal force to come up relationship between the angle phi which is the angle that the block left with. And the speed we then use um conservation of energy to write that speed in terms of an angle. And then we combine those two together to basically calculate what that angle would be. We then repeated the calculation. In this case, we focused just on the con uh conservation of energy and added in the work done by friction. And we saw that the speed would actually decrease. And this led us to come to the conclusion that the angle would actually have to increase or to fly off. So I hope that this has been useful and I'll see you in the next video.

(III) A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface.
(b) If friction is present, does the skier fly off at a greater or lesser angle? <IMAGE>

Key Concepts

Newton's Laws of Motion

Centripetal Force

Energy Conservation

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