The small mass m sliding without friction along the looped tra...

Question

The small mass m sliding without friction along the looped track shown in Fig. 8–47 is to remain on the track at all times, even at the very top of the loop of radius r.

(a) In terms of the given quantities, determine the minimum release height h.

Accepted Answer

Welcome back. Everyone in this problem. A group of students are experimenting, they place a loop at the foot of an incline and release a ball from rest along the incline. The goal is to keep the ball attached to the track throughout its trajectory. Given that the mass of the ball is M and the radius of the loop is R evaluate the minimum release height of the ball in terms of the given quantities that achieves the goal. Assume that friction here is negligible for our answer choices. A says that it's two RB 2.5 RC two plus 9.8 R all multiplied by R and D that is 49 R squared. Now, if we're going to think about the minimum release height of the ball, then we can first try to understand what forces are at work here. Now, if we think about the, the, the situation, we're releasing a ball from rest that goes into a loop and leaves the loop. So we also have energy involved. OK. And recall that by the conservation principle of energy, the energy that goes into the system is going to be equal to the final energy of the system. In other words, the energy cannot be destroyed, it can only be changed or transferred. Now, with this idea in mind, let's look at our system and try to find an initial and final point and understand what's going on at both of those points. Now, at the minimum release height of the ball, we can call this the initial point in our system and in our system at the initial point because it is released from rest. That means there is no kinetic energy, only potential energy is in the initial stages. However, in the final stages, let's say that our final stage is when the ball is at the topmost point of the loop. And this is a pretty convenient uh point for our final stages. OK? Because no, if we consider that to be the final stage, OK, then at this point, our ball is going to have kinetic energy because it's going to be moving and it's also going to have a potential energy because it, it, it has a height that's off the ground. OK. Now, based on the principle of conservation of energy, then this tells us that our initial potential energy, OK. Pe is going to be equal to our final kinetic energy and our final potential energy. OK. What do we know about those energies? Again, recall that potential energy is equal to the mass multiplied by the acceleration due to gravity multiplied by height and kinetic energy. Equals half MV squared. OK. So using those ideas, using those definitions in our problem, then this tells us by the conservation of energy or initial potential energy MG multiplied by our initial height which is H OK is going to be equal to our final kinetic energy. In this case, half MVF squared, OK, plus our final potential energy which is going to be equal to MG multiplied by the height of the top of the loop here. Because notice that height is going to be, let's call that height yt No. In this case, there are few things that we don't know. We know that the mass is capital M, the radius of the loop is R. We want to figure out H but we don't know the final velocity. OK. And we don't know the height of our loop. So how can we figure that out? Well, the height of our loop is a bit straightforward because our loop, our loop has a radius of R. So this height is going to be R plus RR two R. OK. So we know that YT equals to R, that one is that one, we can figure that one out. But now how can we find the final velocity? In other words, what we do, we know about the velocity at the topmost part of the loop. Well, in the loop, again, the ball is moving in a circle. And hence, that means the net forces central PTA if the ball is released from the minimum release height edge, the ball stays attached barely to the track at the topmost point of the loop. So, if we think about the forces acting here, we're going to have the ball's weight MG acting downwards and we'll also have normal force, the force exerted by the surface onto the ball being equal or are also going in the same direction as the ball's weight. Ok. Let me put it up here to make it a bit more visible. OK. But what do we know about that normal force? Well, we know our normal force is zero. So we can resolve our forces here, we can resolve our centripetal forces and our normal forces to eventually get an expression for the velocity at the top of our loop. OK. So with that in mind, you know what I should have probably put that in blue, but here solving for our final velocity, OK. Then we know our centripetal force is going to be equal to our normal force plus our weight MG. But what else do we know about centripetal force? Well, again, recall that for a cent interpret force based on Newton's law, it's equal to its mass multiplied by the ax, the radial acceleration which in this case, the acceleration around uh or for a centripetal force is V squared divided by R. OK. So FC equals MV squared divided by R. OK. And this velocity V is going to be our final velocity that we want. So that's all of that divided by R, we know that our normal force is zero. OK. So that tells us then that MG is going to be equal to MVF squared divided by R. We can cancel M from both sides that tells us G equals VF squared divided by R. So that means VF is going to be equal to the square root of G multiplied by R. Guess what? Now that we know our final velocity, the velocity at the topmost part of the loop, we can put it into our formula that we have using the principle of conservation of energy to eventually solve for it. Because now, OK, let's make some space here. You know what I should move this down here. Well, no, let me leave it. OK? Because we can make some space here. OK? Because know what this means is that using it back in our formula, let me put this in red. That now means MG is going to be equal to half of M multiplied by that final velocity, which is the square root of GR all squared. OK. Plus MG multiplied by the height of our loop, which we know to be two R know that we have that we can start to simplify. OK? Because first notice that we can cancel M from all of our terms. And the next we can also we can square the square root of Gr So when we do that, we get GH to be equal to a half of gr, OK? Because the square cancels the square root. So a half of Gr plus G multiplied by two R or two. Gr again, we can cancel G from all of our terms. And then that's going to leave us with H equals a half of R plus two R which is 2.5 R or 2.5 R. Thus the minimum release height of the ball in terms of R is 2.5 R B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

The small mass m sliding without friction along the looped track shown in Fig. 8–47 is to remain on the track at all times, even at the very top of the loop of radius r.

(a) In terms of the given quantities, determine the minimum release height h.
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Key Concepts

Centripetal Force

Conservation of Energy

Minimum Speed at the Top of the Loop

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