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.

(b) If the actual release height is 2h, calculate the normal force exerted by the track at the bottom of the loop, then

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An engineer is designing a water slide. He thinks of making it a looping water slide. The sliders are to be on the slide path at all points through their trajectories. He calculates the minimum height from which the slider should be released to be H equals 2.5 multiplied by capital R where R is the loop radius. In reality, the actual release height would be equal to three H three multiplied by each calculate the normal force exerted on a slider of mass capital M when it is at the bottom of the loop as shown in the figure ignore friction. OK. So our end goal it appears is ultimately, we're asked to figure out what the normal force exerted on the slider of a mass M when it's at the bottom of the loop as shown in the figure below. So we're trying to figure out what the normal force exerted is. That's the final value we're ultimately trying to solve for. So looking at our figure that's provided to us in the prom itself, we can see that the top of the slide is at the far right of our figure. And from the bottom, which is the ground, which the ground is indicated by this dotted black line to the top of our water slide is three H. And as we could see, we have our little stick figure of a person lying flat on their back sliding down and it'll go a loop de loop. So it'll go up around the circle and then come down to the very bottom at the end of the slide on the far left of our figure. So the motion will go side down from the top up into this loop all the way around and then be spit out at the end on the left. Awesome. So it's going from right to left or motion. So now that we have an idea of what's going on in this problem, let's look at our multiple choice answers to see what our final answer might be. A is six multiplied by M multiplied by GB is 11 multiplied by M multiplied by G C is 14 multiplied by M multiplied by G. And finally D is 16 multiplied by M multiplied by G Awesome. So with that in mind, first off, let us draw a more detailed diagram of our provided figure to see all the forces at work. In this particular problem. So when we do that, we will get a figure that looks something like this. So as we could see in our new figure to help us visualize the forces that are at work we have on the far right, our height, which is three multiplied by H and to keep things simple. Instead of drawing a stick figure, each time we're just gonna pretend this person is this cir this circle, the gray circles represent the person. So the initially the ball or the person is at the top of the slide and then it will slide down, go through the loop. But when it's at the bottom of this loop before it goes to the top of the loop. So focusing on the bottom of the loop, using a red arrow, we have our normal force pointing up and then we have our weight force pointing down which we know that the weight is mass multiplied by the vector G or gravity, which is our weight force. And then finally, once it comes out of the loop, we have the person at the end of the slide. Awesome moving right along here. Let us also note that inside of the loop, the slider will move in a circle, it's a circular motion. Therefore, the net force will be centripetal. Let us also note that the bo that at the bottom of the loop, the magnitude will be the normal force FN. Let us also note that the mass of the slider is capital M and capital R is the radius of the loop. And let us also note that V subscript B is the velocity inside the loop at the bottom. Considering these factors, we could write that the centripetal force FC is equal to the normal force FN minus the mass multiplied by the acceleration due to gravity which is equal to, to keep things simple. Just say M multiplied by VB squared, which is the velocity at the bottom of the loop divided by capital R. And let us call this equation one. So now at this point, we must recall and apply the principle of conservation of energy. Since we are told that we can ignore friction, there will be no non conservative forces. Now let us use the subscript I to denote the initial quantities which correspond to when the slider is at the top of the release point. So when it's at the top of the slide, when they are about to slide down it, so that's the initial point, the release point height is what we're gonna call it. Let us also use the subscript F to denote the final quantities which correspond to when the slider is at the bottom of the trajectory point off the loop. Let us also note that. So let's make a note here that V I the initial velocity is 0 m per second. Let us also note that Y I our initial distance is equal to three multiplied by H which in this case, we are told that it's that H is actually equal to 2.5 R. So we could write that three multiplied by 2.5 multiplied by capital R is all equal to 7.5 R fantastic. And we know that VF the final velocity is equal to VB which is equal to some unknown value because that's what we do not know this value as of now. And we also know that YF our final distance is equal to zero meters because it's not gonna be moving or going anywhere. So if we consider that the initial energy, so let's write this in blue, if we consider that the initial energy is equal to the final energy. As stated in the principle of the conservation of energy, we can write that one half multiplied by M multiplied by V I squared plus M multiplied by G multiplied by Y I is equal to one half multiplied by M multiplied by VF squared plus M multiplied by G multiplied by why? F Awesome. So, and let's note that each of these MS here are the capital MS which denotes the mass of our slider here. So with that in mind, we now need to plug in all of our known variables. And when we do that, we will get zero plus M multiplied by G multiplied by 7.5 R which is equal to one half multiplied by M multiplied by VB square plus zero. So now rearranging this equation to isolate and solve for VB will give us that VB is equal to the square root of 15 multiplied by G multiplied by R. And let's call this equation two. Awesome. So now if we substitute in this value for VB as we found in equation two, and we'll plug it into equation one, we will get that FN is equal to M multiplied by 15, multiplied by G multiplied by R all divided by capital R plus capital M multiplied by G. And when we simplify this equation, we will get that FN is equal to 15 multiplied by M multiplied by G plus M multiplied by G. And when we simplify one last time, we will get that FN, the normal force is equal to 16 multiplied by M multiplied by G. And that's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter D 16 multiplied by M multiplied by G. And that's it. We did it. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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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.
(b) If the actual release height is 2h, calculate the normal force exerted by the track at the bottom of the loop, then

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