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.

(c) by the track at the top of the loop, and (d) by the track after the block exits the loop onto the flat section.

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. Some students are trying to build a self powered roller coaster. The idea is to drop a cart from rest down into a looped path. The cart is to maintain contact with the loop path throughout its motion along the loop. The minimum height drop from which this can happen is found to be H equals 2.5 multiplied by R where capital R is the radius of the loop. If the car is instead dropped from a height of three multiplied by H calculate the normal force exerted on the cart by the path at the top of the loop on the horizontal portion of the path after leaving the loop, ignore friction. OK. So apparently it appears that we're asked to figure out what the normal force exerted on the card is by the path at the top of the loop loop and on the horizontal portion of the path after leaving the loop. So we're asked to solve for two separate answers. And those two separate answers will be the final answers that we're ultimately trying to solve. For our first answer is we're trying to figure out the normal force value. And we're also asked to figure out what the normal forces on the horizontal portion of the path after it leaves the loop. Awesome. And then the first answer is considering the at the, considering the normal force on the path at the top of the loop. OK. Now that we have an idea of what we're solving for, let's quickly look at our diagram that's provided to us by the problem itself. As we could see on the far right hand side of our figure here we have our cart that initially starts from rest at the top of our roller coaster path and it's at a height of three multiplied by h from the ground where the dotted line indicates the ground and the cart will then be released from rest and heading from right to left. It will roll down this track into the loop and it will go all the way around the loop without falling down in a circular motion. And then it will be spit out on the end of our roller coaster track and come to rest eventually on the other side on the horizontal path. So it's gonna be starting from like a vertical upwards. It's gonna be like think of like a cart being up on top of a hill and it's gonna roll down the hill, same kind of logic. Awesome. So now that we have a visualization of what's going on in this particular problem, let's look at the multiple choice answers and read them off to see what our final answer pair might be. A is at the top of the loop 10 multiplied by M multiplied by G on the horizontal path zero B is at the top of the loop 10 multiplied by M multiplied by G. And on the horizontal path, M multiplied by GC is at the top of the loop 11 multiplied by M multiplied by G on the horizontal path zero. And finally, for D at the top of the loop 11 multiplied by M multiplied by G and on the horizontal path M multiplied by G Awesome. So now, first off, let us note that inside of the loop, the cart will be moving in a circle. Therefore, the net force will be centripetal. Let us also note that at the top of the loop, the magnitude will be the normal force FN. Let us also call M to be the mass of the cart and capital R to be the radius of the loop. And that VT is the velocity inside of the loop at the top. Now, considering all of these factors, we can write the following relation or equation, we could write that the centripetal force FC is equal to F and the normal force plus capital M, which I'm just gonna say M from now on to keep things simple, multiplied by G where M is the mass of the car and G is the acceleration due to gravity is equal to M multiplied by V T which is, is the velocity of the cart in the loop at the top of the loop. And don't forget that this value is squared all divided by capital R. Awesome. So now if we consider the downwards direction to be positive, we could write that. And let's call this equation one, we can write that FN the normal force is equal to M multiplied by VT squared divided by R minus M multiplied by G Awesome. So now we must recall and apply the conservation of energy principle. Since we are told to ignore friction, we know that there will be no non conserv forces. Let us use the subscript I to denote the initial quantities which correspond to when the cart is released at the top height. So at the beginning and let us also use the subscript F to denote the final quantities which correspond to when the car is at the top point of the trajectory inside of the loop. Let us also take the zero potential level to be at YF thus, we could write that the initial velocity V is equal to 0 m per second. We also can write that why I the initial distance is equal to three multiplied by H minus two, multiplied by H which is equal to three, multiplied by 2.5. Because that's what we're told that H is equal to 2.5 R. So plugging in are known variables. So 2.5 multiplied by R minus two, multiplied by R which is equal to 5.5 R or 5.5 multiplied by R to be precise. And since YF is considered to be the zero potential level at the value for height, this could be calculated with respect to height. Therefore, we could write that VF the final velocity is equal to VT which is equal to some value that we do not know yet. We haven't solve for this value yet. We can also write that YF the final distance is equal to 0 m because it's not gonna be moving anymore or going any distances once it's at the very end of its trajectory. So if we recall that the initial energy, so let's write this in blue. So this is the conservation of energy. So we know that the initial energy is equal to the final energy. And because we know that we could write that one half multiplied by capital M multiplied by V squared plus capital M multiplied PG multiplied by Y I is equal to one half multiplied by capital M multiplied by VF squared L M multiplied by G multiplied by YF. OK. So now when we plug in our known variables, we will get zero plus M multiplied by G multiplied by 5.5 R. So 5.5 multiplied by R to be precise is equal to one half multiplied by M multiplied by the T. Since we know that VF is equal to VT, so VT squared plus zero. So solving for VT, so isolating and solving for VT all by itself will give us. So using a little bit of algebra, we will get that VT is equal to the square root of 11 multiplied by G multiplied by capital R. And let us call this equation two. So now if we substitute the value of VT which we just found in equation two and plug it into equation one, we will get that the normal force FN is equal to M multiplied by 11 multiplied by G multiplied by R all divided by R minus M multiplied by G. So now if we simplify this equation, we will get that the normal force FN is equal to 11 multiplied by M multiplied by G minus M multiplied by G. And if we simplify one last time, we will find that FN is equal to 10 multiplied by M multiplied by G. And that is our final answer for part one, I should say. So at the top of the loop, so to keep things simple, I'm just gonna denote it as tol. So top of loop. So this is our answer for the top of the loop. So now moving right along here, so fur on the horizontal path. So now we must consider the condition where the car is on the horizontal surface. Let us draw a free body diagram for this case, which will look something like the. So looking at our free body diagram, which looks something like this, we can see to keep things simple. Instead of drawing a cart, we'll just pretend that this gray circle is our cart. And as we could see when it's on our horizontal plane or the horizontal path, we have the normal force denoted it in a blue arrow pointing up. And then we have the weight force pointing down representing represented by a red arrow. And we know that the weight force is the mass of the object multiplied by the acceleration due to gravity. So vector G awesome. So looking at our free body diagram and considering the forces at work here. So since the cart is on a horizontal surface, there will be no circular motion, the only time circular motion is in effect is when we're in the loop itself. So the cart will have no vertical motion at this condition either. Therefore, the net force will be zero. Therefore, we could write that FN the normal force is equal to mass multiplied by gravity. And that is our final answer for the horizontal. And to keep things simple, just put a capital H for horizontal pack. So how we got this answer is if we remember considering our conditions that we can see in our free body diagram, we know that the normal force minus mass multiplied by gravity equals zero because the net force is zero. So rearranging this equation to isolate and solve for our FN, our normal force will give us FN equals mass multiplied by gravity. And that's it we solve for this problem. Hooray. So the correct answer has to be the letter B when we look at our multiple choice answers. So B is at the top of the loop is 10, multiplied by M multiplied by G. And on the horizontal path, it's M multiplied by G and that's it we solve for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

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.
(c) by the track at the top of the loop, and (d) by the track after the block exits the loop onto the flat section.
<IMAGE>

Watch next

Motion Along Curved Paths practice set

Struggling with Physics?

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. (c) by the track at the top of the loop, and (d) by the track after the block exits the loop onto the flat section.<IMAGE>

Watch next

Motion Along Curved Paths practice set

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.
(c) by the track at the top of the loop, and (d) by the track after the block exits the loop onto the flat section.
<IMAGE>