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Denduium consists_ 2.8 kg stone swinging on 4,4 m string negligible mass The stone has speed of 8.2 m{s when passes its lowest point (a) What the speed when the str...

Question

Denduium consists_ 2.8 kg stone swinging on 4,4 m string negligible mass The stone has speed of 8.2 m{s when passes its lowest point (a) What the speed when the string at 61 the vertical? (b) What the greatest angle wlth the vertical that the string will reach during the stone'$ motion? (c) If the potentlal enerey the pendulum-Earth system taker be zerc the stone Olest Doint Uhat the tota mechanic? enerqy system?(a) NumberUnit(b) NumberUnit(c) NumberUnit

Denduium consists_ 2.8 kg stone swinging on 4,4 m string negligible mass The stone has speed of 8.2 m{s when passes its lowest point (a) What the speed when the string at 61 the vertical? (b) What the greatest angle wlth the vertical that the string will reach during the stone'$ motion? (c) If the potentlal enerey the pendulum-Earth system taker be zerc the stone Olest Doint Uhat the tota mechanic? enerqy system? (a) Number Unit (b) Number Unit (c) Number Unit



Answers

A pendulum is made by tying a $1.33-\mathrm{kg}$ stone to a string $3.82 \mathrm{~m}$ long. The stone is projected perpendicular to the string, away from the ground, with the string at an angle of $58.0^{\circ}$ with the vertical. It is observed to have a speed of $8.12 \mathrm{~m} / \mathrm{s}$ when it passes its lowest point. (a) What was the speed of the stone when projected? (b) What is the largest angle with the vertical that the string will reach during the stone's motion? (c) Using the lowest point of the swing as the zero of gravitational potential energy, calculate the total mechanical energy of the system.

Okay, so we're gonna be looking at problem number 20 from chapter eight, and this is a pendulum problem. So we have a pendulum here, and it is attached to a negligible it is. No, it is attached to a stone, and it is going to be swinging. Okay, so there's going to be gravity acting downwards, and it's going to be swinging this way, right? Swinging this way like this is gonna be an arch. Sorry. This arch is really weird. Okay, so this mass here weighs two kilograms, and we know that the string length l designated and l is equal to four meters. Okay? And we know that when the string passes through this point, it has a velocity of eight meters per second, and it's at the bottom here. So this is at a degree of zero degrees. This is at zero degrees. And so a A wants to know What is this speed when the string is at 60 degrees to the vertical? Okay, So less first, find out how much kinetic energy there is when the string passes through zero degrees. Right? And kinetic energy is found by this Formula 1/2. Um, be square. And so if you find the kinetic energy here, that's also that's equal to the mechanical energy off this of this system at 60 degrees. So whatever this value is, you know what? Let's actually just find that two kilograms and then the speed meters per second square. Okay, so it's 64. So 64 jewels, right? And let's use this equation after all, because energy conservation so mechanical energy, mechanical energy, mechanical energy is equal to 64 jewels and it's equal. So mechanical energy is equal to the sum of the potential and kinetic energies. So we have, um, basically mg Alan, this case and then whatever angle this is, they, uh, one minus the coast sign of that and then plus 1/2 em the square. And all you have to do now is basically substitute than the chemical energy, which is equal to 64 jewels into here. So you have 64 jewels there. 64 jewels is equal to the mass of the rock. Uh, which is to kill a grounds two kilograms gravity 9.8 years. Because I can square length. What is it? Four meters times one minus 60 co sign of 60 co sign 60 and then plus 1/2 em the square solve for Why did I want right? Um is to sow substitute in your values, right? And then all you have to do is basically solve for the sulfur. This V So what do you do? Right, you do subtract this to that sign. Whatever this quantity comes out to be subtracted to this site. Sulfur V and that will be your velocity at the bottom. All right, now, let's clear this out. Clear this out. Okay, Now, part B B is asking for what B is asking for. The greatest angle that the vertical Oh, the greatest angle with the vertical that the stream will reach. All right. What does this mean? Okay, well, let me explain. This means, like, basically what is? Well, let me just let me just do it. Four meters. Yeah, is over here. Right. So be We know the energy. Mechanical energy. Mechanical energy is equal to M g l. One minus co signed data. This is the equation we've been using. Right? And then we know that mechanical energy is still equal to 64. It's still equal to 64 jewels. So it's looking for basically what is the angle when it reaches the top right when it reaches the top and when it reaches the top here, we know that velocity is equal to zero because momentarily lost is there on the top. So we have 64 jewels, is equal to everything else, literally stays the same 9.8 meters per second square. Your length doesn't change, because why would it one minus co sign of data and then this is zero. So all you have to do is solve here for co sign. And how do you do that? Okay, you divide this entire thing over here. So the C two times for eight. Father, this is eight. So eight divided by 9.8 is equal to one minus co sign. Fade up so you can do minus 18 times 9.8 minus one is equal to negative coastline. They divide by negative one. Did my button I give one, and all you have to do is our coast on this side on that side, and then you'll get your angle of data. And now the last question is, if the potential energy of the pendulum Earth system has taken to use. There are stones. Lowest point was a total mechanical energy of the system. Well, we already where do you know that? You know, for this entire problem, this has been the lowest points. Been zero. So we know our mechanical energy is 64 jewels, and that's all for this question.

Problem. Seven point seven two. We have a small office mask. Went one to Clarence. It's fastened to a nationalist string with length point eight meters toe for a pencil. And so the pendulum swing so make so it makes a maximum agonal forty five degrees with the vertical. If a resistance this inevitable first were asked was speed Iraq when a string passed through the vertical position So here only to define our potential energy of gravity to be zero at the base, that one. So we have just gravitational potential at the top here, and it's going to turn into all kinetic energy based here. So we have gravitational potential. You go to K two, write down This says point one, this is point two. So then we find out what those are so we have m g times some height. Alright, is why that's going to equal to one half. Uh, mess lost e squared. Oh, media Lee can start on this of what we defined. Our is why in order find water speed is at the bottom. Okay, so we need a little geometry here. So, uh, the pendulum forms a triangle between our length when It's at forty five degree angle. And with the vertical, that's gonna be a right triangle. So I need to find what this is. This is why. So I know that this whole length is L. But in order funding with this, like this awaken into is say, l minus this length is going to be. Why? So we find this length. I know that half of you says l haven't angle. I'm fine with the Jason English. So that's a no cosa forty five the race. So then why equals our full length hell Minus this? Personally, I'll co sign but Greece. Okay, so then I get substitute that info. Why? So I have a g times l I immediately just pull that out. And I have one minus co sign forty five Reese. That's good. Equal to one half once his work. Okay, so then I get V is equal to square root. Love to she Bill one minus coasting. Forty five, crease. Okay, so then all that's remaining here. His numbers to ten point one times zero point eight meters times one nice. Co signed forty five so that I get B is equal to two point one or four meters per second. That's a hard day, Part B. We're fine. Ass defying was attention Spring what makes an angle of forty five years with the vertical? Okay, so for this, we need to set up a free by ever free by diagram. So free by a diagram basically just takes our point. And we're their point here, and we signed all the forces that are on it. So as forty five degree angle, you have our attention. And director straight downward is our best times gravity. So what we do is we need to find, uh, what dimension is this spring there in the string there. So since these air only forces acting on it, well, I'm going, Tio, find what tension is by finding out what's equal opposite tube because we know that the strength is not stretching or they're pulling in mass it all it's forming a circle. So the acceleration is in this direction. That means that the tensions in equal opposite to whatever component of gravity is in that direction. So that is this going to be the same thing before we dine before with you co sign for five degree angle right there. So green. Right? Attention. We can write down our self forces. Him g co sign of the forty five is equal to attention. Okay, then that's going to equal to are zero point eight three humans. Okay, then put forth for point. See, we're the same thing for questions. See, we do same thing at point two. We have our attention and we have gravity. Three downward. We're not accelerating upward on. That means that we attention is equal is equal opposite to the mass times acceleration that gravity. So attention go to them, G. And it's getting equal to zero point two times nine point one one point. Go on, eat and it's going to be a solutions.

The storm that is traveling in a circular part on DA. It's connected through a string on one end of the strain is fixed. So we consider this as our fixed end and the other end is where the stone is attached and that is creating a circular part. Now we can immediately see that intense a circular part, and the stone is following that path. That means there is some acceleration that is directing towards the center. And we can call that a radial exploration which is equal to V squared over are these the velocity and with the stone is spinning and then our is the radius of the cell feel apart. And the only force that we have here is the tensile force during the stream and no other forces there on this direction. Now, if we actually try to, uh, draw the free body diagram the stone. So here we're considering that the stone is on the right side of the circular part. So that means let's say the stone is around here and, ah, if we consider the radial direction as our ex direction, so this will be the ex direction for us, and if we consider the line, or let's say the point that it's coming towards us Assauer positive wide direction and we can mark that like so So this direction right here is coming towards us. So that's the positive direction. That's the convention that we're choosing to choosing for this problem now if we look at the forces that's acting on the stone here. So for this ex direction, we have being, Ah, radial acceleration. And then for the forces, the only force that's acting on this direction on the ex direction is thie tense. I'll force of the attention due to the strength so I can call that tea. And for the right direction, we have the gravity that's acting downwards. So that means the force will be masters gravity and to counterbalance. That will have the, um, equal and opposite normal force, which should be which mean save n right. And this is from Newton's third Law that these two forces should be equal if the system is stable. So by stable, I mean that there's no movement in why direction for that stone, so that means these two forces air balancing each other. So that's why there's no movement. But of course there's movement in extraction that Andi That's why we have to figure out the tensile force that's been acted. Ah, on the stone. So yeah, so that's the everybody diagram that we want and then to solve for the Mac or maximum velocity that the stone can have before it breaks. We need the extraction of force because that's where the 10 side forces, so for extraction will have f X equals M times X s o. That's the radial exploration for us. And from here we can write effects as the tensile force which should be equal to mass times a eggs. And for our case, it is. The X is a radio and that is be spread over our And from there, if we solve for C will get square root off tee times are over m. And now we can put all the values and salt for V. So that's for tension. We have 16 mutants. Then we have for radius on for the length of the string, we have 0.9 meter. Then for the mass, we have a kilogram 0.8 kilograms. Sorry. And ah, from here we can get the velocity as eight find to do meters per second. Now, we actually we should actually cross check whether our units are correct or not. And to do so, we can actually expand this unique Newton on. We know from this force aggression that Newton should be the unit of mass times the unit of acceleration. So that's that should be ah, so for our case, that should be kilograms. We're using the unit s o. That's ridiculous. Grand times. Ah, leader for a second squared and then we have a meter over here, and then we divide that a kilogram, So then we can get rid of the kilograms. And since we have a square root over here, so let's not forget the square root. So we have meet us great over second spread. And if we take the square of that, we get meters per second. So that means our units are consistent. So we should actually do this little calculation on the site. So to check whether the the units they were using is actually consistent with the units that were given. Thank you

And part of this problem, whereas to draw free body diagram of this stone, we know that the radius of the string is zero point nine meters. The mass of the stone is you're appointing eight kilograms and a maximum attention. We could have been a string of sixteen Newton's before it, Ricks. So if you imagine our stone here, we know there's going to be gravity pulling it down of magnitude and times G. Now it doesn't actually move down because there's gonna be a normal force by the table. And then there's also going to be a radio force because of the string. And so I'm assuming that's being world around by, say this point here. And so that's the three body diagram. There's only three forces here, and that completes party for Part B. We need to find the maximum velocity that we can have before the string breaks. And so for this, we're just going to notice that the radio force is equal to the tension. Now we know the radio acceleration. It's given in the book as b squared over R, and we also know that the radio force is equal to mass times the radio acceleration. And so it's equal to the mass times velocity squared over R. And so we can take this expression here and plug it in for the left side. Here we do that. We get that the tension is equal to don't be squared over R. Now we can plug in our max tension to figure out what our Maxwell Long City is. We also have deployed under Massena rays, which are givens in the problem. We knew that we were sixteen unions is equal to zero point eight kilograms. It's on speed squared over zero point nine meters. Now we consult us for B and when we do that we have is eight point two three meters per second and that's the final answer.


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