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A uniform spherical shell of mass M =45 kg and radius R = 8.5 cm with moment of inertia HMR? can rotate about a vertical axis 0n frictionless bearings. massless cor...

Question

A uniform spherical shell of mass M =45 kg and radius R = 8.5 cm with moment of inertia HMR? can rotate about a vertical axis 0n frictionless bearings. massless cord passes around the equator of the shell, over & pulley of rotational moment of inertia [ = 3x 10 kg . m? and radius r = 5 cm; and is attached to a small box of mass m 0.6 kg: There is n0 friction 0n the pulley's axle; the cord does not slip 0n the pulley: Use conservation of energy to answer the following: what is the speed

A uniform spherical shell of mass M =45 kg and radius R = 8.5 cm with moment of inertia HMR? can rotate about a vertical axis 0n frictionless bearings. massless cord passes around the equator of the shell, over & pulley of rotational moment of inertia [ = 3x 10 kg . m? and radius r = 5 cm; and is attached to a small box of mass m 0.6 kg: There is n0 friction 0n the pulley's axle; the cord does not slip 0n the pulley: Use conservation of energy to answer the following: what is the speed of the box when it has fallen 82 cm after being released from rest? (Hint: the speed of the box is the same a8 the speed of a point 01 the equator of the sphere which is v Rw. It is also the same as the speed of a point on the outer radius of the pulley rw) M, R



Answers

A uniform spherical shell of mass $M=4.5 \mathrm{kg}$ and radius
$R=8.5 \mathrm{cm}$ can rotate about a vertical axis on frictionless bearings
(Fig. $10-47 ) .$ A massless cord passes around the equator of the shell,
over a pulley of rotational inertia $I=3.0 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}$ and radius
$r=5.0 \mathrm{cm},$ and is attached to a small object of mass $m=0.60 \mathrm{kg}$ .
There is no friction on the pulley's axle; the cord does not slip on
the pulley. What is the speed of the object when it has fallen 82 $\mathrm{cm}$
after being released from rest? Use energy considerations.

Yeah this problem was the concept of the conservation of mechanical energy. And to solve this problem we need to apply the concept of preparation of energy. We consider this situation as the Jiro potential energy position At this point. The potential energy is zero. So when the mass move a distance H. Brother point we can write from the conservation of energy. The final energy or the national energy is the current to the final image. We can write the initial energias uh as zero. And the final energy is the rotational inertia of the sphere. That is half of the rotational inertia of the shell into omega square of shell plus half of the rotational inertia of the fully into. Oh my God is corrupt the police plus half of the translational kinetic energy. That is half of M. V. Square for dogs minus MG into H. Let's say this as equation one the rotational inertia of the shell equals two upon three M. R. Respect. And for no sleeping condition, omega after bully equals we upon the release of the police and omega shell equals the speed of the uh string upon the radiation of the ship. So from this we can write the equation. One is zero equals half of two up, 1 three M. R. Square and two. We square up on our square plus half of I we square upon our square plus half of M. V square minus MGH. And by rearranging this with we can write these square equals MGH upon mm upon three plus I am born who are squared plus M upon to or the speed vehicle squared off MGH upon mm 13 plus I upon to our square. Let's came upon two.

So in this question, we have given and one equals to 0.4 to Katie and two equals two. 0.85 kg capital M mass of the police is 0.35 kg in a radius of the police are one equals to 0.2 m Outer areas of the police are two equals to 0.3 m coefficient of kinetic friction Muche equals to 0.25 Initially, speed we are equals to zero point 82 m per second and displacement of the hanging block. Delta X equals two 0.7 meats. Okay, so moving to the first part in which we have to calculate the final speed. So using candidate energy conservation, principal or conservation of energy principle, we can right gave one initial plus care to initial plus you one initial plus you do initial plus k. Our initial minus F k multiplied by Delta X equals two gave one final plus Scared to final, we'll escape our final plus you one final plus you to finally Okay, this will be the equation for the energy conservation. Say this is equation number one. So now we will calculate all the values. So initial kinetic energy of the hanging block, given initial will be equals 21 by two and 13 initially square. So putting the lose one by two, multiplied by 0.42 multiplied by 0.82 Jiro, 0.82 squared. This will be equals to 0.141 See, this is equation number two. Okay, so the initial potential energy of the hanging block you one initial will be close to mg and one G multiplied by Delta X. So M one is 0.4 g is 9.8 and the Delta X is 2.7. So this value will be good. Point 88 Jule, this is a question number three. Okay, so next kinetic energy off sliding blow gave it away. Equals 21 by two m two, we initially squared. This will be equals 21 by two, multiplied by 0.85 multiplied by 0.82 square. And this will be equals to zero point 2862 This is the equation number four. Okay, so now the initial potential energy of the sliding block will be equal to zero because it is not moving. And the final potential energy of both blog will be equal to Geo, as we have taken it as a reference. Okay, so this is our equation. Say, for example, five. No. The initial rotational kinetic energy K rotational initial equals 21 by two. I am Eva initial square. And no moment of inertia capital I of the disc will be equal to one by two. Mm. Are one squared plus R two square multiplied by we initially developed by our cool holy square. This is the value of omega. Okay, so after putting values care, rotational initial will be equals 21 by two bracket When by two capital M s 0.35 our initial is 0.2 square plus 0.3 square multiplied by re initial is 0.82 8 to 0 point it simply zero point it. Okay, so 0.82 um uh, so 0.8 It was 0.8 Joe. When Joe it okay? And no, this is zero point it 0.2. There are two is 0.3 0.3 Holy square. So after solving it, we get care. Rotational initial it calls to joke Prank 085 0.85 Mhm. Okay. See, this is equation number six and kinetic Friction Force on the sliding block will be equals. Two f equals to M. U K. And this will be equals two mu k multiplied by M. G. So from here, milky will be equals two mhm f k will be equals to 0.25 multiple herb I m two m two is 0.85 multiplied by 9.8. So this will be equal to 2.83 Newton said. This is a question number seven. Now we will put all these equations from 227 into one. Go get Oh, okay. So now we are putting values. So 0.141 plus 0.286 plus two point double eight plus zero plus 0.85 minus two point g 083 multiplied way 0.7 This will We equals 21 by two multiplied by 0.42 multiplied by three Finally square plus one by two multiplied by zero point 85 We finally square plus one by two. Break it one by two Capital M R one is square plus r two square multiplied by We finally square divided by our two square. Okay, so after solving this, we get one point 96. Yeah, close to geo point 21 We finally squared plus 0.4 to 5. 4 to 5. We finally square plus one by two multiplier way 2.275 Mike played by turned to deport minus food multiplied by we finally squared Divided by r two is 0.3 square. Okay, so after solving this equation, we get 1.936 close to Jill Blink 7613 Final square. So from here we finally will be equals two under root of 1.936 They were by 0.76 Okay, so after solving this, we final will be equals to 1.59 m per second. This is the answer for the first part in which we have to asked to calculate the final speed of the block now moving to the part B in which we have to calculate the angular speed at the same moment and angular is field Omega final will be equals two we finally developing our and we have we finally equals to 1.59 and r two s 0.3 So from here, mega final will be equals two 53.1 radiant per sector. This is our answer for the sake.

So there's a drawing in the book but I'm just going to draw a side view of it. There's a table right here. There's a pulley on the end. There's a mass down here and then there's a mess up here and I think that's all there is to it. Okay, let's go ahead and see what's going on here. There's a sliding block of mass .850. Okay. And this mass down here is .4. And this is a cylinder mess of solid cylinder .350 kilograms and a radius of .0300 meter. Since it's a solid cylinder then moment of inertia for a solid cylinder Is going to be 1/2 M. R squared. All right. Keep reading. There is a coefficient of kinetic friction Of .25. Oops. I don't want to write it there. I'll just right over here. So that makes me think that I should draw a free body diagram, friction, force, tension, normal force. And then I'll draw a free body diagram here. They'll draw a free body diagram here. Mm. All right. There's no friction on the axle of the pulley. So the velocity of the whole system really is 0.8-0 meters per second at the photo gate. And now we want to go a distance of seven tens of a meter away. All right. We're supposed to use energy methods. Oh man. Yeah. Okay, sir. I'm going to say that the potential energy initially is M. Two G. Time's D. So it hasn't traveled that distance D. Yet which would cause M. Two to go downward. D. So it has that potential energy and then it's got kinetic energy which would be M one M M. And M two are both moving at a velocity. That would be M one plus M two moving at V squared Plus we've got 1/2 I omega squared. So that would be the energy initially. In the end. This is going to equal well the potential energy will will go away but this does not go away. Um One plus M two V. I'm gonna call this V two V in the end squared Plus one half. I omega two squared. All right. So, we need the speed in the end. So V equals omega our or rather I'd like to write omega is V overall. So, let's substitute some things into this equation. I want to leave em to G. D alone. We'll leave this next one alone too. I wait a minute. I forgot my one half. One half Mv squared. One half. One half man. U. K. I I is gonna be one half M R squared times. Omega squared. But instead of omega I'm going to write V squared over R squared equals one half M. one plus M two V two squared plus one half. I I hasn't changed one half. M R squared V two squared over R squared. Okay, let's see what we can cancel out here. First of all are squared over. R squared goes away. R squared over R squared goes away. So let's keep going. M. two g. D. Plus. Now I'm going to subtract. No I'm not never mind. We're going to let it go. 1/4 mm V squared 1 4th. MV Squared is going to equal yeah 1/4. M V two squared. Okay so V two. I'm going to finish this here. It's going to be the square root. Oh M two G. D. M. Two. Okay. Yeah plus one half M one plus M two V squared. Wait a minute. I'm gonna factor out the V squared And I'm gonna do that on the other side as well. So it's 1/2 I am one plus M two plus 1/2 of them. Okay Because one half times one half is 1/4 and then there are all multiplied by V squared over one half. Um One plus M two plus one half. And Because again 1/2 times one half is 1/4. All right. I think that's all I can do with this. So let's go ahead and put it in the calculator using dez most graphing calculator so I can declare my variables for one thing and one is point 85 mm is .35 are is plenty or three you is 0 to 5 but I didn't even use mu that's not good. Yeah M two is 0.42. Um V. Is point 82 E. Is 0.7. Oh man mm Man I did not use um you at all and there's definitely going to be some energy loss due to friction but um I'm gonna say work equals the force of friction times the distance we know D. So the force of friction is going to be um you times the normal force And the normal force is going to be M. one G. Times the distance. So um as it moves we're gonna lose energy. So energy initial is going to be greater then the energy final. So I should put on my final side plus W. Because the initial energy, like I said it's going to be greater than the final energy because we're going to lose energy. All right, initial energy greater than my final energy. Just thinking this through. Yeah because we lose energy. So I need to add some to the final energy. All right so let's adjust this a little. So at the end of this I need to have plus double you still at the end and then here M. G. D. Plus this. Plus this was supposed to this I still need to have a plus W. In the end. All right now mm Now I need to put a minus double you right here subtracting W on both sides will give me that and now it should work. So um let's go ahead and do this. I'm going to calculate W. First is new. M one G. D. 1.46 jules. Okay, now putting this into the calculator, V two is the square root. Oh M two GD plus one half. I'm one What them two plus 1/2 of em aims V squared. Right? It's double you over one half. I'm one plus M two. Just one half and that gives me 1.63 meters per second. Oh, I hope this is right. Yes, it is. Woo. Okay. Be with the angular speed. That is easy. Now, omega is V over R so V over our the two overall, I mean, Which is going to be 54.2 radiance per second. Thank you for watching.

In this exercise, we have the system that shown here in the figure it consists of a block that's lining on a surface, and the water has a mass and one of 0.85 kg that is connected by a string, ah to another block that has a mass in two equal to 0.42 kg. And this block, this counterweight hangs from the from the string, and the string passes through a pulley that has a mass of 0.35 kg and has us a shape of a hollow cylinder within a radius of 0.2 m and an outer radius of 0.3 m. The block and one is initially sliding with a speed V zero of 0.82 m per second and the proficient friction. The friction coefficient between the surface and the block, and one is your 10.25 and we want to know using conservation and methods in question. A what is the Speed V of the block after in ah, it travels a distance 0.7 m, So in order to use conservation, energy conservation of energy, we need to remember that the initial potential energy you I plus the initial kinetic energy ke I is equal to the final Put the show energy You f was the final puttin show Connecticut GF plus the work done by the friction on the block and one uh, no notice that initially I'm going to choose the zero the potential energy for block and 1 July exactly at the center of massive lock in one. So it doesn't the petition idiot doesn't change, but the potential energy for block and two changes And I'm gonna choose the zero the potential energy to lie exactly at the initial ah point that of the center of massive block to So the initial potential energy is just zero. The initial Connecticut energy is equal to one half of m one times V zero squared, plus one half of them too. Times e zero square noticed that they are connected by a string. So, uh, which ever speed with them have the other has as well plus one half of I times Omega square. I is the moment of inertia off the buoy and this is equal to the final. Ah, potential energy noticed that m one traveled 0.7 m. So this means that M two descended 0.7 m. So the variation in potential energy is minus m two g times d. The final kinetic energy is one half of him one V squared, plus one half of them to be squared. Plus, I Omega squared over to actually in the left hand side. I should have written these omega zero square plus the work. And the work is just the normal force times the the distance traveled and the normal forces m one g Ah, and naturally, it's normal force times mute. So it's a no normal force, which is m one g times, mu times the This is the total work done. And now I'm gonna put this equation on hold in order to calculate the moment of inertia I off the of the pulley. The moment of inertia is equal to the massive apuuli times the inner radius squared plus the outer radius squared, divided by two. Okay, so keep this in mind, I'm gonna try to organize the terms a little bit, so have also, in order to organize the terms, I'm gonna need to remember that easy uh, Omega zero is people too. V zero, divided by r two and omega is V divided by r two. So we have one half of m one plus m two times v zero square, plus one half off I. The zero squared divided by r two square is equal to minus m two. I'm gonna write it like this. G de m one mu mine is in to plus the square times over two times m one plus in two. Plus I the square over to our two squared So you have V zero square times one half of m one plus m two plus one half of I divided by r two squared is equal to g d. I am one Mu min is in to plus the square times one half of m one plus m two plus I over to are two squared so v zero squared. Actually, I'm gonna just gonna go ahead and divide both sides by ah, this term here. So everything gets a little simpler. Have zero square is equal to G d. M. One mu minus m two divided by one half of m one plus in two plus I over to our two squared plus V square. So we're left with V is equal to the square roots of V zero square plus G d times in two miners in one mu divided by one half of em. One was in two. Thus I over to our two squared and I can soups into the numbers we have. So V is equal to the square root of 0.82 squared meters per second square plus G, which isn't went 8 m per second square times d is your 0.7 m times m two. She's zero point 42 minus and one that 0.85. Actually, I should have written that kilogram here. Just get it drank. So have kilogram. Mine is your 0.85 kg times zero points, 25 divided by one half of 0.45 was 0.85 kg plus hi and I, as we saw before, ISS em capital limb, which is just 0.35. So it's, uh is your 0.35 times divided by two and then and much applied by 0.2 You should have your program have your winter 2 m square. Let me just give it a little tighter 0.2 m square plus 0.3 m square. So we actually this is I. So we should still divided by two times are it's usually two times is there 103 m swear And from this we get a V is equal to one point 59 meters. We're second in question be. We have to find what is the angular speed of the Pooley And the angular speed is just the leaner speak V divided by the outer radius are too. So this is 1.59 m per second, divided by 0.3 meters which is equal to 53 radiance per second and this concludes our exercise.


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