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16. A cart of mass m is released from a height h above the bottom of a loop of radius R such that it just makes the loop: After completing the loop the cart launche...

Question

16. A cart of mass m is released from a height h above the bottom of a loop of radius R such that it just makes the loop: After completing the loop the cart launches off the right end of the track, which is a distance H above the ground. The track is frictionless Express your answers algebraically below in terms of the given quantities m, R, H, and g, as needed. (a) [6 pts:] Draw the free body diagram (FBD) for the block at the top if it just makes the loop-the-loop: Use the FBD and knowledge of

16. A cart of mass m is released from a height h above the bottom of a loop of radius R such that it just makes the loop: After completing the loop the cart launches off the right end of the track, which is a distance H above the ground. The track is frictionless Express your answers algebraically below in terms of the given quantities m, R, H, and g, as needed. (a) [6 pts:] Draw the free body diagram (FBD) for the block at the top if it just makes the loop-the-loop: Use the FBD and knowledge of circular motion to determine the speed at the top of the loop. Hint: just make the loop means that the normal force instantaneously vanishes at the very top. (b) [6 pts:] Find the minimum height h to just make the loop-the-loop. Show your work (c) [6 pts ] What is the speed of the cart on the right side of the track after completing the full loop; but just prior to launching off of the edge? (d) [7 pts ] Find the distance d where the cart lands, measured relative to the bottom of the ledge as shown (e) Bonus: [5 pts:] Sketch a velocity versus time graph, with a sep- arate labeled curve for the horizontal and vertical velocity compo- H nents for the motion of the cart just after leaving the end of the track to just before hitting the ground.



Answers

Figure $9-44$ shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass
$m_{1}=0.600 \mathrm{kg}$ and its center is initially at $x y$ coordinates $(-0.500$
$\mathrm{m}, 0 \mathrm{m} ) ;$ the block has mass $m_{2}=0.400 \mathrm{kg},$ and its center is initially at
$x y$ coordinates $(0,-0.100 \mathrm{m}) .$ The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move
until the cart hits the pulley. The friction between the cart and the air
track and between the pulley and its axle is negligible. In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function
of time $t ?$ (c) Sketch the path taken by the com. (d) If the path is
curved, determine whether it bulges upward to the right or downward
to the left, and if it is straight, find the angle between it and the $x$ axis.

For this problem. On the topic of angular momentum. We are showing a ballistics card on an incline in the figure. An incline makes an angle theater with the horizontal. The cart has a mass big M. And a moment of inertia for each of the two wheels is little M. R. Squared over two. We went to first find the acceleration of the cut along the incline, then the amount delta X. By which the ball overshoots the card and finally the distance deed at the ball travels along the incline. Now we'll consider motion starting from rest over distance X. Along the incline and by the conservation of energy we have the translational kinetic energy plus the rotational kinetic energy plus the gravitational potential energy. Initially less work than against resistant forces. Down to E. Must equal the final translation of kinetic energy plus the final rotational kinetic energy plus the final potential energy. And so we know initially the cart has gravitational potential energy which is M. G. X. Sign theater. There's no work then Against Resistant Forces or Delta E. zero. And finally it has translational kinetic energy. A half I am V squared plus rotational kinetic energy Which is two times seven M. Little M R squared times omega squared which is the over our squared. And so from here we can see that two I am G X. Sine Theta is equal to big M plus two little M. Multiply by V squared. Now we know that the acceleration is constant. So we can use the equation re squared is equal to the I squared plus to A X. And this is equal to zero blessed to A X. Since the cards down from start and rest. And so we can see from our equation above that to M. G. X. Sign theater is equal to um Plus two Little M. Times to A X. Which means by rearranging we can find the acceleration of the card. A. And it's big M. Times G times Sine theta divided by Big M. Blessed two Times Little M. As required. We'll do part C. Next. And in part C we want to show the distance that the ball travels along the incline. Now suppose the boil is fired from a cart at rest, It moves with acceleration. G. Sign theater which is A. X. Down the incline and minus G. Co sign theater, which is a Y perpendicular to the incline. So for its range along the lamp, along the lamp, we have why am minus why I. Is the Y. I times t -1g. Co signed data, T squared. And so if we rearrange this equation, we get the time T and T. is equal to two V. Y. I. Since the y displacement is zero over G times the co sign of theater. Similarly along the X direction x minus X. I. Is the X. I plus a half A. X. Times T squared. So the X direction is along the incline. So this distance is D. And this is zero plus a half A. X. Is G. Sine theta times T squared and T. We get from above. This is for the why I squared over G squared Coulson squared data. So simplifying, we get this distance that the bowl or that the this is that the ball measured along the incline is equal to two V. Y. I squared sign theater divided by G. Call science squared data as required. Next, we'll do part B of the problem. And here we want to find the amount delta X. That the ball will overshoot the cut. So in the same time the cart moves a distance x minus X. I. Which is the exciting times, T plus a half A X T squared. And so the distance of the cart moves will call it D C zero plus half into G signed data, times M over big M. Blessed to times little M times T squared which is for V Y I squared over G squared. Cosine squared of theater. So the distance that the cart moves simplifies to V. Y. I squared. Sign data times big. M over G. Into the M Plus two times little M times Call science squared theater. So the ball overshoots the cart by a distance delta X. And data X is equal to d minus D C. Which is to be Y I squared scientific to divided by G cole sine squared theta minus two V Y I squared scientific data times M over G. Call science squared data into em Plus two times little m. And so this becomes two V Y I squared sign of data times M plus four V Y I squared. Sign off data times little M minus to be Y I squared sine of data times big M. All divided by G Cool science squared theater into big M plus two little M. And so we get delta X to be for times little M times V Y I squared sign of theater divided by big M plus two times little M into G Cool science squared theater.

Hi, everyone. This is the problem based on well, a stick part. Here it is, given our card. Last of all off. More thing. So here and is the mask off the board Capital Emmys Mosque off Sakhar having a moment off, Energia Twice off. Moment off in reserve the veal. That is a Marie script. In the first part, we have to fire the acceleration off the card in the horizontal direction. Toby M plus to him G sign theater. Toby proved in second part. Show that yeah, over suit dot card. And but who am upon am plus to him in tow? Sign off. Theater upon causes square tittle, Be right upon the here. Be by Is the velocity or court initial velocity off? All in the vertical direction. See part show that the distance traveled along the climb. Mm. By the bar is mhm toe Levite respect by a jeep. Sign off theater upon courses square the duties that distance a longer inclined plane. Let us here. Okay. All right. Yeah. Considered This is Toby XX is And this is Toby by axis. Yeah, first part. Considered Muslims starting from rest for dispense X over dispense X Hello and climate A plane contribution off energy. Right. Translate. Three kind of technology. Oh, vocational kind of technology. Gravitational potential. Energy initiative. Changing energy. Kate. Prostrated care reticent? No. Finally, Jiro plus zero m g x sent it. Er plus you half MV squared bliss there a modest script more so he can write to N g X. Sign up data. Yeah. Yeah, I am. Presto him into the script. Here. I have to make a correction. This is this morning. Small damage mass of the weed, the people right. Acceleration is constant, so we can apply the square. It's goingto be a square has to a x initial velocity zero. This is too. I am here. I have to make a correction. This is capital, um, to m g X Sign off theater upon M plus to wear Jiro plus tau X So acceleration you will get. And I'm mg Sign off Phaedra. And plus to him, she's the answer off a part. Mhm. Right. First people psalter C part. Let the ball is fired from a card and addressed it Moves build acceleration. Geese are not eaten down to the inclined plane. Okay? And minus G costs opted out perpendicular and client played for the reach along the ramp. You may also called Dreamed wise cult. Oh, being white, I not take plus half cheap because off the time to do you script that is zero. So these curto toe by upon G corset hands of distance Privett X minus Excise Pepto e x I t yeah. In tow to be by cheap, caustic chi squared on solving it d You will get to be white, I exclaimed, Sign off theater upon g causes square theater for see Pardon. Sorry. Be part X minus X I was scared. Toe v x, I t X t square. This is far. Motion off the card. Uh huh. D card. You will get zero plus. How? Jeez, I knocked it up in tow. AM upon I am plus to him into toe be by eye upon g corsetry square. So on solving it it is to be square. Sign up data upon uh huh de Paris. I am blessed to him. Question Square theater. So the wall bill over suit by that distance, do you minus dizzy? Oh, substituting the well unsolved ing it you will get for em diverted by. Right? That's so thanks for watching it

So we've given. These are two scenarios and for both off those two scenarios, we know that the initial velocity is a seven meter per second and here we know that this Ah, on the first ram, the final velocity would be zero. So for the first trap for part A, we can simply say that kinetic energy initial plus potential energy in the show his equal to kinetic energy final plus potential energy final. And the initial kinetic energy is 1/2 M times seven squared. The initial potential energy is Europe. The final kinetic energy is zero, and the final potential energy is MG times each. We can cross out the two m's in there and substitute jeez equal to 9.8 salt, for each comes out to be equal to 2.5 meter. So heart A was pretty straightforward. Pretty simple. Now let's go to part B in part me, we have to actually break this problem down into two parts. So first we have to find what is the speed at the end off each one. So that's the first part that we're going to solve. We do not know the speed at the end of the first part and in order to find that we are just going to apply the law of conservation of Energy kinetic energy initial plus potential energy initial is equal to kinetic energy final plus potential energy final. The initial kinetic energy is 1/2 m times seven squared. We can take the potential energy at the bottom to be called to zero. The final kinetic energy is 1/2 m times the unknown velocity and the potential energy is mg times H one. So we're told that each one is equal to 1.25 meters. So solving this for V gives me these equal to 4.95 meter per second. Now we start with this and what we realize is that beyond this point, the motion off the object is going to be projectile. And since the motion off, the object would be projectile, What I can do with this velocity is break it into two components. There's going to be a vertical component on the horizontal component. This would be Vico sign of Thera and at the highest point, the velocity would be vey course Sign off. Vera and Vera were told is 50 degrees. So with that, we can go back and again say a kinetic energy initial, plus potential energy initial is equal to kinetic energy final plus potential energy final. Now the initial kinetic energy in this case would be 1/2 M times 4.95 squared, and we can take the potential energy to be zero. The kinetic energy at the highest point would be 1/2 em times 4.95 course sign off 50 degrees squared plus M g. H. Two. We could cross out all the EMS and then we can grab a calculator Substitute Jeez, equal to 9.8 salt for H two h Do comes out to the 0.73 meters and therefore the total height Each one plus H two would be 1.25 plus 0.73 which is 1.98 meters.

Your investigation. In Part eight, we have contravention of momentum for the totally in monastic policemen. Under vision of momentum gives us empathy in diarrhea is the worst M three plus and C indu beer. Now we can put the value here 60 kilogram multiply with We're just working with your per second typical toe 1 20 kilograms, plus 60 kilograms in two years. From there, we can find a below FIA that once 13 tree weapon. Second, I get no in part B here to obtain the Fourth of friction. The first Consider Newton's second Law in the Y direction from Mission Apple by from mutual effort by physical 20 Which gives us that, and it's equal to minus 60 and were deployed and minus 60 point online. A 60.0 kilogram multiply with 9.8 meter particles that is equals to zero or we know that and musicals to 5 88 Newton and the Water Friction Event Order. Friction is here. Ask a. The mission off asking is because to Mucha in tow n now, when barrel put the value, the Rock one for the road zero multiplied with 5 88 Newton we will get to 35 Newton, the values And here now from here Afghan Afghan go through minus 2 45 And I guess now in bars the change in the person's momentum equals with the impulse or B I plus I here plus I is equals through pdf. Now, from here on, the I plus f will be Is it quick stroll and me plus at lofty is going through and be off here We can put w angry. I am miss 60 60 kilogram multiply with four meter for sticking minus 2 35 Newton into T that is equals through 60 kilogram multiply With 1.15 Mitchell was taken here we get the value of P, which is the euro 0.680 seconds now in part B. Yes, don't change in a moment in which the person is, um, we have that is minus FBI final velocity minus initial velocity when we will put the value 60 kilogram multiply with 1.33 miners for I kept meter for second. That is final minus initial velocity that is a close to minus 1 60 Newton Second, I guess the change in momentum off. The card is when people change them Went. Um here we will get 1 20 kilogram. Multiply with 1 20 multiply 1.33 meters per second. Major particular minus zero Here. That physical through plus 1 16 Newton. It was taken. I get now in body E. We have access. Is it minus excited? If it was through one by two. We I i plus b s motive live in HD because it was 21 by to the I is for plaster. Yes is 1.33 metre per second multiplied the TV which is 01680 seconds. Here we get the value 1.81 meter excess minus excite No in part FB have here exit minus excite Xs minus x. I is a close to one by two, we have lost three F and 30 that is the bulls won by through the zero plus 1.33 Mr Paul second in 201680 sections. That is equals True. The rope want 454 454 meter. Now, in Bardi, we have half off and and left with awful MBf squid minus half off and b I squared. It's equal stool. Then people put value half or on the 60 kilogram multiply. 1133 meter per second squared minus powerful 60 kilogram masses. Same text. Regular grounds. Multiply with here MBf and B I is I am the ah 60 kilogram multiply with four meter per second squared here. Regard. Have a look that is minus 4 27 rules. No, in part that we have powerful M V s quit minus Prospal's MB Ice quit is equal. Struth, one by two months is 1 20 kilogram multiply with 1.33 meter partaking for minus zero that is 107


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