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Puck of mass m 1.40 kg slides in circle of radius cm on frictionless table while attached to hanging cylinder of mass M 3.70 kg by cora through hole in the table_ W...

Question

Puck of mass m 1.40 kg slides in circle of radius cm on frictionless table while attached to hanging cylinder of mass M 3.70 kg by cora through hole in the table_ What speed keeps the cylinder at rest?NumberUnitsthe tolerance is +/-59f

puck of mass m 1.40 kg slides in circle of radius cm on frictionless table while attached to hanging cylinder of mass M 3.70 kg by cora through hole in the table_ What speed keeps the cylinder at rest? Number Units the tolerance is +/-59f



Answers

A puck of mass $m=1.50 \mathrm{kg}$ slides in a circle of radius $r=20.0 \mathrm{cm}$ on a frictionless table while attached to a hanging cylinder of mass $M=2.50 \mathrm{kg}$ by means of a cord that extends through a hole in the table (Fig. $6-43 ) .$ What speed keeps the cylinder at rest?

For this problem. On the topic of force and motion. As we can see the figure, we have a pack of mass 1.5 kg sliding in a circle of radius 20 cm on a frictionless table Is attached to a hanging cylinder which has a mass of 2.5 kgs. Through a cord that goes to the hole in the table. We want to know the speed of the park that will keep the cylinder at rest. Now for the cylinder to remain address the magnitude of the tension force T. Of the card must equally gravitational force MG on the cylinder. The tension force supplies the centripetal force that keeps the puck in its circular orbit. So the tension force T. Is M. V. Squared over our and therefore the weight of the cylinder MG must equal um V squared over R. Where the game is the massive cylinder and it'll end the mass of the park. We can then rearrange this equation and solve for the required speed. So we is equal to the square root of big. M. Times G times are over a little M. All of these values are known, so we can substitute them in This is the square root of the mass of the cylinder, 2.5 Kg, and the acceleration due to gravity 9.8 m/km2 Times The radius of the park's orbit, 0.2 meters Divided by the mass of the park, 1.5 Kg. And so calculating, we get the speed of the park that is required to keep the cylinder at rest Is 1.81 m the second.

Here we have a hanging mass of capital M. And it is hanging here to support the circular motion of this little mass that is spinning around up top. This is a frictionless table so the only forces acting on this are going to be the centripetal force inward caused by this hanging mass. We want to know the speed that this thing needs to be moving at in order to keep this in equilibrium to keep this thing suspended. And so the the net force is going to be the centripetal acceleration acting on this mass. And that has to equal the weight force of this mess because the only thing um the only force acting and this is the tension force. So this tension force has to equal the tension force up here which is going to balance with the net force acting on M. A. Subsidy. So the this force is equal to the tension force, the weight force down here. Capital M. Will also have to equal the tension force energy, attention force in the opposite direction. And so we can set those equal to each other so little then hey sexy. It's equal to capital M. G. The weight force on this mass is equal to the centripetal acceleration. Uh, the centripetal force on this little mass. Well centripetal acceleration can be written as v squared over R where r is the radius of this circle and the is what we are looking for. So if we plug this in here, we're gonna get em V squared over. R tickled to capital M cheap will multiply through by our and then divide by M little M. And these will cancel this cancel. And we got B squared is equal to mgr over em. And then if we square root both sides, we're going to get an answer be equal to the square root of capital. M G r All over little mm. And then once we know some numbers, we can throw those values in here. We can figure out the actual speed

We're gonna be looking at problems 39 of chapter five. The Physics fifth. The question says the disk of mass M. On a frictionless table is attached to a heavy sand of capital M By a cord through a hole in the table. D Figure 5.42 on the speed with which the disc was moved in a circle of radius after the senator stay at rest. So asked for us to stay at rest. The tension above it should equal the mass of the cylinder multiplied by G. And the question that uses capital M. For the mass of the cylinder. Um and if it's OK. Sam and so now let me know that the forces on the disk we have um MG going downwards, we have the normal force coming out the board. We have friction small left going in the opposite direction to the tension which is defined by the center of two acceleration. So the tension is equal to the friction was equal to M. V squared of our therefore we can say that the mass of the cylinder is equal to the mass of the disk house I. V squared over. R. So therefore we squared is equal to therefore we squared is equal to um mm The mass of the cylinder G. R. Over the mass of the disk. And so the velocity regularly square initiative that that is our final answer

In this exercise, we're going to talk about the circular motion. So let's remember that when a body is under circular motion, we're basically saying that the rest of force on this body is a radio force that is going to be equal to its mess time, its velocity squared, divided by our Okay, this is Newton's second law for a circular motion. Okay, so in this exercise, we're going to work with the following set up. So we have a puck that has a mass m that I'll come to right here in green, and we know the mass of the puck to be, which is 0.2 kg. So we have this 0.2 kg park that is rotating in a circular motion in a freak frictionless horizontal table, and it's rotating in a orbit off radius. The off 0.5 m. The king and, uh, we know that the bug is connected by a string and that connected by the string. There is also a mass mm of 1.2 kg hanging below the table. And in this exercise, we want to find what must be the velocity off the park of rotation off the park such that the mass, um, remains to hanging at rest. Okay, So to find this, we can first try to draw the fourth diagram for our system off book and mess. So first, let's look at the puck. We know that since the puck is moving in a circular motion, the resident force off the puck should be pointing towards the center off off the orbit. And should be eagle toe MV squared, divided by R. Okay, so let me write it here. F R. So we have this force. Okay. Now, if we look at the mass capital M, we have a gravitational force pushing the mess downwards. So the weight and we're going to have attention on the wire t uh, pushing the mass upwards, competing against gravity. And we want to find the limit on which D is going to be equal should the gravitational force. Because in this case, we're basically saying that the mass is hanging still. Okay, now we can see that attention here is going to be equal to the to the radio force off the off the park. Okay, so we have basically that this is going to be equal to f R. So if we, uh, just rewriting, we have that. Sorry. Uh, the mass. The weight off the mass capital M is going to be equal to the priest centripetal force off the book. So MV squared, divided by r And since we want to find V, we can just isolated and we're going to have that V is the square root off mgr divided by the mess of the book. So we just be stood the values we're going to have that V is the square root off 1.2 kg times 9.81 m per second, square time, 0.5 m divided by the mass off the park. And this is going to be equal to five point 42 meters per second.


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