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A stone with a mass of 0.80 $\mathrm{kg}$ is attached to one end of a string 0.90 $\mathrm{m}$ long. The string will break if its tension exceeds 60.0 $\mathrm{N} .$ The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Make a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without breaking the string.

center of the circular path

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

Hope College

University of Sheffield

University of Winnipeg

Lectures

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In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

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In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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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

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