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A stone is tied to a string (length 1.10 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 15.0% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

8.48 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 5

Dynamics of Uniform Circular Motion

Newton's Laws of Motion

Applying Newton's Laws

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03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

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In this problem, we have to find the velocity of Iraq being world around a circle in two different scenarios. First scenario has the circle that it's being world in horizontal parallel to the ground and the second scenario is critical. So let's start by drawing free body diagrams for our two scenarios. In our first case, in which we have a ball here attached by some spring string parallel to the ground, there are two forces acting on a bulb. First there is the four sporting downward f g. And then there's the force pointing towards the center of the circle detention. In this case, there's only one force that is pointing towards the centre of our circular motion and therefore that must be the only force responsible for us centripetal force. Okay, that means that we can write that t is equal to M V squared over all our equation for the centripetal force after you doesn't contribute because it's pointing straight downwards. Now our second scenario is a little more complicated. We're gonna look at two specific points 20.1 the bottom of the circle and to the top of a vertical circle now at 0.1 right. We have a force pointing downward F g and a force pointing upward towards the center of the circle T. So that means that in this case it's the chip. It'll force must be equal to the difference between these two. It's t minus FT. The second point. The top of the circle are two. Forces are pointing in the same direction. We have t toward the center of a circle and F g as well downward toward the ground. So f c it goes t plus after. Now this problem tells us that the, uh the velocity rather the tension that the um, vertical ball experiences at its maximum tension is 15% larger than the tension anywhere for the horizontal circle. So where's the tension in scenario to a maximal Well, based on the two scenarios that we have here, if we solve working in the first case, we have that tea equals FT plus F. C. In the second case, we have that it is equal to a F C minus after you. Therefore, we know that case one must be our maximum scenario, and that allows us to now set up inequality between the two tensions that were given based on the information that we have. So he said that for the horizontal case, T must be equal to M. V squared over R everywhere. In the second case, our maximum tension is mg plus and B squared over r. And the first quantity for the horizontal case is multiplied by 1.15 for the 15% greater all market. This is horizontal and this is a critical case. So now we can begin solving for a problem by simplifying and then play in the quantities that we know. All right, so we can cross at our EMS to get that 1.15 b squared over r equals G plus v squared over R. Now we want to solve for G in this case, and when we simplify, we find that 1.15 these squared over R minus v squared over R must be equal to G, which means that 0.15 b squared over r equals G. And now we solve. How about our final equation for B G times are over point 15 Now G is, of course, 9.8 meters per second squared That's just the acceleration of gravity. And our radius, which were given in the problem description, is 1.1 years and that is gonna be divided by 0.15 constant, which you're going to give us. That V is equal to 8.48 meters per second, which is going to be our final answer to this problem.

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