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A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass $m_{1}$ attached to the center of the rod and a particle of mass $m_{2}$ attached to the outer end of the rod. The inner section of the rod sustains a tension that is three times as great as the tension that the outer section sustains. Find the ratio $m_{1} / m_{2}$ .

$\frac{m_{1}}{m_{2}}=4$

Physics 101 Mechanics

Chapter 5

Dynamics of Uniform Circular Motion

Newton's Laws of Motion

Applying Newton's Laws

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Lectures

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.

03:30

A rigid massless rod is ro…

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A rigid, massless rod has …

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A rigid rod of mass $m_{3}…

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The particle of mass $m$ i…

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A rod of length $\ell$ and…

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and this problem, we have to find the relationship. Looking to masses on a rotating staff relationship between M one and M two were also given a relationship between the two tensions acting on the two masses. So using those tensions, let's set up a free body diagram. And instead of some forces for each particle, let's look at the first particle and one first. As we can see from our free body diagram, there's two forces acting on it, t one pointing in one direction which will take me the positive and tea too. Those forces together must be equal to the centripetal force. I'm one Omega squared R Um, Since the forces the bar is rotating the net force must be what is causing its motion, so it must be equal to the Temple Force. Now, T one is equal to three t too, which means that this side and total is equal to 2 to 2 em. One omega squared over R. So, since we're specifically looking for contributing the masses, we can rewrite this so that we can say that M one is equal two, two times, T too. Over Omega squared four. All right now Let's look at our second particle and two, as we can see from our free body diagram here, there's only one force acting on it, causing with triple motion. It's tea too. So here we have t two equals m. I'm to Omega squared. And this case we're working with two are since we're measuring from the distance to the other end of the bar and this bar is located at the right end. So here we can say that I am too is equal two, two times or excuse me, it's tea, too. Over to Omega Squared R All right now, let's really are two masses like the problem asked us to You want a sulfur and one over m two that's gonna give us too tea, too. Over Omega squared are over T to over to omega squared. Daughter. When we solve this out, all are constants are going to cancel our T twos or are in our omega squared. So let me solve this. We're gonna get that M one over Mt was equal to two times 2/1, which is of course, equal to four. So that's going to be the relationship between our two masses

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