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The horizontal rod $O A$ rotates about a vertical shaft according to the relation $u=10 t,$ where $u$ and $t$ are expressed in $\operatorname{rad} / \mathrm{s}$ and seconds, respectively. A 250 -g collar $B$ is held by a cord with a breaking strength of 18 N. Neglecting friction, determine, immediately after the cord breaks, $(a)$ the relative acceleration of the collar with respect to the rod, $(b)$ the magnitude of the horizontal force exerted on the collar by the rod.

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Physics 101 Mechanics

Chapter 12

Kinetics of Particles: Newton’s Second Law

Newton's Laws of Motion

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

McMaster University

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.

09:37

Isaac Newton (4 January 1643 – 31 March 1727) was an English mathematician, physicist, astronomer, theologian, and author (described in his own day as a "natural philosopher") who is widely recognised as one of the most influential scientists of all time and a key figure in the scientific revolution. His book Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy"), first published in 1687, laid the foundations of classical mechanics. Newton also made seminal contributions to optics, and he shares credit with Gottfried Wilhelm Leibniz for developing the infinitesimal calculus.

06:20

A 1 -kg collar can slide o…

02:53

A 1-kg collar can slide on…

14:49

A small 250 -g collar $C$ …

05:48

Rod $O A$ rotates counterc…

01:46

The $10-lb$ -uniform rod $…

So for this problem, the first thing to note is that the force acting on the collar due to the spring is equal to the first constant off the spring K into our minus the position of a r A. Now for pot eight, we have that the teeter component of the force is equal to zero and at a the radio component off the force. So this force always only has radio component, but a this radio component is minus if due to the spring. And this force is equal to zero as this spring as the colors held constant. So if the resultant force on the spring is zero, we have that the acceleration in either component is about zero. So the radio component off the acceleration off the color at a zero and he Peter component off the acceleration off. The color is also zero. So there is no acceleration off the color at this point. So that's our first answer onto for part a part B. We use the fact that he some off the radio component of all the forces must equal to I am a are where a a R is the radio component off the acceleration. So what this means is that he force, due to the spring, is equal to M times our double dot. So this is a minus are peter dot squared. So am into D to r D T squared minus are de deter t t squared. And we note that the acceleration off the color relative to the rod is simply due to our GT Square, the second derivative of the position or the first derivative off the acceleration. And so at a we have that zero is equal to am into the acceleration off the color relative to the rod minus are detailed squared, which is 150 millimeters times. 16 radiance. The second squid and hence we get the acceleration off. The color relative to the rod is equal to 38,000 400 millimeters per square second, and so in. Yes, I units. This is 38 0.4 meters per square second. So that's the acceleration off the collar turn to the rod and eight. So, finally we need to solve Potsie, and we want to find the transfers component or the teeter component of the velocity of the color at point B. So after the court is cut, the only horizontal force acting on the color is due to the spring, and thus angular momentum about the shaft is convert and conserved. So using the conservation of angular momentum we get the following are a times the mass of the collar m times the transfers component Offense Velocity V A theater is equal to our position. B 10 a.m. times It's trans verse component of velocity. We beat Peter Mhm and remember that the Peter component off the A is equal to are a peter dot not the initial angular speed data dot and so, from here we can see that weakens, rearranges and find. He translates component off the color at point B, and this is rearrange the equation. 150 millimeters. Over 450 millimeters are a over RB into 100 50 millimeters times detail dot, which is 16 radiance for second kitted up. Not, and hence we get the transfers component off the collar at point B to be 800 millimeters the second, and this translates to 0.8 meters. The second

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