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A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the track in 360 s. What is the magnitude of the centripetal acceleration of the car?

0.79$m / s^{2}$

Physics 101 Mechanics

Chapter 5

Dynamics of Uniform Circular Motion

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

McMaster University

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all rights and this problem were asked to find the angular acceleration of a car as it goes around the track. And the first piece of information that we know is that the radius of that track is 2.6 kilometers. Now. The first thing that we want to go ahead and do is convert our radius in two meters. And that's because whatever we're working with units off. Whenever we're working with angular acceleration and angular velocity, we're gonna be working in units of meters per second. We want to make sure that we're always checking on units that we get the right answer. Now, the second piece of information that we know that the car goes around the track in 360 seconds and we have a name for the quantity that describes how long it takes one object to complete one full rotation on, and that is the period that is given by this letter Capital T. All right, so the equation for angular acceleration answer that we know is these squared over are. And if you don't remember, that comes from the fact that force is equal to mass times acceleration and the centripetal force equals M B squared over R. So therefore, centripetal acceleration must equal. He's weird. Where are you? Okay, so that means that we have to solve for B in terms of our given quantities or NT. Now we know another equation for tea, which is to pi over omega, which is our angular velocity. So we can also rewrite Omega in terms of regular velocity. And that's given by Omega Is V over our. So let's plug that back into our equation for the period. And we're going to get that t is equal to two pi r over me. So from there we can solve for V. And then all that we have left to do is plug it into our equation for Alfa. So now we can say that the Spiegel Q two pi R already All right, now let's plug that into our equation for Alfa. Great. So we can say that Elsa equals two squared pi squared r squared over t squared and all of that divided by our I'm actually gonna switch to a new screen was getting a little bit crowded here. So that is then going to be equal to four pies four pi squared r squared over t squared and all that for our now r r squared And this order in here I'm gonna cancel out and we're gonna be left with four pi squared r over t squared. And so now all that's left to do is plug in our original numbers And so we're gonna get four pi squared times 2000 600 meters all over 360 seconds. Squint! And when we plugged it into a calculator, we're gonna get that's equal to 0.79 Q. And the last thing that beginning to check is our units. So we can see from this. There are units. We're gonna be equal two meters over second square, and that checks out because those are, in fact, the units of angular acceleration.

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