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Ch 04: Kinematics in Two Dimensions

Chapter 4, Problem 8

A toy train rolls around a horizontal 1.0-m-diameter track. The coefficient of rolling friction is 0.10. How long does it take the train to stop if it's released with an angular speed of 30 rpm?

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Welcome back, everybody. We are told that a student launches a ball with an initial angular speed of 85 rpm minute. Now it is launched onto a flat circular path and we are told that the radius of this path is 35 cm or .35 m. Now, there is rolling friction here and we are told that the coefficient of rolling friction is .25. We are tasked with finding how long it will take before the ball comes to complete rest. So this means that our final angular speed will be zero. We have a formula for this. It is our final angular speed minus our initial angular speed all divided by our angular acceleration. Now, before we can start plugging in terms, we need to make sure everything's in the right units. I'm gonna go ahead and put our initial angular speed into the right units here, which we need to be radiance per second. So we have one revolution is two pi radiance and every one minute we have 60 seconds. This gives us an initial angular speed of 8. radiance per one second. Great. So I'm just going to move this over a little bit so that we're not too cramped here. We have our final and initial angular speeds, but now we need to find our angular acceleration and we're gonna have to work with a couple formulas here. So we know that our angular acceleration is just the tangential acceleration divided by the radius. But what is our tangential acceleration are tangential acceleration is the negative of the rolling friction force divided by our mass. I'm gonna just step aside here. Our rolling friction force is going to be equal to our coefficient of rolling friction times our normal force. But the ball is just sitting on the flat circular path. So the normal force is just going to be equal to the force due to gravity acting on the ball. So let's now sub in this value up here. And we get that this is equal to negative new, our times our mass times acceleration due to gravity divided by mass. These masses will cancel out giving us a tangential acceleration of negative new R G. So now we can plug this value back into this formula and we will find that our angular accelerate is equal to negative new R G divided by our radius. We have all of these terms. So let's go ahead and plug them in and find this we have that this is equal to negative 0.25 times 9. divided by our radius. Of 0.35 giving us an angular acceleration of negative seven point oh one radiance per second squared. So now we have found our angular acceleration. So we are ready to use this formula up here. So let's go ahead and do that. So we have that our time is equal to our final angular velocity of zero minus our initial angular velocity of 8.9 divided by our angular acceleration of negative seven point oh one. Giving us an answer of 1.3 seconds for the ball to come to a complete stop corresponding to answer choice. A thank you all so much for watching. Hope this video helped. We'll see you all in the next one.
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A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in FIGURE EX4.13. The ball's position is shown at 1 s intervals until t = 3s. At t = 1s, the ball's velocity is v = (2.0 i + 2.0 j) m/s. (b) What is the value of g on Planet Exidor?
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A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in FIGURE EX4.13. The ball's position is shown at 1 s intervals until t = 3s. At t = 1s, the ball's velocity is v = (2.0 i + 2.0 j) m/s. (a) Determine the ball's velocity at t = 0 s, 2s, and 3s.
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