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Ch 12: Rotation of a Rigid Body
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All textbooksKnight Calc 5th EditionCh 12: Rotation of a Rigid BodyProblem 12

Chapter 12, Problem 12

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

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Hello fellow phy today we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A 10 kg fox terrier dog runs and jumps through a hoop onto a rotating circular platform in a circus scene. The platform with a mass of 65 kg and a radius of 1.25 m rotates at 43 rotations per minute around its frictionless central axis. The dog is trained to run at a speed of 2. m per second in the same direction as the platform's rotation and tangent to the platform. The dog lands exactly on the platform's edge, determine the angular velocity of the platform after the dog has landed on it. So that is our end goal is to determine the angular velocity of the platform after the dog has landed on it. OK. So we're given some multiple choice answers. They're all in the same units of radiant per second. So let's read them off to see what our final answer might be. A is 1.4 B is 2.3 C is 3.8 and D is 5.4. So first off, let us treat and model the circular platform as a disk. Now let us find the momentum or moment of inertia. I should say the moment of inertia for the platform. So the moment of inertia for the platform, let's call it I subscript P is equal to one half multiplied by the mass of the platform multiplied by the radius of the platform squared. So now let us consider the system formed by the platform. And the dog note that since there is no external torque applied to this system, we can conserve angular momentum. Thus, we can state that the initial angular momentum is equal to the final angular momentum. So we can say that L I is equal to LF OK. So the initial angular momentum which is L I is equal to the sum of the angular momentum of the platform plus the angular momentum of the dog. So then we can go on to say that L I is equal to LP plus LD. So L I which is the initial angular moment is equal to this to the angular momentum of the platform plus the angular momentum of the dog. So LP, the angular momentum of the platform is written as LP is equal to IP multiplied by Omega I Omega subscript I. So the dog's angular momentum is vector LD is equal to MD multiplied by RD vector RD pro vector BD which simplifies to LD is equal to MD multiplied by RP multiplied by VD multiplied by sine alpha. OK. So note that RD is equal to RP since the dog runs tangent to the platform, and that alpha is the angle between RP and V since the dog runs tangent to the platform. So note that alpha equals 90 degrees. Thus the initial angular momentum L I can be written as L I is equal to IP multiplied by Omega I plus MD multiplied by RP multiplied by VD. Let's call this equation one. And we can say that L S is equal to if multiplied by Omega F which we can also simplify to say that LF is equal to IP plus ID multiplied by Omega. Yes. OK. So then we can go even further to say that in this equation two, that LF is equal to one half multiplied by MP multiplied by RP squared plus MD multiplied by RD swear all multiplied by Omega F. Now we can set equations one and two equal to each other and solve for the angular velocity. The final angular, the final angular velocity which is Omega F. OK. So let's do that. So one half multiplied by MP multiplied by RP squared multiplied by the initial angular velocity Omega I plus MD multiplied by RP multiplied by VD is equal to one half multiplied by MP multiplied by RP squared plus MD multiplied by R P squared multiplied by omega F the final angular velocity. So when we use a little bit of algebra to rearrange and just isolate and solve for the final angle angular velocity. And we get that omega F, the final angular velocity is equal to one half MP multiplied by RP squared multiplied by omega I plus ND multiplied by RP multiplied by VD, all divided by one half multiplied by MP MP multiplied by RP squared plus MD multiplied by RP squared. So now at this stage, we can plug in all of our known variables to solve for the final angular velocity. So let's do that. So Omega F is equal to one half multiplied by the mass of the platform which was 65 kg multiplied by the radius of the platform which was 1. m squared multiplied by the angular the initial angular frequency which was 4. radiance per second plus the mass of the dog, which was 10 kg multiplied by the radius of the platform again, which was 1.25 meters multiplied by the velocity of the dog which was 2.0 m per second. OK. All divided by one half multiplied by the mass of the platform which was kg multiplied by the radius of the platform squared, which is 1.25 m squared plus plus the mass of the dog, which was 10 kg multiplied by the radius of the platform squared, which was 1.25 m squared. So when we plug that all into a calculator, we should get our final angular velocity to equal 3.8 radians per second. So this is the angular velocity of the platform after the dog lands on it for ray, we did it. So let's go look at our multiple choice answers to see what our final answer is. That means the final answer has to be the letter C 3. radiant per second. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.
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