A 140-g baseball, with a diameter of 7.5 cm, is pitched at 145...

Question

A 140-g baseball, with a diameter of 7.5 cm, is pitched at 145 km/h. It spins at 1200 rpm. If the baseball is assumed to be a uniform solid sphere, what fraction of its kinetic energy is rotational?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A car tire with a mass of 9.6 kg and a diameter of 38 centimeters is moving at a constant speed of 160 kilometers per hour. It has a uniform rotational speed of 1400 revolutions per minute, determine what the fraction value of its total kinetic energy that is due to its rotation will be consider the tire as a hoop. So that's our end goal is ultimately, we're trying to figure out what the fraction value of the total kinetic energy that's due to rotation is and that's what our final answer is gonna be is whatever this fraction value of the total kinetic energy due to rotation. Awesome. So let's look at our multiple choice answers to see what our final answer might be. A is 0.14 B is 0.28 C is 0.39 and D is 3.6. 1st off let's write down all of our known variables. We know that the mass of the tire is equal to 9.6 kg. We also know that the diameter is equal to 38 centimeters, which is also equal to 0.38 m. We also know that the radius R, which we're gonna say diameter is D and radius is R and mass is M. So we also know that the radius is equal to the diameter divided by two. So we know that 0.38 m divided by two is equal to. So they make this look nicer here. So we have 38 me. So it's 0.38 m divided by two is equal to 0.19 m, which is our radius. Now, we need to note that the linear speed, which we're gonna denote as V is equal to 160 kilometers per hour. But we need to convert kilometers per hour to meters per second. So we need to take 160 kilometers per hour. So you take kilometers per hour and we need to multiply. So now we're gonna be using dimensional analysis to do this conversion. We need to note that in one hour there is 3600 seconds. We also need to note that there is 1000 meters in one kilometer. Awesome. So when we plug that into our calculator, we will find that it's equal to 44.44 m per second. Awesome. And that is for our linear speed, we also know that our rotational speed, which I'm gonna denote this as capital R, capital S. So our rotational speed is equal to 1400 RP MS or revolutions per minute. And we can easily solve for angular velocity, which we're gonna denote angular velocity as omega. And to do that, we're gonna have to take our rotational speed and we're gonna have to convert it to radiance per second units. So we're gonna take 1400 RPMs which were gonna write it as revolutions per minute. So we can easily see our units and cancel them out as we go along. So using dimensional analysis, we know that there's two pie radiance in one revolution. Awesome. And we also need to convert minutes to seconds. So we need to note that in one minute there is 60 seconds. Awesome. So when we plug that into our calculator, we will get 140 divided by three P or multiplied by pi I should be precise. So it's 140 divided by three, multiplied by pi and this is in radians per second units. Now. Awesome, we did it. So now at this point, we need to note that we can approximate the car tire like a hoop because that's the hint that's given to us in the prom itself. At this point, we now need to recall and use the moment of inertia equation to solver the inertia of the tire. So we need to note that I the moment of inertia is equal to M multiplied by R squared. And plugging in our known variables gives us 9.6 kg and this is multiplied by our radius squared which is 0.19 m square. And this is equal to let me plug it into our calculator, 0.34656 and its units are going to be kilograms multiplied by meters squared. Awesome. So now at this point, we need to recall and use the equation to solve for the linear kinetic energy, which we're going to denote linear kinetic energy as ke subscript L. So the linear kinetic energy is equal to one half multiplied by M multiplied by V squared. And this is equal to when we plug in our known variables, one half multiplied by 9.6 kg. Awesome. And don't forget your parentheses multiplied by our speed V which is the linear speed. So it's 44.44 meters per second squared. Awesome. So when we plug that into our calculator, we will get nine thou so 9479 0.585. And this is gonna be in units of jewels. Awesome. Fantastic moving right along here. So now we need to recall and use the equation to solve for the rotational kinetic energy which we're gonna denote as K E subscript R. So this is the rotational kinetic energy and this is equal to one half multiplied by I multiplied by omega. And this is squared. So it's omega squared and this is equal to plugging in our known values one half multiplied by our value for I, which is the moment of inertia, which we found to be zero point 34656. And don't forget that its units are kilograms multiplied by meters squared. And then this is multiplied by our value for Omega which we found Omega to be 140 divided by three, multiplied by pi and don't forget that its units are radians per second. Awesome. And don't forget that our Omega value is squared. So when we plug that into our calculator, we will find out that the rotational kinetic energy is equal to 3724.447 jewels. Awesome. And that's when we round to three sign so to three digits to three decimal points. OK. So when we're rounding the three decimal points, OK. Moving right along here. So let us recall that the total kinetic energy is equal to. So we're gonna denote the total kinetic energy as ket is equal to ke L plus A er Awesome. So therefore, the fraction of the total kinetic energy due to its rotation will be the ratio of the rotational kinetic energy. So the rotational kinetic energy to the total kinetic energy which we can write the fraction of the kinetic energy that is rotational as drum roll K E are. So the rotational kinetic energy divided by the total kinetic energy which is equal to ke the rotational kinetic energy divided by the kinetic energy of the linear kinetic energy. So Kel plus K er the rotational kinetic energy. So moving right along here. So when we plug in our known values, we will take that we know that K er is 3724.447 jewels and this is divided by 9479 point 585 Jules. Awesome. And this is plus and so we don't get confused here. Make sure we have our j looking nice here. So it doesn't look like a number and then we need to add this by 3000 724.447 Jules. Awesome. So let me plug that into our calculator. We will find that this is all equal to when we round two decimal places, 0.28. And we did it we solve for this problem. Hooray. So looking at our multiple choice answers, the correct answer has to be the letter B 0.28. Thank you so much for watching. Hopefully, that helped and I can't wait to see in the next video. Bye.

A 140-g baseball, with a diameter of 7.5 cm, is pitched at 145 km/h. It spins at 1200 rpm. If the baseball is assumed to be a uniform solid sphere, what fraction of its kinetic energy is rotational?

Key Concepts

Kinetic Energy

Rotational Motion

Moment of Inertia

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