A 750 g disk and a 760 g ring, both 15 cm in diameter, are rol...

Question

A 750 g disk and a 760 g ring, both 15 cm in diameter, are rolling along a horizontal surface at 1.5 m/s when they encounter a 15° slope. How far up the slope does each travel before rolling back down?

Accepted Answer

Hey, everyone. Let's go through this practice problem. The speed of a cylinder and a spherical ball rolling along the horizontal table is 2. m per second. When they come across a ramp making an angle of 30 degrees with the table, the cylinder has a radius of five centimeters and a mass of g. While the sphere has a radius of six centimeters and a mass of 500 g. Find the distances traveled up the ramp by one, the cylinder and two, the sphere just before they start rolling down the ramp. We're given four options to choose from option a 37 centimeters for part one and 35 centimeters for part two. Option B 37 centimeters for part one and 17 centimeters for part two, option C 74 centimeters for part one and 35 centimeters for part two and option D 74 centimeters for part one and 69 centimeters for part two. The first thing I'm going to do is draw a diagram of the situation. So here is our ramp. It makes some angle fata with the ground and both the cylinder and the ball are reaching the bottom of the ramp from the ground and then roll up the, the ramp at some distance. This distance is what the problem is asking us to find for both the cylinder and the ball. And I'm going to label it. D I'm also going to label the ball and the cylinders vertical height from the ground. I want to call this Y and the reason why this is helpful to us is because of trigonometry because the sign of beta, in other words, the sign of the ramps angle is equal to opposite over hypotenuse. So if we think of the wedge as a right triangle, then the sign of the angle is going to be equal to the vertical distance Y divided by D, the distance we're trying to find. And if we rewrite this equation algebraically to solve for that distance, we can see that the distances we want are equal to the vertical heights divided by the sign of the angle. This means that we can use energy concepts to analyze the ball in the cylinder. Because energy, the energy equations will allow us to analyze the vertical heights of the ball or the cylinder from the ground. So, and if we can find the height, then we can find the distance. So let's get started by remembering the law of conservation of energy which states that in an isolated system such as this one, presumably the total change in mechanical energy is zero. In other words, no energy is leaving or entering the system, but it's just being converted between kinetic energy and potential energy. You can alternatively write this in a more expansive way as the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. Now, let's use what we know about the individual energy formulas to expand this out even further in order to relate all the different variables that were given. However, before we do so, it's important to recall that because that we're dealing with objects that are rotating upper ramps rather than projectiles that are just being slit up the ramp. We have to consider the two different components of the kinetic energy. There's just the standard kinetic energy, the one half MV squared, but there's also the additional kinetic energy term from the fact that it's rolling. So the usual kinetic energy term is one half multiplied by the mass of the particle or the object multiplied by the square of its speed. But there's also the rolling kinetic energy and this is equal to one half multiplied by the moment of inertia multiplied by the square of the angular speed. So now let's write out a fully expanded version of the energy conservation equation that we wrote earlier. So first, the initial rolling kinetic energy one half multiplied by the moment of inertia multiplied by the initial angular velocity squared plus one half M V sub I squared plus then the potential energy term. So gravitational potential energy is M multiplied by G multiplied by the vertical position. So why, why should I this is equal to all the final versions of these same energies? So one half multiplied by the moment of inertia multiplied by omega sub F the final angular velocity squared plus one half M V sub F squared plus M G Y sub F. So this is our full conservation of energy equation as it pertains to this problem. However, right off the bat, there are a number of things that we can simplify to make this equation less unwieldy for us to use in this problem. First, the initial potential energy term for the sake of the problem, I'm going to define all of the initial variables to relate to the, to the cylinder or the sphere at the bottom of the ramp just before the ball startss moving, rolling up the ramp. And I'm going to use the subscript F the final variables, the final state to refer to when the ball or the cylinder reaches the top of its motion, when it's at rest for just a moment before it starts rolling back down. Now, because we can define the coordinate system. However, we'd like, I'm going to make this very easy on us by assuming that the bottom of the ramp is where Y is equal to zero. And if we do with that, then that means that we can cancel out the initial potential energy term entirely because Y sub I is equal to zero. So that term goes away. Another thing we can do to simplify this is to keep in mind that since in our final state, the ball or the cylinder is at rest and not moving, that means it has no kinetic energy and it doesn't even have any rolling kinetic energy either. It's completely at rest there. So the final kinetic energy terms both go away. So the energy conservation equation we're left with is much simpler than the one we had a moment ago. It's just one half I omega sub I squared plus one half M V, sub I squared equals M G Y seven F. And that's it. And that's much simpler. So now let's actually look at the individual parts of the problem and see what we can learn about this for the cylinder and the sphere. Let's start with part one of the problem which specifically asks about the cylinder. So for part one of the problem, look, let's look at the cylinder. And the first thing we need to know is the moment of inertia of the cylinder. Now, you'll probably have a chart that you can look up what the formulas are for the moment of inertia of different shapes. And you'll see that the moment of inertia of a cylinder is equal to one half multiplied by the mass of the cylinder M subi multiplied by the square of the radius of the cylinder. So we'll want to plug this in for our moment of inertia in the full conservation of energy equation. Another thing we'll need to know is that since our conservation of energy equation includes omega and angular velocity, but the problem never gives us any values for angular velocity. We want to cancel out that angular velocity while we, as we simplify this moving forward. So for that end, it's useful to recall that the tangential speed of an object is equal to the radius of the rotational speed multiplied by the angular speed. So if we want to get rid of the angular speed in our energy equation, let's solve this for omega by dividing both sides of the equation by R. And we find that the angular velocity is equal to the tangential velocity divided by the radius of the rotational motion. So now any time we have an omega, we can just substitute this formula in for it. And we won't need to worry about having a value for the angular speed. Now, let's get to actually plugging all these terms into our energy equation. So first let's start from the left one half I omega sub I squared. So one half multiplied by and then we're multiplying it by the moment of inertia for the cylinder. So as we just discussed a moment ago, that's one half M subi R subi squared. OK. And then this is being multiplied by omega sub I squared. As we just discussed that Omega sub I can better be written as V divided by R. So that's the initial speed V knot or the sub I divided by the radius of the cylinder are subi and all of that is squared. And this is plus the kinetic, the, the normal kinetic energy term that's plus one half M Suby V sub I squared. And this is equal to M G or rather M sub G Y sub F. Now there are a few things we can do to simplify this massive equation. First of all, since every single term has M sub sill the mass of the cylinder, they can all cancel out. So this equation does not depend on the mass of the cylinder. We can also cancel out the R sub sill squared since that's being canceled out by a denominator here. So now let's write this out in a more simple term. So one half multiplied by one half A s equal to 1/ multiplied by V sub I squared plus one half visa I squared. And that is equal to G multiplied by Y sub F. So now we can simplify this even further because we have two terms on one side of the equation that have the sub I squared in it. So we'll just have to do some fractional addition here. One half plus 1/4 is the same thing as saying 1/ plus 2/4. So that can just be written as 3/4 multiplied by V sub I squared. And then we want to solve this for the final position of the vertical height. So we just solve this for Y sub F by dividing both sides of the equation by G. So what we're left with is three vs of I squared divided by four G. And now we have what we need to find the vertical height of the cylinder, the final vertical height of the cylinder. So if we plug in the values given to us, Y Sabbath is equal to three, multiplied by the initial speed of the cylinder, which the problem gives us as 2.2 m per second. And we divide this by four multiplied by G, the gravitational acceleration, which of course is equal to 9.81 m per second squared. We put this into a calculator, we find a height of about 0.370 m. And as we discussed way earlier, if we want to find the actual distance the cylinder traveled, we just want to use the equation we wrote way earlier stating that the distance it traveled in the sub is equal to the final height divided by the sign of the angle which is 30 degrees. So in other words, D subi is equal to 0.370 m divided by the sign of 30 degrees. And when you put this in your calculator, make sure your calculator is in degrees mode in order for this to work. But we find a value of about 300.74 m or alternatively written as 74 centimeters. And that is our answer to the first part of this problem. And just from that, we can guess that we're on the right track with this problem because if we look at our multiple choice options, two of the options include 74 centimeters as a potential option or the answer to part one of the problem. But now let's move on to part two of the problem which asks about the sphere. This will be the same general process. But once again, we'll need to know the moment of inertia for the spherical ball. This turns out U B equal to 2/5 multiplied by the mass of the ball multiplied by the square of the radius of the ball. Now let's use the exact same process we use for the cylinder to simplify the equations out and solve for the distance the ball traveled we're using the same process. So that's one half multiplied by the moment of inertia. So that's 2/5 multiplied by the mass of the ball multiplied by the square of the radius of the ball. And this is being multiplied by the angular speed. So that's the final speed divided by the radius of the ball. And this whole little fraction here is being squared and then we're adding one half the, the regular kinetic energy term. So one half multiplied by the mass of the ball multiplied by the square of the speed. And this is equal to mass of the ball multiplied by G multiplied the final vertical position. So now we'll do the same kind of simplifying we did before the mass the balls cancel out because that's in every single term, the radii cancel out in the leftmost term. And then we can multiply out the fractions. So we end up with one half multiplied by 2/5 or alternatively two tens multiply to B V sub I squared plus one half the sub I squared. And that's equal to G multiplied by Y sub F. Once again, we'll do some fractional addition here. So 2/10 plus one half is equal to 7/ or 7/10 multiplied by V sub I squared. And then again, to solve for Y sub F, we just divide both sides of the equation by G. So Y sub F is equal to seven V sub I squared divided by 10 G. Now all that's up for us to do is plug this into a calculator and then do the same thing we did in part one for finding the actual distance along the ramp that we traveled. So plugging in the values now, Y sub F is equal to seven multiplied by the same initial speed of 2.2 m per second. All that's squared and multiplied by 10, multiplied by the gravitational acceleration, 9.81 m per second squared. And if we put that into a calculator, then we find a final height of 0.345 m. Now, all that's set to do is plug that into the distance equation. So the D for the ball is equal to the horizontal position divided by the sign of the angle or in other words, 0.345 m divided by the sine of 30 degrees. And if we put that into a calculator, then we find a distance of about 0.69 m or alternatively 69 centimeters. And that is our answer for part two of the problem. And if we look at our multiple choice options, you can see that option D agrees with both of the choices. We found 74 centimeters for part one and 69 centimeters for part two. So that is our final answer to the problem. So I hope this video helped you out if it did and you want want more practice, please consider checking out some of our other videos which will give you more experience with these types of problems, but that's all for now. I hope you all have a lovely day. Bye bye.

A 750 g disk and a 760 g ring, both 15 cm in diameter, are rolling along a horizontal surface at 1.5 m/s when they encounter a 15° slope. How far up the slope does each travel before rolling back down?

Key Concepts

Moment of Inertia

Conservation of Energy

Rolling Motion

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