A 4.0-cm-diameter disk with a 3.0-cm-diameter hole rolls down ...

Question

A 4.0-cm-diameter disk with a 3.0-cm-diameter hole rolls down a 50-cm-long, 20° ramp. What is its speed at the bottom? What percent is this of the speed of a particle sliding down a frictionless ramp?

Accepted Answer

Hey, everyone in this problem, we have a ring with an inner radius of two centimeters and an external radius of five centimeters that's released on an inclined surface making an angle of degrees with the horizontal. OK. This incline is one m long and we're asked to do two things. Now, we're gonna focus just on the first thing for now. And that is to find the speed at the end of the incline. We're told to assume that the incline is frictionless. So let's draw out what we have here. We have an incline. It makes an angle of 25 degrees with the horizontal. It's one m long and then we have our ring, ok. This odor radius and inner radius that is gonna roll down that incline. Now, because we have a ring as it moves down this incline, it's gonna be rotating and it's also gonna be moving linearly along. And so we have rolling motion or sometimes called a free wheel situation. Now, the initial angular speed omega knot and the initial linear speed von those are both going to be zero, zero m per second and zero radiance per second because we're told that this ring is released on this incline. OK. So it's released. So initially, it's at rest. Now, we also have some initial height H knot and this height can be found because we know the length of the incline which, which acts as the hypotenuse and we know the angle, OK. And this is gonna be related through sign of the angle because we're talking about the opposite side. So we get that hypotenuse of one m multiplied by sine of 25 degrees. Now, when it reaches the end of the incline, OK. The final angular speed, we don't know the final speed or linear speed is what we're trying to find. OK. They ask for the speed at the end of the incline and they just ask for speed, not um angular speed or rotational speed. And so we are looking for that linear speed and the final height H F is gonna be zero m. I have the initial height is H knot. That means that we're assuming that the bottom of the incline has a height of zero m. OK? So that final height is zero m. So we have some height, we have some speed. Let's think about our conservation of mechanical energy. So we can write out our conservation of mechanical energy. One more thing to do before we get started in that equation, the radiuses, we were given two centimeters and five centimeters. We wanna convert those to our standard unit. We're gonna need to use those in our calculations and it's easier to do that conversion first so that we don't have to do it every single time that those appear. OK. So that inner radius R I is two centimeters. We're gonna multiply that by one m divided by centimeters because we know that there are 100 centimeters in every meter, the unit of centimeter divides out. And what we're doing is essentially dividing by 100 to go from centimeters to meters. You get 0.2 m. Now the radius of the outer ring we're gonna call R not or R zero R O that is five centimeters. And I get multiplying by one m divided by 100 centimeters. The unit centimeters divides out and left with 0.5 m. OK? So we've written down everything this question has given us, we are ready to get started with part one. And again, we're using a conservation of mechanical energy. So we have that the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. OK. Knot plus U knot is equal to K F plus U F. Now, our kinetic energy is gonna have two components and we have the kinetic energy from the rotation, the rotational kinetic energy as well as the regular linear kinetic energy as this thing moves along that ink line. So we're gonna have K knot, which is just gonna be our linear kinetic energy plus K R knot. That rotational kinetic energy plus eu G not because the potential energy we have is due to grav gravity. OK. We have gravitational potential energy, we don't have any springs or anything like that acting. So that's the only potential energy we have to worry about. So K knot plus K R knot plus U G knot is gonna be equal to. And the same thing in the final case K F plus K R F plus U G F. Now we're called that our linear kinetic energies are gonna be given by one half M V squared. The rotational kinetic energy is very similar. It's given by one half I omega squared. And our gravitational potential energy is given by M G H. So filling in all of these terms, what we're gonna have is one half M V not squared plus one half I omega knot squared plus M G H knot is equal to one half M V F squared plus one half I omega F squared plus M G H F. Now, this looks really messy, but we're gonna simplify it. OK. We have a lot of values that were zero. So we can eliminate some of these terms. Now, initially, both Omega knot and B knot, the angular and linear speeds are zero. So we're gonna have no initial kinetic energy. The first two terms on the left hand side are going to go to zero. In the final case, we have this final height of zero m, which means that the last term on the right hand side is gonna go to zero as well. What we're left with is M G no is equal to one half M V F squared plus one half I Omega F squared. And what we're looking for again is V up the speed at the end of the in. Yeah, Omega F is something that we don't know either. OK. So how can we calculate V F? If we don't know Omega F, we'll recall that we can relate V F and Omega F through the following. We know that the speed V at the center of mass is equal to R multiplied by Omega. What this tells us is that we can assume that final velocity, OK? Is gonna be the velocity or the final speed is gonna be the speed at the center of mass of that ring. And so it's gonna be equal to R Omega F which tells us that we can write Omega F as V F divided by R. So we can write Omega in terms of the app which is what we want to find an R, which is gonna be the outer radius or not of this ring, which we know. OK. Let's put that into our equation. We have M G H knot is equal to one half M V F squared plus one half I V F divided by R O squared. No, we don't know the moment of inertia I yet. And we don't know that his mass is OK. We do know that the moment of inertia is going to have a mass term in it though. So let's start by calculating the moment of inertia and maybe that will help us get rid of these mass terms as well. So the moment of inertia I for this ring, OK, we have an outer radius and inner radius for this ring. You can look this up in a table in your textbook or that your professor provided. And the moment of inertia is gonna be one half M multiplied by R I squared plus R O squared. OK? So the inner radius squared plus the outer radius squared multiplied by half of the mass. Now we know our inner and outer radiuses. So let's substitute those in, we get one half and multiplied by 0.2 m squared plus 0.5 meters squared. And simplifying this gives us 0. m squared multiplied by the mass M. So we have our equation M G H knot is equal to one half M V F squared plus one half I V F divided by R O all squared. We're gonna take that equation, we are gonna come back to it and we are gonna substitute in this moment of inertia that we found. So we're going back to this equation, we have the mass M multiplied by the acceleration due to gravity 9.8 m per second squared multiplied by the initial speed V knot or sorry, the initial height H knot. And we found that to be one m multiplied by sine of 25 degrees. OK. And this is equal to one half M V F squared plus one half I which is 0.145 m squared multiplied by M multiplied by V F squared, divided by R O squared, 0.5 m squared. And now we can see that we have this mass term M in every single one of these terms. OK. We have this factor of M. So we can divide the entire equation by M and that is gonna divide it on the left and the right hand side. So we didn't actually need to know the mass of this ring in order to solve this problem. Simplifying on the left hand side, we have meters per second squared multiplied by meters. And so what we're gonna get here is 9.8 m squared per second squared, multiplied by sine of 25 degrees. That is gonna be equal to one half V F squared plus 0. V F squared. Simplifying on the right now, we have one half V F squared plus 0.9 V F squared. That's gonna give us 0.79 V F squared. So we're gonna divide by that 0.79 and we get that V F squared is gonna be equal to 9.8 m squared per second squared multiplied by sine of degrees divided by 0.79. We take the square root and we simplify now, when we take the square root, we're gonna get the positive and the negative route in this question, what we're worried about is just the speed, not the velocity. And so we just want the magnitude. So we're gonna take the positive route and that's gonna be 2. m per second. That is the answer for part one. OK? That is the speed of this ring at the end of the incline. Now, if we go up to our answer traces, we can see that they have two significant digits. So rounding our answer to two significant digits for part one, we get 2.3 m per second. Now we can already eliminate answer choices A and B because they have 2.9 m per second instead. So we're looking at answer choice C or D. Now we want to move to part two and part one is done. Part two is asking us to find the ratio R of the speed of the sliding object to the rolling ring, ok? At the end of the incline. Now we know the speed of the rolling ring, ok? We've just found it. So now what we wanna do is calculate the speed if we just had a sliding object. OK. If that object was replaced by a cube or something that was just gonna slide down the incline instead of roll, now we're gonna do the same thing. We're gonna look at our conservation of mechanical energy. So once again, we have K knot plus UK naught is equal to K F plus U F. Right. In this case though, we don't have any rotational kinetic energy. So we don't have to split up that kinetic energy into two pieces. That means what we get is one half M V knot squared plus M G H knot is equal to one half M V F squared plus M G H F. Now, once again, the initial speed is zero. OK. This is released. So it's initially at rest, initial kinetic energy goes to zero on the left hand side and on the right hand side, we had a final height of zero. And so that second term on the right hand side goes to zero. OK. In addition to that, we can divide by the mass on both sides. OK. That's gonna divide out and we're left with G H knot is equal to one half V F squared. Substituting in our values 9.8 m per second squared, multiplied by one m multiplied by sine of degrees. OK? We have the same incline, we have the same height and this is equal to one half V F squared. We're gonna multiply both sides by two to isolate V F. We get 19.6 m squared per second squared, multiplied by sine of 25 degrees. And if we take the square root again, taking the positive route, because we want the speed, which means the magnitude, we get 2.878 m per second. Now we're gonna add a subscripts here. So this is gonna be V F S the final speed for the sliding object in our previous V F which was 2.29 m per second. We're gonna call V F R for the rolling object. OK? So we have these two final speeds but the question was not asking for the speed of the sliding object for part two, it was asking for the ratio of the two. So what we want to calculate is R which is equal to V F S divided by V F R, OK? The ratio of the sliding object to the rolling object, this is equal to 2.878 m per second, divided by 2. meters per second. And if we work this out, we get this ratio R is equal to 1.257. So now we are done with part two as well. We can go up to our answer traces and compare what we found we had already eliminated. Option A and B based on the first part of this problem. So we're left with C and D and the correct answer is gonna be option D, the final speed at the end of the incline is 2.3 m per second for the rolling object. And the ratio of the speed of the sliding object to the rolling ring is 1.26. That's it for this one. Thanks everyone for watching. See you in the next video.

A 4.0-cm-diameter disk with a 3.0-cm-diameter hole rolls down a 50-cm-long, 20° ramp. What is its speed at the bottom? What percent is this of the speed of a particle sliding down a frictionless ramp?

Key Concepts

Moment of Inertia

Conservation of Energy

Rolling Motion

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