A uniform marble rolls without slipping down the path shown in Fig. 10.59 , starting from rest.

A uniform marble rolls without slipping down the path shown...

Key Concepts

Conservation of Energy

This concept involves equating the initial gravitational potential energy of the system with the sum of translational and rotational kinetic energy (or solely translational kinetic energy in the case of sliding) at a later point in the motion. It is fundamental in determining the speed and kinetic energy distribution of an object as it descends a height, enabling the calculation of necessary conditions to overcome constraints such as remaining above a pit.

Rolling Without Slipping

Rolling without slipping is a condition where the point of contact between a rolling object and the surface has zero velocity relative to the surface. This relationship (v = ?r) connects translational and rotational motion, ensuring that the division of energy between linear and rotational parts is properly accounted for when applying conservation of energy.

Moment of Inertia

The moment of inertia characterizes the distribution of an object's mass relative to its rotation axis and determines its resistance to angular acceleration. In problems involving rolling objects, it plays an integral role in how potential energy divides between translational and rotational forms, but under certain conditions, such as efficient energy conversion, the dependence on the radius may cancel out in the final result.

Comparative Analysis of Rolling versus Sliding

This analysis involves comparing the dynamics of an object that both rotates and translates (rolling) with one that only translates (sliding without friction). The energy partitioning in a rolling object (into translational and rotational kinetic energy) typically requires a higher starting height to achieve the same translational speed as a sliding object, where all potential energy is converted solely into translational kinetic energy.

Transcript

00:01 Okay, so in this question, we have a marble at the top of a hill, and it's going to roll down the hill, and we want it to be able to, it'll roll off the hill and over a pit, and we want it to be able to roll as it kind of flies over the pit, we want it to be able to reach the other side of the pit without falling into it.

00:27 And so we have some measurements in the book.

00:33 There's a diagram in the book, a picture of the whole situation.

00:36 We have our height h, which is how far above the cliff it all roll off of that the marble starts.

00:48 And then the left side of the pit has a height of 45 meters and the right side has a height of 25 meters and the right side has a height of 20.

00:55 Meters and the pit has a width of 36 meters so we have you know all these different dimensions but in part a we're trying to figure out what the initial height off above the cliff on the left side that this marvel has to start in order for it to fly over the pit as it after it rolls off the cliff so for this we're going to be assuming that as it reaches the left side of the pit, it is moving perfectly horizontally as it rolls off.

01:32 So all of its velocity is horizontal.

01:35 And from there, it free falls over the pit.

01:41 And we have to figure out first how much velocity it needs in order to clear the pit.

01:49 So to do that, we need to do some calculations with kinematics.

01:53 First, we need that the change in height so we're going to call this delta y from the left side to the right side in order for it to clear it has to fall no more than 20 meters before it reaches the right side of the pit and this is because on the left side it's 45 meters up but on the right side it is 25 meters from the bottom of the pit so its change in y has to be 45 minus 25 meters so it's so it should change in y can't be more than 20 meters otherwise it won't clear the pit so um that's the first thing we need to do and then we can set up the kinematics equation that delta y is going to be equal to one half a t squared plus uh v y t so here we're going to use this to solve for the time that it takes for to travel over the pit um so because as it rolls off the left side of the pit we're assuming that its only its velocity is entirely horizontal, its velocity in the y direction is going to be zero.

03:04 So this second term cancels.

03:07 And then its acceleration, it's just going to be that due to gravity.

03:10 So we know that a is going to be equal to g.

03:14 And so then we can rearrange this equation to solve for t.

03:19 So we do that by multiplying both sides by two, dividing both sides by the acceleration, and then taking the square root.

03:24 So we get an equation for time, the time it takes to travel over the pit is going to be equal to the square root of two delta y divided by g since the acceleration is that due to gravity so we know delta y and we know g so we can go ahead and calculate this out so we can say that t is going to be equal to two times 20 meters divided by 9 .8 meters per second and we take the square root of all of that and if you calculate that you should get that our time is going to be equal to 2 .02 seconds um and then we can use this to figure out how fast the marble has to be as it falls as it travels off the left side of the cliff in order for it to clear the pit and land on the right side so for that we can say that its velocity in the x direction, which is its total velocity, is going to be equal to how far it has to travel the distance d divided by t.

04:32 And so this distance d is the width of the pit, which is 36 meters, and t is the time that we just calculated.

04:39 So the minimum velocity in the x direction that it needs as it rolls off the left side of the cliff or the left side of the pit has to be equal to 36 meters divided by 2 .02 seconds.

04:57 And so it has to have this velocity or higher in order for it to actually clear the pit and land on the right side.

05:05 So if you calculate that, you should get that its velocity as it rolls off the cliff on the left side.

05:14 At the bottom of the height h, this velocity in the x direction has to be at least 7.

05:21 Point 82 meters per second and now we just need to do one more step and now we can to do this we need to analyze the kinetic energy and potential energy of the marble so to do this we'll say that the left side of the pit the top of the left side um of the pit we'll call that we'll set that as our zero for our gravitational potential energy.

05:55 So when it rolls off the cliff on the left side, it has zero gravitational potential energy.

06:02 And so that means when it, at the very top, where it just starts rolling, where it starts from rest, that means that its energy is initially going to be mgh, because it's starting at rest, so it has no kinetic energy.

06:18 And then its gravitational potential energy is just, you know, due to its height h, since we set the bottom of that height equal to zero.

06:30 So as h above our zero point, the height is h above our zero point.

06:35 So its potential energy is just mgh.

06:37 And then as it rolls off the cliff, its potential energy is zero, since its difference between our zero, since it is at our zero point vertically.

06:46 And so its energy is just its kinetic energy, which means that it has its linear terms, so 1 1�mv squared.

06:54 And so this is the center of mass, but this is the same velocity as our vx from before.

06:59 So we can say that that is vx.

07:01 And then we need a rotational term.

07:03 So we have one half times the moment of inertia times the angular velocity omega.

07:11 And other things that we need here to note here are that it is a uniform marble...

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