A solid sphere of radius R is placed at a height of 30 cm on a 15° slope. It is released and rolls, without slipping, to the bottom. From what height should a circular hoop of radius R be released on the same slope in order to equal the sphere's speed at the bottom?

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Hey, everyone in this problem, a physics teacher uses a ramp that makes an angle of 20 degrees with the horizontal to show students about rolling motion of different objects. OK. So the first case she releases from rest, a disk of diameter D from a height of 50 centimeters above the floor, then she releases a uniform spherical ball of diameter D from a height of H. The two objects reached the end of the ramp at the same speed and were asked to find h they were told that these objects roll without slipping. We're given four answer choices. Option, a 32 centimeters, option B 42 centimeters, option C 47 centimeters and option D 56 centimeters. Now, let's go ahead and draw what we have. So we're gonna draw the first case in blue and we have a ramp. OK. That makes a 20 degree angle with the horizontal. We have a disc and this is our disc case and the diameter of the disc is D which tells us that the radius of the disc is gonna be D divided by two, all right. And the height of this is centimeters. OK? The high end of that ramp. So that's our disk case. And then we are going to draw in red. The other case that we have, we have the exact same ramp makes a 20 degree angle with the horizontal. But this time, we have a sphere again with a radius of D divided by two. But the height is unknown in this case, has a height H And what we're trying to do is figure out what this height is given the fact that the two objects reached the end of the ramp at the same speed. OK. So let's go ahead and write out the variables we have, OK. We have Omega, we have V and we have H OK, because we have rolling motion here, we have to consider the angular speed Omega as well as the linear speed V. And we're gonna start with the subscript knot D. OK? And this is gonna to indicate that initial time period for the disc. Now, for the disc initially, it's not moving. OK. They, they start from rest and so Omega knot is gonna be zero m per second. Oops and this should be meters per second. It should be radiant per second. OK. V. Knot. Again, it's not moving. So we have zero m per second. It's not rotating and it's not moving downwards and the initial height is gonna be 50 centimeters. Now, we're gonna convert this centimeters into meters. OK. We're gonna have gravity acting here at some point. So we're gonna be using that 9.8 m per second squared. So just have consistent units. We're gonna convert to meters even though we'll have to convert back at the end. That's ok. So we have 50 centimeters. We're gonna multiply by one m divided by centimeters. The unit of centimeters divides out. We're essentially dividing by 100 when we get 0. m. So this is the initial case for a disc and then we have the final case and actually let me move the 0.5 m underneath. So we have some more room and at the end of this ramp, OK. We have some angular speed, you have some linear speed and we have a height of zero m. OK? So we have Omega F D V F D and H F D and we have two unknowns there now for the sphere very similar. OK. Again, it's starting from rest. So Omega dot S is gonna be zero radiance per second. V nought. S is gonna be zero m per second and the initial height H not S is gonna be H OK. And that's what we're trying to find in the final time point Omega F S and V F S. We don't know and we know that H F S that final height is gonna be zero m. OK? Because that sphere reaches the end of the ramp. So what we have in each of these cases is a conservation of mechanical energy. OK. As the object moves from the top of the ramp down to the bottom of the ramp. So let's start with the disc and let's look at our conservation of mechanical energy. OK. So for the disc which we're continuing in blue, OK, we'll always have the disc in blue that's severe in red. The conservation of mechanical energy tells us that the initial kinetic energy, what the initial potential energy is gonna be equal to the final kinetic energy plus the final potential energy. Now recall that kinetic energy is given by one half M V squared. OK. When we're talking about linear motion, so when we're talking about this rotational motion, we have one half by omega squared. So our kinetic energy in this case has to account for both of those terms because we have rolling motion. This is both rotating and moving along that incline. And we talk about the potential energy here. We don't have any springs or anything acting that will give us some spring potential energy. But we do have the gravitational potential energy because these objects are starting at a height. OK. So expanding this equation, we have one half MD V, not D squared plus one half ID omega, not D squared, four MD G H not D is equal to one half MD B F B squared. That's one half ID omega F DS squared OS MD G H F D OK. So that's really, really long, but this is gonna simplify quickly when we take a look at these variables, we've written down now those initial speeds, both the angular speed and the linear speed are zero. So the initial kinetic energy, both of those terms are going to zero. So on the left hand side, the first term and the second term go to zero and we're left with just MD G H not D, OK. So MD multiplied by G multiplied by H not D. On the right hand side, we know that the final height is zero. So the last term on the right hand side goes to zero and I left with one half multiplied by MD multiplied by V F D squared. Who now we need to figure out what this moment of inertia ID is going to be OK. And when we go to this sphere equation, we're gonna have to do the same for the sphere. So let's go ahead and do that. Now, let's calculate these moment of inertia so that we have those ready to plug into our equation. We're gonna take a little detour, we're gonna go back up to our variables that we've written down and we're gonna write the moment of inertia. I for each of these cases. Now, for the disk, you can look this up in a table in your textbook or that your professor has provided the moment of inertia. And let me write it up with the diagram. The moment of inertia of this disk is gonna be equal to one half MD multiplied by R D squared. And for the sphere, and if we look at the moment of inertia for a sphere, it is rotating like this one is we get 2/ MS R S. So getting back to our equation for the conservation of energy, we have MD multiplied by G multiplied by H not D is equal to one half MD multiplied by V F D squared plus one half multiplied by that moment of inertia, which is one half multiplied by MD, multiplied by R D squared. OK. And all of this is multiplied by Omega F D all script. OK. Yeah, you might be thinking we don't have the mass of the disk. What are we supposed to do here? Because we have that term. Well, the good news is the factor MD shows up in every single term. So we can divide this entire equation by that mass MD. And it will divide out of every term and we don't need to worry about it. And what that tells us is that the result of this problem is actually gonna be the same no matter what the mass of that disc gets. OK. So that's really interesting. Now, we're gonna substitute in everything else. OK. On the left hand side, we have G which is 9.8 m per second squared multiplied by the initial height of 0.5 m. This is equal to one half multiplied by V F D squared. OK. Pull us one quarter, multiplied by R D squared. Now, we aren't given information about the radius, but we are given information about the diameter. OK? By calling it D. So we're gonna write this as D divided by two squared. Now, we have two unknowns. OK? We've mentioned this before. We have the velocity V F D and we have the or the speed and we have the angular speed Omega F D. OK. Now, we're told that these have the same speed at the bottom. What that means is they're talking about linear speed. So if they have the same linear speed, what we really wanna do is compare this V F D term. Hm. In order to solve for V F D, we need to do something with Omega F D. Recall that the velocity of the center of mass V C M is equal to our Omega. So we can assume that that speed, the center of mass is the speed at the end of this ramp. And that tells us that Omega final is gonna be equal to the final velocity V F or the final linear speed divided by the radius R OK, which is equal to V F divided by D over two, right? So back to our equation. Now, we can substitute that Omega F D is gonna be equal to two V F D divided by the diameter D all squared. Yeah. So this started really messy and slowly and slowly we're getting to clean this up. Now, we aren't given the diameter other than we're calling it D but you can see on the right hand side that now we have ad squared divided by D squared. OK. So those D squares are gonna cancel up on the left hand side. Simplifying gives us 4.9 m squared per second squared. On the right hand side, we have three quarters V F D squared. This tells us that V F D squared, it's gonna be equal to 6.53 repeated meter squared per second squared. OK? By dividing by that three quarters. And if we take the square root, we get that V F D is equal to 2.556 m per second. All right. So we found the speed at the bottom of the ramp for the disc. And, but remember what we're trying to find. We're trying to find that height H in order to do that, we needed to know what that speed at the bottom of the ramp was because the sphere has the exact same speed. So now we can switch over to the sphere, right? And for the sphere, we have the conservation of mechanical energy as well. And I'm gonna save us some steps and because this is gonna work out in the exact same way we know that the initial speeds angular and linear are both zero. OK. So the initial kinetic energy is zero. The only term we have on the left hand side is the gravitational potential energy which is given by MS multiplied by G multiplied by H not S. On the right hand side, the co um gravitational potential energy is going to be zero because the height is zero, we're left with just those kinetic energies. We have one half MS multiplied by V F S squared plus one half multiplied by the moment of inertia I s which we found to be 2/ MS multiplied by R S squared multiplied by omega squared. OK. So we were able to skip a couple of lines there just based off it being the exact same as that desk equation. OK. So we've already simplified it before. We don't need to go ahead and do it again. Now, the mass MS is gonna divide out just like it did. In the disc case, we're left with G multiplied by H not S is equal to one half V F S squared plus 1/5 multiplied by DS divided by two, all square K writing that radius as a diameter divided by two. And then if we write our final velocity or final angular speed, Omega F as V F divided by R again, we can write this as two V F S divided by are oops divided by D. Yes, all squared fix the D. So it's clear. And again, that diameter DS is gonna divide out, substituting in the values we have 9.8 m per second squared multiplied by the height we're looking for H no S and let's just write this as H now. All right, that's how it's given in the problem. This is equal to one half. Now, we know that the velocity at the end of the ramp, that final velocity or the final speed, linear speed is the same for the disc and the sphere. OK. So V F S is equal to V F D and we found V F D it's 2.556 m per second. OK. So one half multiplied by 2.556 m per second, all squared plus 1/ multiplied. OK. The two squaress will divide out, the DS squares will divide out and we're just left with that velocity 2.556 m per second, all squared. And we're gonna simplify on the right hand side and then we're gonna divide by 9.8 m per second squared to isolate for the height H and we get that the height H is gonna be equal to 0. m. OK? And that is that final answer we were looking for. So in order to figure out that height, we needed to first find that speed at the end of the ramp. OK? Because we knew that, that was the same for the disc as it was for the sphere. Once we knew that speed, we were able to use our conservation of mechanical energy to find the height that the sphere was released from. If we go up to our answer choices and compare this to what we found, we can see that the correct answer rounding to the nearest centimeter. OK. Is gonna be option C 47 centimeters. Now our answer was in meters to convert it. We need to multiply by 100. Ok? So we get 46.7 centimeters and again, rounding to the nearest centimeter gives us option C 47 centimeters. Thanks everyone for watching. I hope this video helped see you in the next one.

A solid sphere of radius R is placed at a height of 30 cm on a 15° slope. It is released and rolls, without slipping, to the bottom. From what height should a circular hoop of radius R be released on the same slope in order to equal the sphere's speed at the bottom?

Verified Solution

Key Concepts

Conservation of Energy

Moment of Inertia

Rolling Motion