(a) A yo-yo is made of two solid cylindrical disks, each of ma...

Question

(a) A yo-yo is made of two solid cylindrical disks, each of mass 0.050 kg and diameter 0.075 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0050 kg and diameter 0.013 m. Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its 1.0-m-long string, if it is released from rest. (b) What fraction of its kinetic energy is rotational?

Accepted Answer

Welcome back. Everyone. In this problem. A composite object is made up of two solid cylindrical discs connected by a light rod of negligible mass and it starts from a position of rest. The smaller disk has a mass of 0.025 kg and a radius of 0.05 m. While the larger disk has a mass of 0.15 klos and a radius of 0.1 m. This setup is arranged to rotate around the common axis of the cylinders and is dropped from a height of 2 m without any initial rotation in a situation where air resistance is ignored. First, we want to use the principle of conservation of energy to calculate the linear speed of the system just before it hits the ground. And second, we want to know what fraction of its kinetic energy would be rotational just before impact. For our answer choices. A says its speed is 2.5 m per 2nd and 29% of its kinetic energy is rotational. B says it's 3 m per 2nd and 78% respectively. C says it's 4.9 m per 2nd and 75% respectively. And D says it's 5.1 m per 2nd and 45% respectively. Now, let's just quickly try to visualize what's going on here. So we're talking about two solid cylindrical disks, OK? That's connected by a light rod of negligible mass. So we could say that our discs look something like this. OK? And for our composite object, it has been dropped 2 m above the ground. OK? And now we want to know what its speed would be just before, OK, just before it hits the ground. And as our problem says, the principle of conservation of energy is going to be very important to help us figure that out. Now, based on the principle of conservation of energy, we know it tells us that the initial energy of the system is going to be equal to the final energy in the system. Now let's ask ourselves what type of energy is involved in both scenarios? Well, initially, OK, we know that there's going to be a gravitational potential energy you because from the top here it starts from a position of rest. So that would be the only energy for our system initially. But on our final energy or or final energy, the energy of the system just before it hits the ground, we know that it's going to have kinetic energy that's both linear and rotational. OK. So we're gonna have kinetic energy that's translational and kinetic energy that's rotational. So what that tells us then is that for our system, for our system, where based on the principle of principle of conservation of energy, the initial energy equals the final energy, then we know that our initial potential energy. OK, which in this case would be the total mass of the system. Let's call that M total multiplied by the acceleration due to gravity. G multiplied by the height is going to be equal to our translational kinetic energy, which is going to be half MV squared, where this is the mass of the total system plus our rotational kinetic energy, which is going to be a half of our moment of inertia multiplied by omega squared. And now from this problem, we are trying to figure out the final speed of our system. Now we can figure that final speed out first by rewriting our angular velocity omega recall that our angular velocity. Let's let me make a quick note here call that angular velocity is equal to our speed divided by the radius. So in this case, then that means our equation is going to become MGH equals half mg sorry half MV squared. OK, plus half of our moment of inertia or a system multiplied by V squared divided by R squared. Now here notice that we can factor VV squared from both terms OK. And then divide it by that factor. So if we were to do that here, then we're gonna have V squared multiplied by a half of M total. OK, plus a half of our moment of inertia divided by R squared. So I could rewrite this then to be our moment of inertia divided by two R squared. And no, that means our speed, OK. If we were to divide by this term and the square root both sides, then our speed or linear speed is going to be equal to the square root. Let me write that properly here. The square root OK? Off or initial potential energy divided by a half of M plus E. Well, oh, this should have been a half as well on our previous term for I divide. Well, oh no, I rewrote it. I forgot that I rewrote it my bad. So we can factor out that half and says that the mass plus a moment of inertia divided by R squared. So it's good to keep track of what you're doing, right? So now we have an expression here for our speed. And let's see if we know everything. Let's see if we have all the information. Well, we know the mass of both disks so we can figure out the total mass. OK. We know our acceleration due to gravity, we can take it to the 9.8 m per second squared. We know the height it fell from, we can figure out the moment of inertia because we know the moment of inertia for a solid cylindrical disk is half Mr squared, we don't have that value exactly here. So let's see if we can work it out. Let's see if we can figure this out and then we can substitute it into our formula. OK. So now when we do that, OK, when we do that, let me make some space over here to the right, the moment of inertia for our system. Well, recall, sorry that a moment of inertia for a solid cylindrical disk is a half of Mr squared. So in this case for our system, where if we take M to be the mass of the smaller disk and common R to be its radius, while capital M and capital R to be the mass and the radius of the larger disk respectively. OK. So let me just write those properly here. Then using what we know the moment of inertia for the system is going to be a half of Mr squared plus a half of Mr R squared where one is for the larger disk and one is for the smaller dis, we can factor it a half and then we can substitute our values. OK. So that's gonna be a half of a 0.025 kg multiplied by 0.05 m squared plus the mass of the larger disk, 0.15 kg multiplied by its radius, 0.1 m squared. OK. And so that would be the moment of inertia for our entire system. OK. And when we work that out, we get that moment of inertia to be equal to 0.000781 to 5 kg square meters. And we also know that our total mass is going to be the sum of both masses, 0.025 kg and 0.15 kg, which would be equal to 0.175 kg. So now that we have all this information, we have our moment of inertia, our total mass we can substitute them to solve for our speed. Because now our speed then is going to be equal to the square root of 0.175 kg, multiplied by 9.8 m per second squared, multiplied by a height of 2 m all divided by a half of our total mass. 0.175 kg. OK? Plus our moment of inertia for our system, 0.0078125 kg square meters divided by R squared. So that's 0.05 m squared. Now when we go ahead and solve for our speed, we put this expression into our calculator. You should find that you get it to be 2.5409 m per second or to two significant figures, 2.5 m per second. So that means then that the linear speed of the system just before it hits the ground is 2.5 m per second. And now in our second, the second part of our problem finding our rotational speed as a fraction of the kinetic energy. Then that means we're going to be looking for our rotational speed. Let's call that Kr divided by our total kinetic energy. Now how can we figure that out? Well earlier, remember we said our initial energy equals our final energy and our initial energy was our potential energy. So really this is going to be our rotational energy kinetic energy divided by our initial potential energy. Now we know that this is going to be a half of I omega squared divided by MGH. And if we rewrite omega as V squared divided by R squared, this expression is going to become IV squared, OK. IV squared divided by two R squared. Mtmr total mass multiplied by G multiplied by H. Now we can go ahead and substitute our values that we know. OK. Which if we come down to this line, it's going to be 0.0078125 kg square meters divided sorry multiplied by our speed. We just found 2.5409 m per second. OK? And all of this is divided by two multiplied by 0.05 m squared, multiplied by 0.175 kg by 9.8 m per second squared, multiplied by 2 m. OK? And like we did before, we can put, just put this expression directly into our calculator. Now, when we do that, OK, we should get it to be 0.2941 or approximately 29%. So our rotational kinetic energy is 29% of our total kinetic energy. Therefore, with both of these answers, if we go back to our choices, that means a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Conservation of Energy

Kinetic Energy

Moment of Inertia

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