CP A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. (b) With what speed does the bucket strike the water?

Question

Accepted Answer

Welcome back everybody. We are taking a look at a uniform disk here. Now, around this uniform disk is a tightly wound string. And at the bottom of this string is a jewelry box. Now we are told a couple of different things about this system. We are told that the mass of the box is one kg. We're told that the mask of the, sorry the mass of the disc is six kg. Were told that the radius of the disk is 10 centimeters or 100.1 m. Now we let go of the box initially, it's all the way up here, Right? But we let go of the box and we are told that the box starts at rest and then it falls about one m right before hitting the ground. And we are tasked with finding what is the speed of the box right before hitting the ground. Well, in order to find this, let's use a kid a Matic equate. Here, we have a enigmatic equation that says that the final velocity is equal to the square root of our initial velocity squared plus two times our tangential acceleration times the distance covered. We know the distance covered. Be one. We know the initial velocity to be zero. We need to figure out what this tangential acceleration is. Now, what does tangential acceleration even mean here? Right, well, I'm actually going to draw out a free body diagram of our jewelry box. So we have acting on this jewelry box right here, we have a tension of the string pulling upwards and the weight of the box pulling downwards or the, you know, force due to gravity here are tangential acceleration is going to be acting like this, it is the acceleration of the box going down. And I say tangential because it is tanja to the uniform disk at that point. So that's what we need to find here in order to find that we're going to have to relate the centripetal forces to the linear forces here. So here's what we're gonna do. We're going to observe and have to come up with a system of equations to solve for our tangential acceleration. So let's look at this. We know that the sum of the centripetal forces or in this case the torque right, is going to be the moment of the inertia for the uniform disk times it's angular acceleration. We also know that the sum of all forces in this direction, in the up and down direction is equal to mass times acceleration. So from this will be able to solve for our tangential acceleration. And let me show you how. So first, let's look at this formula right here, our torque is just going to be our force due to tension right, Which is pulling downwards times the radius of our disc. This is going to be equal to the moment of inertia for a uniform disk, which is one half times the mass of the disk, times the radius squared times our angular acceleration. Well, I want to relate angular acceleration to our tangential acceleration and we do that simply by dividing by the radius. So we can see here that one of the powers of this radius in this radius cancel out. And then if we divide by the radius on both sides, these terms cancel out as well. So we have that are tension force is equal to one half times the mask of mass of our disk times our tangential acceleration. Now we don't know what our tension force is, which is why we're going to use this second formula or Newton's second law. Right. Another variation of it to find out what our tangential acceleration is. So let's go ahead and break this down now using our force body or our free body diagram, right, we know that the summation of forces is going to be the mass of the box times the acceleration due to gravity minus the tension and the reason minus although it's going up are tangential acceleration is going down. So I'm gonna make the downward direction positive explaining these signs here, this is equal to our mass times are tangential acceleration rearranging some terms. Here we get that our attention is equal to the mass of our box times the acceleration of gravity minus the mass of the box times the tangential acceleration giving us the mass of the box times g minus A. So now, as you can see we have two equations for the tension. So now we can set them equal to one another and then solve for our tangential acceleration term. So we have one half times the mass of the disk times are tangential acceleration is equal to the mass of our box times. Gravitational acceleration minus are tangential acceleration using some algebraic manipulation here we are able to achieve that are tangential acceleration is equal to M B times G. All over em B plus M D over two. Let's go ahead and plug in all those values here we have one times the acceleration due to gravity which is 9.81, all divided by one plus six divided by two. When we plug all this in, we finally get that are tangential acceleration is 2. m per second squared. Now that we have that we are good to go back to this equation right here to solve for our final velocity of the jewelry box right before it hits the ground, we have that our final velocity is equal to the square root of our initial velocity squared plus two times acceleration Of 2. times the distance covered, which is just one m and we get that our final velocity is equal to 2.21 m/s corresponding to our final answer choice of B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

Verified Solution

Key Concepts

Conservation of Energy

Rotational Dynamics

Kinematic Equations