CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rads. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. (d) How much work was done in pulling the cord?

Question

CP A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. (d) How much work was done in pulling the cord?

Accepted Answer

everyone in this problem. We have a child's toy. It has a cube that rotates on a level board with negligible friction. Were given some information about the cubes mass. Okay. We're told that the string that is rotating at the end of has negligible mass and a child is going to reduce the radius of rotation. Okay. They're gonna pull on the string. What we want to know is calculate the work done by the child in shortening the string. Okay. And we're told to treat the cube like a point mass. Alright, so what we have is we have the mass of this rotating Cube is 0.35 kg. Now we're told that it's rotating at 3.2 radiance per second. Okay, so its initial speed Is 3. radiance per second. And it's gonna be .45 m from the hall. Okay, so .45 m from the axis of rotation. So it's initial radius is 0. m. And when we're talking about after the child shortens the rotation, Okay, they're gonna pull that string. It's gonna shorten the string. Therefore shorten the radius of rotation. We're gonna have the same mass. The mass hasn't been changed, so 0.35 kg. The omega. Well, we don't know yet. And the radius of rotation at the end we're told is 0.22 m here. Okay. Okay, so this is the information we know right now. We don't know, Michael Finally. Now what we're trying to find is we're trying to find the work done by the child. Okay, so a child is going to be a non conservative force because we have the work done by non conservative forces. Now let's recall. We have the work done by non conservative forces plus the kinetic energy initially plus the potential energy initially equal to the kinetic energy, final plus the potential energy. Final. Okay, this is our conservation of energy. So, our work done by non conservative forces, our initial kinetic energy. And our initial potential energy is going to equal the final kinetic energy and the final potential energy. Alright, so that's our conservation of energy. Now, we're looking at these terms, we know we don't have any potential energy. Okay, we are on a level board. So we're not dealing with any sort of gravity. We don't have any spring forces. Um string has a negligible mass. Okay, so all of these potential energies are going to go to zero. Okay. In this non conservative work is therefore just going to be equal to K final minus K initial. Okay, it's just gonna be the change in kinetic energy. And that is what we're trying to find that that work. Alright, now, let's recall What is the kinetic energy kinetic energy? If you're talking about linear motion is one half mv squared. Okay, When we're talking about angular motion, Well, we have one half I omega squared. Okay, where is the moment of inertia? And omega is angular velocity. So this is going to give us one half. I will call it my final omega f squared minus one half I initial omega initial squared. All right. Now looking at these terms So we need to find these moment of the nurses. I Okay. So let's go ahead and do that. We're gonna do it up here. So if we're looking at the moment of inertia, I initially we're told to treat the cube like a point mass. Okay, so recall you can use your moment of inertia table. Okay. That's either in your textbook or that your professor professor provided and the moment of inertia for a point mass is going to be M R squared. So in this case we have our initial So we're gonna have 0. kg Times 0.45 meters. All squared. Okay. And that's gonna give us a moment of inertia of 0. kg meters squared. Alright, so we have I final what about er sorry I initial what about my final? Well I final is going to equal again the same thing. We're treating it as a point mass. M R squared. Okay. In this case our is the our final The mass is still 0.35 kg. The radius after the kids shortens that string is 0.22 m we have 0.22 m squared. And this is going to give us a moment of inertia of 0. or Okay, kilogram meters squirt. So the moment of inertia by decreasing that radius we've decreased the moment of inertia. Alright, so we know I final we know I initial we know the initial angular speed. Omega. I What we don't know is this quantity omega F. Okay. The final speed in order to find the non conservative that work done by shortening the string. We need to know the final speed. Okay, Alright, so how can we find that final speed? Well, let's recall that we have conservation of momentum in this case we have conservation of angular momentum. Okay, because we have no friction negligible mass of the string. Okay, we have no net external torque. So we have conservation of angular momentum. Alright, conservation of angular momentum tells us that the initial angular momentum L. Is going to equal the final angular momentum. L. F. Recall angular momentum is given by I omega. So I have I initial omega initial is equal to ay final. Oh my God, Final all right, let's move down just a little bit. So we have some more room to work and substituting in these values. We know that the initial Moment of Inertia I is 0.070875. The initial angular speed we were given is 3.2 radiance for a second. The final Moment of Inertia. K. in Omega F. That quantity that we're looking for. Alright, so on the left hand side we are going to get 0.2268. On the right hand side, we still have 0.1694 times omega F. And this is going to give us an omega F. Value Of 13.38843. Okay, radiance per second. Okay. And so this makes sense. Right with our conservation of momentum, we've decreased the moment of inertia by decreasing the length of the string and so the angular velocity has increased. So we've decreased one quantity, the other quantity increases. Alright so we have our oh my God. Now let's go back to our equation for the work done that we are trying to find. Okay we have one half the moment of inertia. Final is 0.01694. The final omega Angular velocity 13.38. That we just found. 13.388 squared. And then I'm just gonna finish this subtraction down here so we don't run out of room on the right hand side. So we have minus one half the initial moment of inertia. That's going to be 0. 0875. And our initial angular speed 3. squared. We're getting this out. We're gonna get 1. 1825 minus 0.36 to 88. Okay this is gonna give us a final answer of 1.15537 in our unit is jules. Okay, So that is our work done by shortening that string. Okay, so that is going to correspond with answer B 1.15 jewels. Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Angular Momentum

Work-Energy Principle

Centripetal Force