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. (c) Find the change in kinetic energy of the block.

Question

Accepted Answer

Hey everyone welcome back in this problem. We have a block that's rotating on a flat bench with negligible friction. It's going to be held by a massless string. The block is gonna be rotating and some radius. Okay. And what we're gonna do is we're gonna shorten the string and we want to find what the change in kinetic energy of the block is. When we do that, we're told to treat the block like a particle. You know when we hear treat it like a particle, what we think of as Okay, we're treating it like a point mass. Okay, So let's go ahead and fill in. The details were before The string is shortened and after the string is shortened. Now before the string is shortened, we have a block of mass 0.02 kg. And that's not gonna change. Okay, after the string is short and we still have the same block. It's gonna have the same mass 0.02 kg. All right, Know before we shorten the string, we are told that it has a radius of 0.35 m. Okay, so big. Our initial is going to be 0.35 m And then it has an angular speed of 1.95 radiance per second. So, Omega I it's gonna be 1.95 radiance per second. Okay? But we shorten the string. We shorten it to a radius of 0.18 m. So our Final will be 0.18 m swoops. And this should be in blue. Our final 0. m that relates to the the after the shortening. And we don't know anything about omega final. Okay, Alright. So let's get started with what we're asked to do. We're asked to find the change in kinetic energy which is delta K. Which is gonna be K final minus K initial. Now recall, we have rotational kinetic energy here. So when we're talking about rotational kinetic energy, instead of having one half mv squared, Like in the linear case, we have one half I omega squared. So for each of these terms one half I the moment of inertia times omega the rotational speed and same thing for K I I the moment of inertia and omega the rotational speed. Okay, do we have our equation here? Well, let's try to figure out what these values are. Okay, so before the string is shortened, our eye initial. Now again we're being told the block is like a particle. So we're we're treating it as a point mass. So when we do our calculation for the moment of inertia, we're gonna use the formula for point mass. Okay, so we have I initial is going to be M R squared. Okay, In this case we have our initial squared So we have 0. Times 0.35 Squared. And this is kilograms times meters squared. So our unit is gonna be kilogram meters squared. Ok. And this is going to give us 2.45 times 10 To the -3 kgm squared. And similarly for the moment of inertia after the string is shortened to that same block that we're treating as a point mass. So we get m r f squared, 0.02, I'm 0.18. All Square. And this time we're going to get 6.4, 8 times 10 to the minus four. And again, kilogram meters squared. Alright, so, back to our kinetic energy equation as delta K that we're looking for. We have I. F. We don't have omega F. We have I initial we have omega initial. So the only thing we need now is to find what is this omega F value. Well, let's use conservation of momentum. And we know that we have no external forces, so we have conservation of angular momentum. Okay, so we can write L initial is equal to l final. Okay, the initial angular momentum is equal to the final angular momentum. Now recall angular momentum. Well, that's given by I. Oh my God, Okay. Moment of inertia times the angular speed. Alright, so we have I omega similarly in the linear case we have envy. Here we have I omega. So I F. Oh my God, F. And again this omega F. That's what we're looking for. So filling in the values using what we just found, we have 2.45 times 10 to the minus three kg meters squared. Okay, omega initial is 1.95 ingredients per second. I final 6.48 times 10 to the minus four kg meter squared. And omega. Okay, so what we'll see, we're going to divide by the 6.48 times 10 to the minus four. And the unit of kilogram meters squared is going to cancel. We're gonna be left with omega final equals If we multiply 2.45 times 10 to the minus three times 1.95 ratings per second, We're gonna get 4. 77, 5 times 10 to the -3. Okay. And let's just give us a little bit more room here too. Right? We're going to divide that by the 6.48 times 10 to the minus four. And our unit is radiant per second. Again, the kilogram meter squared canceled. And we're left with the radiant per second. All right. And this is going to be equal to 7. radiance per second. That's omega final. Okay. And that's the value that we needed. That pink purple value that we need. And when we look at this, the angular speed after, we shorten the rope 7.373. That's bigger than the angular speed that we started with. And that makes sense. Right? We have conservation of momentum. So if we are decreasing the moment of inertia by decreasing the radius, then we're going to increase the angular speed. Omega. Okay, Alright, so back to our delta K. Which we're looking for change in kinetic energy. Now we know omega F. So let's go ahead and plug in our values. We have one half I. F. Which was 6.48 times 10 to the minus four. Omega F. Which was 7.373 squared. Okay. And just for the purpose of space, I'm gonna omit the units here. Just until the end. We're working with kinetic energy. So the unit will be jewel A 2.45 times 10 to the -3 And I'll make it initial 1.95. Okay. And it's a good idea to include the units and just be sure that you have the proper units as you work through your work. Alright, so this is going to give us 0.0176. On the left hand side. 0.00466. Okay, subtracting the two, we get 0. jules. Alright, so that's our change in kinetic energy and our kinetic energy has increased the changes positive. So that means our kinetic energy has increased which makes sense since we are increasing our speed and that is more waited in the kinetic energy equation. So 0.129 jules. That's going to correspond with answer D. I hope this video helped. Thanks for watching everyone

Verified Solution

Key Concepts

Kinetic Energy

Conservation of Angular Momentum

Work-Energy Principle