The 2.0 kg, 30-cm-diameter disk in FIGURE P12.65 is spinning at 300 rpm. How much friction force must the brake apply to the rim to bring the disk to a halt in 3.0 s?

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Everyone in this problem. A 1 kg solid cylinder with a radius of 15 centimeters is rotating at an angular speed of 230 R PM about its central frictionless axis. We want to stop the cylinder in 30 seconds by applying a friction force F to the odor edge. And we're asked to determine the magnitude of F that would do this. OK. So we're given four answer choices all in Newton's option, a 0.06 option B 0.1 option C 2.3 and option D 5.6. Now, when we're dealing with this frictional force F OK. We have rotation, we're rotating this cylinder. And so we want to think about the sum of the torques. OK. That's where that frictional force when we're rotating is gonna come into play. So we think about our sum of torques. We know from Newton's loss that this is gonna be equal to the moment of inertia. I multiplied by the angular acceleration alpha. All right. Now, the torque acting well, we're only gonna have one torque to deal. OK. And that's going to be this torque due to this frictional force. And the axis is frictionless. And so that's the only force we have to deal with. Now recall that the torque is going to be the force multiplied by the radius or the distance from that rotational axis that it's acting at multiplied by sign of the angle between those two. So what we have is that friction force f multiplied by the radius R and because of frictions acting on the outer edge of our cylinder, and we are rotating through that middle, multiplied by sign that beta and this is gonna be equal to I and multiplied by alpha. Now, we've said that this torque is positive. Why we've done that? Well, from the diagram, our cylinder is rotating in the counterclockwise direction. OK. So if we're applying friction, that's gonna oppose the motion, which means it's gonna act in the clockwise direction which we would typically take as negative. But remember in this problem, we're asking for the magnitude of F. So the sign there we are OK to just leave it as a positive for now. Um And we're just taking the magnitude at the end. All right. So we have all of these variables we wanna find F OK, we know the radius of this cylinder. So we know R the angle sign of theta. OK. This angle theta is just gonna be 90 degrees. So th sign of theta is going to go to one because this is acting perpendicularly the moment of inertia eye that's something that we can calculate in alpha, that angular acceleration, we're gonna need to calculate that as well. So let's get started. And let's start with that moment of inertia. I in the moment of inertia, I is going to be given by one half multiplied by M multiplied by R squared. They recall that we have a solid cylinder here. And so the solid cylinder rotating about its central axis is going to have this moment of inertia. OK. And you can look in a table in your textbook or that your professor provided for this equation substituting in our values, we get one half multiplied by 1 kg, multiplied by 15 centimeters. We wanna write this in our standard unit of meters. So we're gonna take our centimeters divide by 100. So we have 0.15 m. We're gonna score that. And what we get is a moment of inertia eye of 0. kilogram meters squared. All right. So we have that moment of inertia. Now we are moving to alpha the angular acceleration. How can we calculate that? Well, we're given information about the initial speed. They were told how fast this is rotating. We're told the time it takes to stop it. So we have some final speed a time we're looking for an acceleration. Let's write all these variables out, see what we have. So we have that the initial speed omega knot is gonna be 230 R pm. OK. The initial angular speed we have that the final angular speed is going to be zero. OK? Because we're stopping this cylinder from rotating. So it's gonna come to rest. We're gonna have zero R PM. We're told that it takes 30 seconds for this to stop and we wanna find the angular acceleration out. OK. So we have our kinematic or U AM equations or rotation. We have three known values here. One that we want to find. So that's great. We have enough information. We can go ahead choose the equation that has these four variables and sell for alpha. Before we do that, we're gonna convert again into our standard units. We wanna go from R PM into revolutions or sorry into radiant per second. OK? So that when we do our final calculation for that frictional force, we're getting the unit of Newton that we want. So for R PM, what we're gonna do is we're gonna take it. OK. So we have 230 revolutions per minute. And we're gonna multiply this by two pi radiant per revolution. We know in every revolution there are two P radiant. You can imagine going around a circle once that's gonna allow those revolutions to cancel. And then we're gonna multiply it by one minute, divided by 60 seconds and we know there are 60 seconds in one minute and that's gonna allow that unit of minute to cancel or divide it. And we're left with the units of radiant per second that we want. So we have 230 multiplied by two pi divided by 60 seconds. This is gonna give us 23 pi divided by three gradients per second. When we simplify. All right now, for Omega final, it's easy because it 00 R PM is gonna correspond to zero radiance per second. There's not much we need to do there. All right. So we have our variables in the correct units. Let's get to our equation. And the equation we're gonna use is that Omega F is equal to Omega nine plus LT Omega F is zero radiance per second. That's gonna be equal to 23 pi divided by three radiance per second plus alpha multiplied by seconds. And we can subtract our alpha multiplied by 30 seconds. So we get negative alpha multiplied by 30 seconds is equal to 23 pi divided by three radiant per second. OK? Dividing by 30 seconds on both sides with this negative to isolate for alpha. OK. We're gonna leave this in exact form with this pi value to avoid some round off error. And we get that alpha is gonna be negative 23 pi divided by 90 radiance per second squared. And we get this negative alpha value which makes sense. We are opposing the motion, we are slowing down this cylinder. So that makes sense. That's great. Now, let's go back up and remind ourselves what we are actually looking for. We are looking for this friction force, the magnitude of this friction force that causes this cylinder to stop. Hey, we found I, we found alpha, we know R we know theta so we can now solve this frictional force act. And let's get back to that. So we said that F multiplied by R multiplied by sine data was equal to I alpha. That means that F is going to be equal to I alpha divided by R sine theta. OK. Now again, theta is 90 degrees. So sine theta is just one. We just have I multiplied by alpha divided by all. This tells us that our frictional force F is gonna be equal to that moment of inertia we found which was 0. kilogram meter squared multiplied by the angular acceleration. We found negative 23 pi divided by 90 gradients per second, divided by the radius which we're told is 15 centimeters. We've already converted this to meters to do our moment of the nursery calculation. So we know that that is equivalent to 0.15 m. Now, when we work all this out, OK, we get F is going to be negative and what we're gonna do is take the magnitude, OK? We were asked for the magnitude. So we're gonna take the absolute value of this frictional force. But what we find is that it's equal to about 0. nos, right. So the magnitude of the frictional force that we need to apply in order to stop this cylinder in 30 seconds is gonna be 0.06 newtons which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

The 2.0 kg, 30-cm-diameter disk in FIGURE P12.65 is spinning at 300 rpm. How much friction force must the brake apply to the rim to bring the disk to a halt in 3.0 s?

Verified Solution

Key Concepts

Angular Velocity

Moment of Inertia

Torque and Friction Force