A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 15 seconds. How much friction force does the brake pad apply to the shaft?

Question

Accepted Answer

Hey, everyone in this problem, a revolving cylinder that transmits power to a spin coder has a radius of two centimeters and a mass of 75 g. One end of the cylinder is attached to the center of a circular plate. The plate has a mass of 600 g and a radius of 15 centimeters. During an experiment, the spin coder is revolving at a constant angular speed of 165 R PM in an emergency. You need to stop that disk. So you shut off the electrical power and apply a tangential frictional force F to the cylinder. The plate stops in eight seconds and we are asked to calculate the magnitude of that force F. We're given four answer choices all in Newton's option A 0.73 option B 1.9 option C 5.8 and option D 6.9. So what we're looking for is the magnitude of this frictional force F. So we're given some information about the motion, the angular speed of this. Um We have some information about the shapes. So let's think about how we can relate those to the and recall it the torque tow can be written as the moment of inertia eye multiplied by the angular acceleration alpha. And this is also equal to the force F multiplied of the radius R multiplied by science data. So we can see that we have this force F that's gonna be that frictional force that we're looking for. And we can write that in terms of I alpha R and sine theta. So we can write this first half because I alpha divided by R sign date. Now we know R and we know the radius of this cylinder. We're told that in the first, we know the angle theta. OK. That makes between the object and the rotation. So we can calculate I and alpha, we'll be able to calculate this force. All right. So let's go ahead and write out all of the known values that we have. We're told that the mass of the cylinder is g. Now, we wanna convert to our standard unit of kilograms. So we're gonna go ahead and divide by 1000 which gives us 0.075 kg. The radius of that cylinder we're told is two centimeters dividing by 100 to get to our standard unit of meters that gives us 0. m. Next we can move on to the plate. OK? And we're gonna use subscript P for the plate. We use subscript C for the cylinder. So the mass of the plate and P we're told is 600 g divided by 1000 to convert to kilograms, gives us 0.6 kg. And then finally, the radius of the plate RP is 15 centimeters, converting to meters by dividing by its 0. m. So those are some of the known values we're given also to that don't make a knot and the speed is 165 A P, that's a constant speeded spinning act before you apply that force to slow it down. And we can convert this into radiant per second. So this is gonna be 165 revolutions per minute, multiplied by two Iranians per revolution, multiplied by one minute per 66 the unit of revolution and minute divides them. We're left with radiance per second and we have 165 multiplied by two, divided by 60 multiplied by pi. And this is gonna give us 11 I divided by two gradients per se. All right. So this is what we were given in the. Now, let's think first about finding and how can we find this angular acceleration? Well, let's think about our equations are kinematic or U AM equations for the angular motion. OK. We know Omega, we just wrote Omega T down. So we know that Omega is gonna be 11 pi divided by two radiant per second. We know that the final angular speed Omega F is going to be zero ratings for a second because this force brings that cylinder to a stop and it brings the plate to a stop. We're told that the time it takes for it to stop is eight seconds. And we want to find what the acceleration was that we apply. So we have Omega no Omega F and T three known values. One thing we want to find out, we're gonna choose the equation with these four variables. So we get Omega F is equal to Omega nine plus alpha T substituting it. Our value is zero radiance per second is equal to 11 pi divided by two gradients per second plus alpha multiplied by eight seconds. You can move our alpha multiplied by eight seconds to the left-hand side. By subtracting, you get negative alpha multiplied by eight seconds is equal to pi divided by two radiant per second, divided by negative eight seconds. We get that alpha is approximately equal to negative two 0.16 radiance per second squared. OK. So we have our alpha value now and it makes sense that it's negative and this is an acceleration that we're applying that is slowing this down to a stop. So that negative acceleration makes sense. Now we need to move, move to the moment of inertia. Ah hm Let's calculate that as well before we get back to our force equation. So the moment of inertia I gonna be made up of the moment of inertia of the cylinder plus the moment of inertia of the place. Now, the moment of inertia cylinder recall is given by one half the mass MC multiplied by the radius RC squared. And similarly for the plate, it's made up of that same kind of shape to just flat instead of being longer. So we have one half MP RP squared substituting in our values, this is gonna be one hack multiplied by 0.075 kg, multiplied by 0.02 m plus one half, multiplied by 0.6 kg, multiplied by 0.15 m squared. If we work this all out, we get a moment of inertia of 0. kilogram meters squared. And that is our moment of inertia. If we go back up, we look at what we found, we found our moment of inertia. I, we found alpha, we know R and theta so we can get back to this force equation and calculate what we wrote. So the force F is going to be equal to the moment of inertia. I 0. kg meter square multiplied for the angular acceleration, negative 2.16 radiance per second squared divided by the radius. And here we're using the radius of the cylinder. OK? Because that's where that force is applied. It's applied to the cylinder in order to stop the plate. So that's gonna be 0.02 liters multiplied by time of 90 degrees. And we're told that this is a tangential force. So this is going to be applied at a 90 degree angle to that cylinder. We have that angle of 90 degrees. Yeah, we work all of this out. We get a force F of negative 0.73062 new. Now, the question was asking for the magnitude of the force. So what we wanna do is go ahead and take the absolute value they define the magnitude. And so the magnitude of the force F is just going to be approximately 0.73 nodes. If we go up and compare this to our answer choices, we can see that this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 15 seconds. How much friction force does the brake pad apply to the shaft?

Verified Solution

Key Concepts

Torque

Moment of Inertia

Friction Force