A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in
FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Question

A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in 
FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Accepted Answer

Hey, everyone in this problem, we have an 850 g toy sled that glides in a circle on a bench, the coefficient of kinetic friction of 0.4 with the help of a 1.5 m long thread, a fan fitted to the toy thrusts air creating a six new force perpendicularly to the thread. We're told that the thread can withstand a force of 90 newtons before breaking, gonna start from rest. And we're asked to determine the number of cycles or circles completed by this toy slide. Before the thread snaps, we have four answer choices. Option A 2.68 revolutions, option B 16.8 revolutions, option C 106 revolutions and option D 8.42 revolutions. All right. So we have a diagram here. We have this circle that this choice slot is making. It's connected to a thread of 1.5 m to the central axis. And then we have the fan causing counterclockwise rotation. Let's go ahead and draw some free body diagrams. So let's start with a top down view, kind of like the diagram that we have if we're looking at this from the top down, we have our sled and the force of the band that is going to push the sled forward. So the force of the fan and then we have friction which is gonna oppose the motion and slow this fan or sorry, the slow this sled down. Now, we also have a 10 force acting towards the center in that centripetal direction and we have a centripetal acceleration pointing towards the center as well. We're gonna call that inward direction towards the center as our positive direction. Now, let's think about what we see if we look at a side view of our toilet, if we have a side view of our toilet, this is resting on a bench. So there's gonna be a normal horse that's going to point upwards. We have a force of gravity pointing downwards and then we have the tension and we're gonna have that pointing to the right again to the middle. Assuming that that sled is kind of oo on the side that it is in the diagram right now. All right. So we wanna know how many circles this sled makes. There's a lot of things we need to figure out before we can do that. OK. And first one is, what is that maximum speed? Hey, this SLED is pointed, speed up and speed up and speed up. What is the maximum speed that we reach before we are going to break the word? All right. So let's start there and that's gonna be step one. OK? So step one is going to be looking at the cent triple forces. OK? We wanna look where we have this tension force. So we're gonna look at the centripetal courses. We know that the sum of the centripetal forces is going to be equal to the mass multiplied by the centripetal acceleration. The only centripetal force we have is the tension. So we get that the tension T is equal to the mass multiplied by the centripetal acceleration. Now recall that this trip acceleration can be written as V squared divided by R and also recall that V can be written as R omega. OK. Remember we wanna find omega this angular velocity. OK. So we want to write this in terms of, oh my God, not in terms of this in triple acceleration, not in terms of the tangential velocity. OK. So we have the tension is gonna be equal to m multiplied by R omega squared divided by. Now, now we're gonna look at the maximum tension so that we can find the maximum speed. OK. So the maximum tension it's gonna be equal to the maps multiplied by the radius. R multiplied by the angular speed omega max squared. All right. So this is gonna give us a way to find that maximum speed that we reach before the thread breaks. Once we know that we can look at how many rotations this makes uh So our attention substituting in our values we're told is 90 noons that maximum force that we can withstand. This is gonna be equal to the mass 0.85 kg multiplied by the length of our string 1.5 m. And then we're multiplying by Omega Max squared. We work this out. OK. Omega max squared is going to be equal to 90 noons divided by 0.85 kg multiplied by 1.5 m. Can we just divide it by everything? On the right hand side, we take the square root and we get that Omega Max, it's gonna be about 8.40 168 radiance per second. OK? We found this maximum speed angular speed that we can reach before the thread will break. How far do we travel? Now, before we reach that speed? So let's think about our kinematics now and let me just label these steps so that they're in order. So we have step one. Now we're moving to step two and that's gonna be our kinematics. OK? Now, we're told that this starts from rest. So our initial speed angular speed, Omega Nat is gonna be zero radiance per second. We know that our final speed is gonna be 8.40168 radiance per second. OK. That's the maximum speed we reach before the thread breaks. We want to find delta, the OK. That's gonna give us the number of radiant that are completed and we can convert that into revolutions. OK. So delta, the is what we are looking for. We don't know the angular acceleration alpha, we don't know the amount of time this takes tea. So now we have a little problem, we only have two known values. In order to use our kinematic equations, we need to have three known values. OK. So we could find alpha or we can find tea and we don't have any other information about tea right now. So let's try to deal with al alpha. So let's move to step three and recall that the tangential acceleration A T is related to the angular acceleration alpha through the following A T is equal to our alpha. Yeah, if we look back at our diagrams that we drew our free body diagrams, we've used the first one, hey, the centripetal view of things, but we also have this second view where we're gonna get this tangential acceleration. OK? And sorry, not from the second picture but from the second direction in the first picture is what I wanna indicate here. OK. So we're looking at this other direction in our first picture, this top down view to get our tangential direction. All right. So let's try to find our tangential acceleration that will allow us to find our angular acceleration alpha and get back to finding the number of circles completed. So just like before we have that the sum of the forces in the tangential direction will be equal to the mass multiplied by the tangential acceleration. The forces we have in this tangential direction while we have the force of the fan in the positive direction, OK, moving the slide forward and then we have the friction force posing that motion in the opposite direction. The force of the fan minus the force of friction is gonna be equal to the mass multiplied by the acceleration, tangential acceleration. Now, the force of the fan is something we're given in the problem. We're gonna leave that alone, the force of friction this SLED is moving. So we're gonna be dealing with kinetic friction. OK? So this is gonna be equal to me. OK? Multiplied by the normal force and this is equal to the mass multiplied by R multiplied by alpha. And remember alpha is what we're looking for right now. Now we know the force to the fan. We know the kinetic friction coefficient. We know the mass and the radius. What we need is the normal force and the normal force is gonna come to play from that second diagram. The second free body diagram we drew OK. Now the sum of the forces in the Y direction, if we're looking at a side view is gonna be equal to zero because this to SLED is not going up or down off the table. It's just staying on this bench, not moving in the vertical direction. So we have in our positive direction, the normal force f in the negative direction, the force of gravity FG so N minus FG is gonna be equal to zero. This tells us that the normal force is equal to the force of gravity, which is equal to the mass multiplied by the acceleration due to gravity. OK. So that is our normal force and that we were looking for there. OK. So one more line with variables and then we're gonna sub in some numbers here. So we have the force from the fan minus the coefficient of kinetic friction. UK multiplied by the normal force which we found to be m multiplied by G is equal to m multiplied by R multiplied by alpha subbing in some numbers six notes minus 0.4 multiplied by 0.85 kg, multiplied by 9.8 m per second squared is equal to 0.85 kg multiplied by 1.5 m multiplied by alpha. Now we can simplify. On the left hand side, we get 2.668 oops, we have our units as kilograms meter per second squared. This is gonna be equal to 0.85 kg multiplied by 1.5 m multiplied my alpha. When we divide by everything on the right hand side, we get that alpha is going to be equal to 2.09 25, 49 radiance per second squared. OK. So this is the alpha value that we were looking for. We needed that angular acceleration in order to use our kinematic equation. So let's now go back to our kinematic equations. We have omega not omega F and now alpha which is 2.09 25, 49 radians per second squared, that's three known values. So we're gonna choose the equation that has those three as well as delta theta. So we're moving to step four here and we're actually gonna solve a kinematic problem. And the equation we're gonna use Omega F squared is equal to omega knot squared plus two alpha delta beta substituting in our values 8.40168 radiance per second, all squared is equal to two multiplied by 2.092549 radians per second squared delta theta. And this gives us a delta theta value of about 16 0.866 565 radius. All right. So we're really, really close. We found the displacement this angular displacement of our Tole. So we know how many radiance it travels. We now just need to convert that into revolutions to match with our answer choices. So let's go ahead and do that. And we have step five, our final step in this problem. OK. So we are gonna take the number of radiance 16.866565. We're gonna multiply and we know that there is one revolution for every two pie radiant one revolution divided by two pi radians, the radiance will divide it and we get approximately 2.68 or four revolutions. All right, we got to the final answer. Let's do a quick recap because that was a long one. Hey, but we just needed to work through it step by step using all of the information. So the first thing we did was use the centrifugal centripetal direction story to find the maximum speed that our toy sled could go without breaking the thread. OK? So that maximum speed right before the thread breaks, once we knew that we knew we wanted to use our kinematic equations to find the dissuasion delta theta. However, we needed another very. So we then used our tangential direction to find the angular acceleration. OK. Noting the relationship between tangential acceleration and angular acceleration, we could then go back to our kinematic equations, use that acceleration to find delta the. And then we just needed to convert it from radiance to revolutions. In that final step. If we compare this to our answer choices, we can see that the correct answer is going to be option a 2.68 revolutions. Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Centripetal Force

Tension in the Tube