The two blocks in FIGURE CP12.86 are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N m. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Question

The two blocks in FIGURE CP12.86 are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N m. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Accepted Answer

Hey, everyone in this problem, we have a 3 kg bag resting on the ground tied to a vertical cord. The cord of negligible mass passes through a cylinder with a groove in which a cord can run. The other end of the cord is attached to a 6 kg ball shown in the figure. Yeah, we'll take a look at that figure in just a minute. The cylinder is four centimeters in radius and has a mass of 1.25 kg. The friction force above the rotational axis produces a torque of one Newton meter. And we are asked to calculate the time needed for the 6 kg ball to hit the ground. If it is released from a height of 1.25 m without initial speed. We have our diagram here. We can see that we have our cylinder a radius four centimeters mass of 1.25 kg. And we have this cord running over it. We have our 3 kg bag resting on the ground and then our 6 kg bag off of the ground by 1.25 m. It's going to get released and it is going to fall downwards and we wanna figure out how long that takes. So a couple things we wanna label on this diagram, we're just gonna label the tensions. We're gonna say that the tension in the 3 kg cord is T one and the tension in the 6 kg cord is T two. We're gonna label those two tensions there. Now we're looking for a time, we have some information about height, an initial speed. So immediately we can think about our kinematic equations. And so if we have our kinematic equations, there are five variables we wanna consider. So we know that the knot is 0 m per second because it's released from rest. We don't know the final velocity when it hits the ground. We don't know the acceleration in the Y direction. And that's gonna depend on the torque. It's gonna depend on the tension. It's gonna depend on all of those forces acting. It's not gonna be subject to gravity only. We wanna find the time tea and we know that we have a height delta Y of 1.25 m. And if we call that vertical distance traveled a positive value, then what we're saying is downwards is being chosen as our positive direction. So we'll make that choice now and then we can use a positive value there. So if we look at this, our kinematic equations, we need three known values and we don't have three known values. So we need to figure out one of these values that we don't know in order to find T and the value we're gonna choose is A Y. And we know that that acceleration is related to the torque is related to the forces. And so we can use some of those other ideas to find this acceleration. And then we'll come back to calculating our time t using our kinematic equations. So let's start with the torque and we know that the summer of the tour is gonna be equal to the moment of inertia. I multiplied by the angular acceleration alpha. OK. This is a rotational version of Newton. Second law. When we think about the torques, we have acting, say we're gonna have torque because of our tension. So because of tension, one, because of tension two, and we also have the torque due to the friction in the cylinder. Now, if we think about the torque from tension one, OK. Tension one is pulling downwards to the left. This is going to cause a counterclockwise rotation or a positive torque. So we're gonna have to one, then we have the torque due to the frictional fork. Now, when we release our 6 kg ball, it's gonna be falling down and we can imagine it rotating this cylinder to the right friction is gonna oppose that motion. So it's going to be acting to the left. So again, it's gonna cause a counterclockwise rotation that's gonna be a positive torque. Wait what the torque due to friction and then we have the friction or sorry, the torque from tension T two. And again, T two is coming downwards. This is gonna cause clockwise rotation or a negative torque. So we have minus Tau two. This is gonna be equal to the moment of inertia. I multiplied by the angular acceleration alpha. OK. So what can we do from here? We have this equation in terms of the angular acceleration alpha. We want this in terms of that linear acceleration A Y that value we need for kinematic equation. Now, the torque equation A recall is gonna be equal to the force multiplied by the distance are that it acts from the rotational axis multiplied by sine of the angle between that position vector and the force. Now, in this case, the angle theta is going to be 90 degrees for each of these values sine of 90 degrees is just one. And so we're just gonna have the force multiplied by the distance to the rotational axis. OK. Now, the distance to the rotational axis, if we take a look at our diagram and that's just gonna be the radius R of our cylinder. And so our torque equation is going to be T one R who was the torque due to friction minus T two R. This is gonna be equal to the moment of inertia I and we can relate the angular acceleration alpha to A. Now what we have to be careful with is our sign here. And if we think about a positive value of this acceleration A Y, that means that this ball is coming downwards, the 6 kg ball is falling downwards, then our cylinder is gonna be rotating clockwise. Hey, that's a negative acceleration. And so alpha and A Y are going to have opposite signs. We can recall a relationship between linear values and rotational values and the linear acceleration will be equal to alpha multiplied by R. In this case, it's going to be negative. And so we get alpha is negative A Y divided by R. And that, that's based off of how we define these directions. We've said that that downward direction is positive for a linear motion that leads to clockwise rotation, which is negative if we're talking about our angular acceleration. OK. So we have this equation, we have this A Y value in our equation. Now, we wanna find that A Y value and we know our, we can figure out the moment of inertia I based off of the shape that we have based off of the cylinder, but we don't have tension T one and T two. So we need to find expressions for T one and T two and then we can come back to this equation. So let's do that. And in order to do that, we're gonna draw some free body diagrams. So we can draw a free body diagram of our 3 kg mass. And we know we have the tension T one acting upwards, we have the force of gravity, we'll call it FG one acting downwards. And similarly, for the ball, the 6 kg ball, we have the tension T two acting upwards and the force of gravity FG two acting downwards. And those are the only two forces acting on each of those. Again, we've chosen down to be our positive direction, we have to stick with that consistent direction. So let's start with the first. OK? We can say that the sum of the forces in the Y direction on this first tension, it is gonna be equal to the mass and one multiplied by the acceleration A Y one. And that's Newton's second law. If we think about the forces active, I have FG one minus T one that's gonna be equal to M one A Y one. The force of gravity can be written as M one G minus T one. It's gonna be what's M one A Y one. Now, if we think about this acceleration A Y one, OK. That means the 3 kg block is coming downwards that's actually gonna correspond to a negative A Y value. And so we can say that A Y one is equal to negative A Y two A, if one is going down, the other is going up and we want that downward direction to be our positive direction. And so this is gonna be negative A Y. OK. So that's that negative downward direction. So that A Y value this downward acceleration that we're forcing to be po and by forcing it to be positive, it forces our A 11 A Y one value to be negative. And so when these questions, it's really important to pick a frame of reference and then think about every acceleration, we write down how is it related to the other acceleration? And in this case again, if one is going down the other is going up, so they're negatively, they're inverses inversely related. OK. So if we use that in our equation, we can write M one G minus T one is equal to negative M one A Y which gives us an equation for T one of M one G plus M one A Y, we are gonna do the exact same thing for this second tension. If we do the exact same thing for the second tension, the sum of the forces in the Y direction on this second tension, this is gonna be equal to M two A Y two A Y two is just equal to A Y. So we get FG two minus T two is equal to M two A Y. And we're dealing with that same ball that is falling, we get that positive acceleration. We can write M two G minus T two is equal to M two A Y or eight T two is equal to M two. G minus M two A Y. OK. So we have an equation for T one in terms of A Y, we have an equation for T two in terms of A Y. If we substitute those into our talk equation, we're gonna end up with an equation where the only unknown is A Y. So let's go ahead and do that. So we're gonna have T one M, one G plus M one A Y multiplied by R lost that torque due to friction to ir minus T two M two G minus M two A Y multiplied by R. This is gonna be equal to the moment of inertia. I multiplied by negative A Y divided by R. We're gonna go ahead and divide both sides by R. We can write this as M one G plus M one A Y plus Tau fr divided by R minus M two G minus M two A Y is equal to the moment of inertia. I now this moment of inertia, we're dealing with a cylinder A rotating about its central axis. So this is just gonna be Mr squared divided by two. And then we're multiplying by negative A Y divided by R square, not just R anymore because we divided the entire equation by R, we get an R square. And what we can see is that this R squared term is actually gonna divide it. Now, we wanna go ahead and move all of the A Y terms to one side and all of the constant terms to the other side. And when we do that, we're gonna have M one plus M two plus what's left over M divided by two A Y. That's gonna be equal to M two minus M one multiplied by acceleration due to gravity G minus that torque due to friction divided by the radius R. OK. So this looks messy. But remember we have every single one of these values except for A Y. Let's go ahead and substitute in now and then we can simplify afterwards. The one we substitute in, we're gonna have 3 kg plus 6 kg plus 1.25 kg divided by two multiplied by A Y. That's gonna be equal to 6 kg minus 3 kg multiplied by 9.8 m per second squared minus one Newton meter divided by 0.04 m. OK? Converting that four centimeters to meters. If we simplify and we divide, we're gonna have that A Y is equal to 4.4 newtons divided by 9.625 kg, which gives us an A Y value of about 0.45714 m per second squared. OK. We've gotten this acceleration value. Let's remind ourselves what we are looking for. We are looking for the time it takes for this ball to drop. In order to do that, we needed another piece of information for our kinematic equation. We now have A Y A 0.457 1 4 m per second squared. And we can use our kinematic equations to figure out this time, the kinematic equation we're gonna choose is the one without the final velocity. Because we don't have information about that. Delta Y is equal to V, not T plus one half A YT squared. Substituting in our values 1.25 m is equal to V nat is zero. So the first term goes to 01 half multiplied by 0.45714 m per second squared, multiplied by T squared. We can divide both sides by the one half multiplied by 0.45714. We take the square root, OK? And we get that T is gonna be equal to 2.338 seconds approximately. Remember when we take the square root, we get the positive and the negative root because we're looking at a time, we're just taking that positive route and that's it. That is the end of this question. So we made it and we had to break this up into a lot of different parts. We had to look at the torque, we had to find those tensions in terms of the acceleration we were looking for. And once we had that acceleration, we could use our kinematic equations. If we round this to two significant digits, we can see that the correct answer is gonna be option D 2.3 seconds. 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

Torque and Rotational Motion

Kinematic Equations