A 12-cm-diameter, 600 g cylinder, initially at rest, rotates on an axle along its axis. A steady 0.50 N force applied tangent to the edge of the cylinder causes the cylinder to reach an angular velocity of 500 rpm in 2.0 s. What is the magnitude of the frictional torque between the cylinder and the axle?

Video transcript

Hi everyone. In this practice problem, we are being asked to calculate the torque produced by the constant frictional force. F in this particular practice problem, we will have a 450 g globe in the form of a solid sphere with a radius of 12.5 centimeter initially motionless. The globe is revolving around its central vertical axis under the influence of a constant tangential and horizontal force. Both with a magnitude of 1.25 Newton. The force is applied along the line around the middle of the globe and the angular speed of the globe 15 seconds after the application of the force is 36 R PM. Assuming that there is a constant frictional force f between the globe and the axis of rotation, we're being asked to calculate the torque produced by this constant frictional force. F. The options given are a 0.4 Newton meter, B 0.9 Newton meter C 0.12 Newton meter and lastly D 0.16 Newton meter. So in order for us to solve this problem, we will have to employ Newton second law in the rotational form. If we have multiple torques acting on a rigid body about a fixed axis. Then the sum of the torques will equal to the moment of inertia multiplied by the angular acceleration. So the Newton second law in the rotational form, I'm gonna write down parentheses, rotational, we will have sigma of the torque or sigma Tau to be equals to the moment of inertia, which is I multiplied by the angular acceleration or alpha. In the problem statement, we will have two different torques where the first one is going to be the torque of the applied force minus the torque of the friction. So the Sigma torque will be torque of applied force minus torque of friction because the torque of the friction will be going in the opposite direction of the torque of the applied force. So torque of applied force minus torque of friction will equals to the moment of inertia multiplied by the angular acceleration alpha. I'm gonna write this down as our first equation which we will refer back to in the end. So now looking at this equation right here, we do not know much about like the torques and also the moment of inertia and also the angular acceleration. So because of that, let's start with actually solving all of them one by one. So let's first look into the torque of the applied force. So looking into the torque of the applied force, we want to recall that to calculate torque, we can multiply force and the distance as long as those two are perpendicular to one another. So in this case, we will have R globe or the radius of the solid sphere multiplied that by F applied. And we are going to assume that um the globe and the F applied are perpendicular to one another based on the problem statement. So therefore, we can immediately just equals this and we can substitute all the information given in the problem statement. First, we have the radius to be 12.5 centimeter and that'll be 12.5 times 10 to the power of negative two m. And then we will have f applied F applied will be 1. Newton which is given in the problem statement and they will give us the torque of the applied force to then equal to 0.156 new then meter just like so awesome. So next, we wanna look into the moment of inertia because the torque of uh produced by the fiction is what we are interested to find. So I or the moment of inertia for a globe or a sphere will follow the uh formula of two divided by five, multiplied by the mass and multiply that by R squared. In this case, we know or we are given that the mass is 450 g and the R is 12.5 centimeters. So we can substitute everything into this formula. To get the moment of inertia. So we have two divided by five multiplied by M which is 50 g. And we wanna convert that into kilograms by multiplying it by 10 to the power of negative three kg. And I want to multiply that again by R R S as we have written previously, 12.5 times 10 to the power of negative two m. And they will give us an I or a moment of inertia of 2.813 times 10 to the power of negative three kg meter square just like. So, all right. So we will assume that the acceleration is constant or the alpha is going to be constant. So we wanna use the rotational kinematic equation to actually calculate alpha. Um In the problem statement, it is given that the angler speed off the globe 15 seconds after the application of the force is 36 R PM. That is the information that we're gonna use to calculate alpha. So it is uh also given that the globe will starts from rest. So Omega not our angular speed initial will be zero radiance per second, but Omega F will equals two um 36 R PM or 36 revolution per minute. And we wanna uh convert that into radiant for seconds by multiplying it with two pi radium for every one uh revolution multiply that again by one minute for every 60 seconds. And that will give us the value of omega F or final angular speed to be 3.77 radiance per second. And then it is given that the T or delta T is seconds in the problem statement. Awesome. So next, we will use the rotational kinematic equation in order for us to get alpha. So in this case, the rotational kinematic equation that we're gonna use as omega F equals to alpha T plus omega knot or omega zero. So rearranging this, you will get an equation for alpha to be omega F minus Omega knot. All of that divided by T substituting the values in that will equal to 3.77 radiance per seconds minus zero radiance per second divided by 15 seconds. And that will give us the angular acceleration of 0.2513 radiant per second squared. Awesome. So now we have all the information needed to solve for the torque cause by the friction by using equation one. So we wanna recall equation one again, I'm gonna write that equation one and equation one will have torque of the applied force minus talk of friction to be equals to I multiplied by alpha. The torque of the applied force is 0.156 Newton meter. Talk of the friction is asked I or the moment of inertia is 2. times 10 to the power of negative three kg meters squared and alpha is 0.2513 radium per second squared and then rearranging everything we will then have torque caused by the friction to be then equals to 0. Newton meter or rounding it up 0.16 Newton meter just like so awesome. So the torque produced by the constant force of friction F is going to be 0.16 Newton meter which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics. That'll be it for this one. Thank you.