Flywheels—rapidly rotating disks—are widely used in industry for storing energy. They are spun up slowly when extra energy is available, then decelerate quickly when needed to supply a boost of energy. A 20-cm-diameter rotor made of advanced materials can spin at 100,000 rpm. b. Suppose the rotor's angular velocity decreases by 40% over 30 s as it supplies energy. What is the magnitude of the rotor's angular acceleration? Assume that the angular acceleration is constant.

Video transcript

Hi everyone in this practice for album, we're being asked to determine the the magnitude of a disc's angular acceleration where the disc will have a tungsten carbide ceramic bearings, which will allow it shafts to spin at speed as high as 10,200 R PM. If a 22 centimeter disk loses 35% of its rotational velocity in 20 seconds. As it does work on a load, we're being asked to determine the magnitude of the discs, angular acceleration taking an account of the angular acceleration being constant. The options given are a 72.1 radiant per second squared. B 53.4 radiant per second squared, C 34.7 radiance per second squared and D 18.7 radiant per second squared. So in order for us to actually solve this problem, we will model the disc as a rigid object with a constant angular acceleration. So we are given the initial and also the final rotational or angular velocities. The initial uh angular velocity which I will represent with an omega omega knot is going to be 10,200 R PM. And the final angular velocity is given when it is given by the information that the disc will lose 35% of its initial rotational velocity. So essentially omega or the final angular velocity is going to be 100% minus 35% of omega zero or the initial angular velocity which will equals to 65% of omega dot or omega zero. So the time that it takes for the disk to experience this is 20 seconds and we are required to determine what the angular acceleration or alpha is. So in this particular practice problem delta theta is unknown. So we wanna pick a rotational kinematic equation without delta theta. So the kinematic equation that we are going to use or rotational kinematic equation that we are going to use is then going to be just like this one. So Omega will equals to Omega knot plus alpha V just like. So, so we actually know what Omega is what Omega knot is and what the T is. So we can pretty much um substitute everything that we know into this rotational kinematic equation right here. So first, what we want to do is to just rearrange this so that we get an equation for alpha. So rearranging this alpha will then equals to omega minus omega knot. And all of that is going to be divided by D. So substituting things, our omega is 65% Omega knot and minus that with an A, the actual omega knot divide that by D which is going to be seconds. And that will give us negative 35% of omega knot divided by 20 seconds. And next, we want to substitute our omega no value, which means alpha will then equals to negative 35% multiplied by 10,200 R PM divided by 20 seconds. However, uh the options given for the angular acceleration is in radiance per second squared. So we wanna convert our R PM into radiant per second. And in order for us to do that, we want to multiply that by two pi radiant for every revolution. So multiplied that by two py radians divided by one revolution and multiplied that again by a one minute for every 60 seconds just like. So, all right, and then actually doing this calculation, we will get a value for our angular acceleration of alpha to be negative 18.7 radiance per second squared. In this case, the negative sign means that alpha is going to be against omega or the angular acceleration is going to be against the angular velocity, which means that it is going to be slowing down. And it doesn't mean uh it doesn't mean that the disk is rotating clockwise or counterclockwise. So the magnitude of our angular acceleration or alpha is going to be 18.7 radiance per second squared, which will actually be the answer to this particular practice problem. So with the angular acceleration of 18.7 radiant per second squared. Option D is going to be the answer to this particular practice problem. So that will be all for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on a similar topic and that will be it for this one. Thank you.