A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: a. The gear's angular acceleration?

Video transcript

Hi, everyone in this practice problem, we're being asked to find a turbine's angular acceleration at the equals six seconds. We'll have a wind turbine with a diameter of 1200 millimeter spinning at an angular velocity of Omega equals 8.0 minus a quarter T squared radius per seconds. The T is going to be measured in seconds and we'll have to find the turbines angular acceleration at T equals six seconds. And the options given are a 0.7 radiance per second squared. B negative 146 radium per second squared C negative 3.0 radiance per second squared and D seven radiance per second squared. So let's consider the wind turbine as a rigid body that is spinning. We need to note that the angular acceleration in this case is not going to be constant, but it's going to be the derivative of the angular velocity that is alpha or our angular acceleration will then equals to the derivative of omega or our angular velocity with respect to time. So alpha equals D omega divided by D T. We were given the, the angular velocity equation Omega equals to eight minus a quarter T squared in radiance per seconds. Uh We want to actually substitute this equation into our angular acceleration equation that we have here. So that in this case, alpha will then equals to D over D T of eight D minus a quarter T squared just like. So, so in this case, we want to recall some principle of derivation. So the D over D X of a constant C is going to just be zero. And the D over D X of a constant multiplied by X squared is just going to be that constant multiplied by two X. So we wanna utilize that or this two principles in order for us to be able to find this or solve this derivation. So alpha is then going to be D over D T of a constant which is eight, which is going to be then zero. And then that will be minus the D over D T of a quarter which is a constant multiplied by T squared, which in this case will follow the second derivation principles or derivative principles. So that will be a quarter multiplied by two T. So that will give us the alpha value of equals to negative 1/2 or negative half T radiance per second squared. So now we are interested to find the turbine's angular acceleration at T equals six seconds. We find or we have found the equation for the angular acceleration as a function of T So what we have to do is just substitute the T with the time period or the time point that we are interested at. So at T equals six seconds, then alpha will equals to negative half multiplied by six radium per second squared, which will give us alpha equals to negative three radium per second squared. And that will be all for this particular practice video. So the answer or the turbines angular acceleration at P equals six seconds is going to equals two negative three radium per second squared, which will correspond to C or option C with the turbines angular acceleration of negative 3.0 radium per second squared. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.