An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev/s^2. (c) What is the tangential speed of a point on the rim of the turntable at t = 0.200 s?

Video transcript

Welcome back everybody. We have a cyclone that is being used for roof ventilation. We're told that this cyclone has a diameter of 60 cm or .6 m. Well that when it rotates it has a pretty steady angular acceleration of 0.2 revolutions per second squared and an initial angular velocity of 0.45 revolutions per second. After a time period of 0.8 seconds have passed. We are tasked with finding the tangential velocity at any point on the outside ring. So we know from formulas in centripetal motion that the tangential velocity is equal to radius times angular velocity. But this case is going to be our final angular velocity. So we need to find this guy and we're going to use this cinematic formula for it. We're gonna use that. Our final angular velocity is equal to our initial angular velocity plus our angular acceleration times, time plugging into values here we get 0.45 plus 0.2 times 0.8 is equal you. 0.560 revolutions per second. But in order for this equation to work, we actually need this in radiance per second. So I'm gonna multiply this by two by radiance for one revolution. These units will cancel out and we are given a final angular velocity of 3.52 radiance per second. Great! Now that we have that we can go ahead and plug this in here. Have that. Our tangential velocity is equal to our radius times our angular velocity Equal to .3 because it's just half of our diameter Times 3.52, which gives us the final answer of 1. m/s corresponding to our answer choice of the Thank you all so much for watching. Hope you all like the video. We will see you all in the next one.