A 200 g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75 rpm. (b) What is the tension in the string?

Video transcript

Welcome back, everybody. We are taking a look at a ball that is rotating around a central axis here. We're told that the mass of the ball is 300 g or . kg. And we are told that it has a steady angular velocity of 18 rpm. We are told that the ball moves in a circle with a diameter of one m. And we are tasked with finding what the tension is in the cord attached to the ball. Let's do this. Let's use Newton's second law in the radial direction. So we're gonna have that the sum of all forces in the radial direction is equal to the mass times radial acceleration. The only force in the radial direction is going to be tension. So this will be equal to mass times our radial acceleration. I'm going to sub in for radial acceleration. The fact that it's equal to V squared over R we have M and we have our, but what is our V? Well, V, we know to be equal to R times R angular velocity and we have both of those. So let's sub this value in here. We have that this M times R squared times omega squared, all divided by R, which gives us that our tension is equal to M R times omega squared. So now let's just go ahead and plug in our values. We have that tension is equal to our mass of 0.3 times our radius, which is going to be diameter divided by two times our angular velocity which is 18 revolutions every one minute. But we need this ingredients per second. So we know in one revolution, my apologies. In one revolution, there are two pi radiance and in one minute there are 60 seconds. Great. We square that entire value. And then when you plug all of this in your calculator, we get 1.1 Newtons corresponding to our final answer. Choice of B Thank you all so much for watching. Hope this video helped. We will see you all in the next one.