A 100 g ball on a 60-cm-long string is swung in a vertical circle...

Question

A 100 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?

Accepted Answer

Welcome back, everybody. We are making observations about a tiny sphere and we are told that the mass of this tiny sphere is 250 g or .25 kg. Now, it is attached to a wire. Let me make this a solid line here. It's attached to a cord or wire with a length of .35. And the other end of this is attached to a servo motor that is rotating like this. Now, initially, the center of the sphere is 1.25 m off the ground. And at some point, this cord is going to break since the servo motor is rotating so fast and the sphere is going to shoot straight up to a height of 2. five m. Now, we are tasked with finding what the tension in this rope is right before it snaps. So here's what we're going to do. We're actually going to use Newton's Second law here and we're gonna do the sum of all forces in the radial direction is equal to the mass times the radial acceleration. Well, the only force that we have in the radial direction is tension and So we will set this equal to mass times radial acceleration. But we can substitute in for radial acceleration, tangential velocity squared divided by our radius. That's going to be this velocity as the sphere shoots straight up in the air after the cord breaks. In order to find this velocity, what we are going to do is we are going to use a kid a magic formula. This cinematic formula states That the final velocity squared is equal to the initial velocity squared -2 times our acceleration times our change in our position. In this case, our y values here. So we need to isolate this initial velocity term. And so I will add to eight times our delta Y to both sides of the equation. And you'll see that, that cancels out on the right here. Then I will take the square root of both sides to get rid of this power. We are then left with that. Our desired velocity is equal to the square root of our final velocity squared plus two times our acceleration times delta Y. We have all these values. So let's go ahead and plug them in. We have that. Our initial velocity is equal to the square root of our final velocity squared. What is our final velocity? Well, our final velocity is gonna be our velocity at this point and once it reaches its peak according to projectile, the velocity at that point is just going to be zero. This will be zero squared plus two times the magnitude of acceleration which is just the acceleration due to gravity. So 9.81 times our change in our Y values. So it goes up to a height of 2.75 and started at a height of 1.25, plugging all of this into our calculator. We get an initial velocity of 5.42 m per second. Wonderful. So now we are ready to find our tension in our rope using this formula right here. We have our mass of 0.25 times 5.42 squared divided by the length of our chord of 0.35. And when we plug on this two calculator, we get a tension of 21 newtons right before the court snaps corresponding to our final answer. Choice of. Thank you all so much for watching. Hope this video helped see you all.

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