A heavy ball with a weight of 100 N (m = 10.2 kg) is hung from the ceiling of a lecture hall on a 4.5-m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?

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Hi, everyone in this practice problem, we are asked to find the tension of our wire holding a marble at the bottom position. When it is oscillating, we will have a student attaching a marble of mass 250 g to the end of a metallic wire off length 1. m, the wire is going to be suspended vertically from the free and on a frictionless piett and the stretch wire and the marble will be displaced by an angle of theta and then released and the system will start to oscillate. The marble speed at the bottom of its motion is 6.3 m per second. And we're being asked to find the tension of the wire at this bottom position. The options given are a 3.8 Newton B 6.2, Newton C 8.6 Newton and D 11. Newton. So in this problem, the marble will undergo a vertical circular motion and the speed of the marble as well as its angular position is going to be constantly changing as it is oscillating, the object will be subjected to two different forces, the weight and also the tension of the wire. So I'm going to start by drawing our system. So we'll have our wire here holding our sphere or our marble. And it will start with or from an angle theta and it will oscillate to the bottom position, to uh to the other side and will go back. So at this bottom part or at this bottom position, the object which is our marble will be subjected to two different forces. One is going to be the weight of the marble which is uh multiplied by G and the other one is going to be the force or the tension force in our wire, which is going to be represented by our data like so, so the weight is going to be vertically downward, whatever uh wherever the object's position is, and the tension T is actually going to be constantly changing. But because we are being asked to find the tension at the very bottom, so the tension at the bottom is going to be directed vertically upward just like so awesome. So what we wanna do is to actually apply Newton's second law at this lowest position or at this bottom position. So Newton's second law, according to Newton's second law, we will then have Sigma F R equals M multiplied by A R S. The Sigma F R is just a summation of all the forces acting on our system, which is our marble. So the two forces are the weight and the tension but the weight is pointing downward downward. So we wanna put a negative sign in front of it. So negative M G plus tension will equals to M multiplied by A R. And you wanna recall that A R is equals to V squared divided by R or the linear velocity divided by our radius. So we want to substitute that or expression will then becomes M G plus D equals M multiplied by V squared divided by R. So next, what we wanna do is to rearrange this so that we can find the tension. So our tension will then be M mo uh multiply by a few squared divided by R plus M multiplied by G. You wanna pull all the MS out from the two terms. So our tension will then becomes M multiplied by parenthesis V squared divided by R plus G just like. So, All right. So now that we found our formula, we can actually substitute our information. So the mass is going to be 0.25 kg because it is 250 g. So mass is 250 g and that will be 0.25 kg. Um That will be multiplied by parenthesis P squared, which is given to be 6.3 m per second at the bottom position Divided by the radius. The radius is going to be the length of the metallic wire which is 1.15 m plus the gravitational acceleration, which is 9.81 m per second squared and then close parenthesis, multiplying or calculating all of this together, the tension will then come out to be 11.1 Newton. And the tension of 11.1 Newton is going to be the tension Off the wire at the bottom position. So that will correspond to option D in our answer choices. So add option D with uh tension off the wire at the bottom of 11.1, Newton is going to be the answer to this practice, this problem. So if you guys still have any sort of question, please make sure to check out our other lesson videos on similar topic and they'll be all for this video. Thank you.

A heavy ball with a weight of 100 N (m = 10.2 kg) is hung from the ceiling of a lecture hall on a 4.5-m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?

Verified Solution

Key Concepts

Tension in a Rope

Centripetal Force

Conservation of Energy