As shown in Fig. 6-70, a 1.34 kg ball is connected by means of two massless strings to a vertic

As shown in Fig. 6-70, a 1.34 kg ball is connected by means...

Key Concepts

Free-Body Diagram

A free-body diagram is a schematic representation of all the forces acting on an object, isolating it from its environment. In the context of circular motion, this diagram helps identify forces such as tensions, weight, and any support forces, which is essential for setting up the equations of motion and analyzing how these forces provide the necessary centripetal acceleration.

Tension in Strings

Tension refers to the pulling force transmitted axially by a massless string, cable, or similar object. In problems involving connected bodies in motion, understanding the distribution of tension forces in different segments of the string is vital, as each segment might contribute differently to satisfying the conditions for equilibrium or providing the required net force for circular motion.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path and is directed towards the center of the circle. It arises from the vector sum of various forces acting on the object and is calculated as the product of mass and centripetal acceleration. This concept is central in analyzing the dynamics of an object undergoing uniform circular motion.

Newton's Second Law in Circular Motion

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. When applied to circular motion, this principle is used to equate the net inward (centripetal) force provided by tensions and other forces to the mass times the centripetal acceleration, thereby allowing for the determination of unknown quantities such as speed or tension.

Transcript

00:01 Okay, so in this problem, we have a mass that is attached to this rotating bar, and this bar is rotating in this direction.

00:11 And this mass is being swung around, and the both strings are taught.

00:21 The strings are tied.

00:23 This is these 1 .7 meters.

00:30 And this length here is also 1 .7.

00:33 So it's 1 .7, 1 .7, 1 .7.

00:36 This is making an equilateral triangle.

00:39 And then we need, and this ball is moving in a circle of radius r.

00:44 So i think the first thing, part a, that we do is we need to draw a force diagram of what's happening with this mass.

00:53 And we've got the weight force of the mass, m .g.

00:58 And we've got two tension forces acting on it.

01:01 So we got, we'll call this one tension one and this one tension two.

01:06 And these two tension forces are providing the centripetal force to keep this moving in a circle.

01:15 The next thing that we need to do is we'll probably need to establish this radius distance here.

01:25 So in this case, we've got this circle or this triangle here where we have a hypotenuse that is, 1 .7 meters and a one side which is 1 .7 over 2 and then we need to figure out what this r is because that's the radius of the circle.

01:42 So let's do that right now.

01:44 We can use this equation a squared plus b squared equals c squared because this is a right triangle.

01:53 So r is one of the sides.

01:57 1 .7 over 2 is one of the other sides.

02:00 So we can just say that r is square root.

02:02 Of 1 .7 over 2 squared, or rather, oops, i wrote that backwards, should be 1 .7 squared minus 1 .7 over 2 squared, since we know the hypotenuse.

02:27 So if we solve this for r, we get a value of 1 .47 meters.

02:33 So that's going to be the radius of our circle, and that will also come in handy in a little bit.

02:39 So first thing we need to do, or rather the second thing, since we've already done the first thing, is we need to find the tension in the lower string.

02:48 We are told what the tension in the upper string is.

02:51 I should have written that here.

02:53 T1 is 20, sorry, 35 newtons.

03:00 And we need to figure out what t2 is.

03:02 So what is t2? and for this, we can use the vertical complex.

03:10 Of these forces because the mass is moving in the horizontal circle, so the vertical components all need to cancel out.

03:19 So we can say that the net force is zero, and we have the vertical component of t1, which is going to be t1 cosine of this angle minus t2 cosine of this angle here, which happens to be the same angle and then also minus the mass times gravity the weight force of it.

03:51 Well, cosine theta, i can figure out what this angle is, but i don't need to because cosine is adjacent over hypotenuse and since adjacent is 1 .7 over 2 and the hypotenuse is 1 .7, coscient it just becomes one half...

Please give Ace some feedback