A small block with a mass of 0.0600 kg is attached to a cord...

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface ($\textbf{Fig. P6.71}$). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed $\upsilon = 0.70$ m/s? (b) What is the tension in the cord in the final situation, when the block has speed $\upsilon = 2.80$ m/s? (c) How much work was done by the person who pulled on the cord? Figure P6.71 (Figure can't copy)

Transcript

00:01 Okay, this is kind of a cool problem, actually.

00:04 I liked this.

00:06 So for the a part, we're supposed to find the tension in the chord when it's at the outermost position.

00:12 So basically, we realize that tension is just the centripetal force inward.

00:18 So force, so yeah, centrifugal inward.

00:24 And so basically that's mv squared over r.

00:28 Oh, come on.

00:32 Mv squared.

00:34 Come on, styling.

00:35 Square over r.

00:44 And you can put in the numbers there, that's 0 .06 kilograms, 0 .7 meters per second.

00:50 You got to square that, and then it's divided by 0 .4 meters.

00:55 So that's how you find the tension in part a.

00:57 And then the tension you find in part b is exactly the same formula, but using the different values, the values of the second point.

01:09 Now, there's going to be more tension in part b, and that should be obvious, just intuitive.

01:17 It should be kind of intuitive.

01:18 You pull it in.

01:19 You're going to have to pull harder.

01:25 But what's not obvious is how you calculate the work in part c.

01:31 So ordinarily, we say, okay, work is the force times the distance.

01:35 Well, the force is the tension of it, and so what do you use the tension from part a or the tension from part b? and the reality is neither of them.

01:43 You have to actually do a bit of an integral here.

01:47 So we're going to set that up and we say, okay, the work that we do to pull it from the first position to the second position, it's equal to the force that you pull with, and that's a vector, and then dot product times the distance that you pull it with in that direction, right? so fine, we're pulling it normal to the direction of motion.

02:15 So we're pulling it in the same direction we want to accelerate in, but normal to the direction that's currently moving in.

02:27 So fine, but we can't really work with this equation because the force, tension is changing.

02:35 So what we really need to do is integrate over the distance we're going to pull it.

02:43 So we're going to start at point four, end at point one, and then integrate it over the tension...

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