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Ch 10: Dynamics of Rotational Motion

Chapter 10, Problem 10

CP A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. 10.10. The pulley is a uniform disk with mass 10.0 kg and radius 30.0 cm and turns on frictionless bearings. You measure that the stone travels 12.6 m in the first 3.00 s starting from rest. Find (b) the tension in the wire.

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Welcome back everybody. We are taking a look at a pulley and a string that is tightly wound around this pulley. Now from this string is hanging. I'm actually gonna make this line just a little bit longer is hanging a metallic block. And we are told a couple of things about this entire system here. Oops! Sorry. There we go. We are told that the mass of the pulley is 8.5 kg. Right? And we are told that it behaves like a uniform cylinder. We're told that the radius of the pulley is 220 mil meters or 0.22 m. And we are told that the block is released from rest meaning our initial angular velocity of our pulley is simply just zero. Now, when the ring is fully unwound, it is, it turns out to be 17 m long. And we are told that it takes 6.5 seconds to do this. Now as this pulley is rotating this way, imagine that it kind of has like an axis coming out this direction we have that are centripetal forces acting outward. So the angle between our axle and our centripetal force is simply going to be that our data, I'll just call it fate upon Crying here is 90°. Right? So we are tasked with finding this is this is a lot of information here but we are tasked with finding what is the tension in our rope or string once it is fully unwound or at this time when it is 17m long. Well, we have this relation here. We have that towel is equal to, let's see here our radius times our tension times the sine of our angle theta prime. And this is equal to our moment of inertia for a uniform cylinder times our angular acceleration. If I divide both sides by r sine theta prime, we are going to have that these terms cancel out on the left hand side and we are just left with that. Our attention is equal to this big thing. Right here we have our radius, we have our Theta prime. But we need to figure out what our moment of inertia is and what our angular acceleration is. So let's go ahead and start with our angular acceleration here. I'm gonna use a kid a magic formula that says our change in our rotational angle. So this is not the same data as state of prime. But our change in theta is equal to our initial angular velocity times time plus one half times our angular acceleration times our times squared. Now we know that our initial angular velocity is zero. So this term is just going to go away. So we just that have uh that our change in theta is equal to one half times the angular acceleration times T. C. We're trying to solve for this. So we need everything else. What is our change in theta? Well, we are told right that 17 m of string is unwound from this pulley. So we can say that that is our delta Y. And if we divide our delta Y by our radius, we are going to get our change in theta. So let's go ahead and calculate that real quick. We have that the string at this time is 17 m long. And we have that the radius of our uniform cylinder, cylindrical pulley is .22. So when you plug this into your calculator, we get that. Our change in data is 77.27 radiance. Now I'm gonna plug this back into this formula up here we now have that 77.27 is equal to one half times are angular acceleration times t squared. I'm gonna divide both sides by this. One half He squared. And we get that are angular acceleration is equal to this. Over here are angular acceleration is equal to 77.27 divided by one half times our time of 6.5 seconds squared. And we get that are angular acceleration is 3. radiance per second. Great. So now we have one of our terms but now we need the other term here, we need our moment of inertia. So let's go ahead and find that we have our moment of inertia for a uniform cylinder is one half times the mass times our radius squared. So let's go and calculate that as well. We have that. This is equal to one half times 8.5 kg times 0.22 squared. When you plug this into our calculator, we get that our moment of inertia is 0.2057 kg meter squared. Wonderful. So now we have all the terms that we need to plug into here to find our attention. So let's go ahead and do that. We have that. Our attention is equal to our moment of inertia of .2057 times. Let's see here. Our angular acceleration of 3.66, all over our Radius of . times are sign of our data prime, which we discussed earlier to be 90°. And when you plug all of this in your calculator, we get a final answer of 3.42 newtons as our tension of our string corresponding to our answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.
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