(II) Suppose the force Fₜ in the cord hanging from the pulley ...

Question

(II) Suppose the force Fₜ in the cord hanging from the pulley of Example 10–10, Fig. 10–22, is given by the relation Fₜ = 3.00 t ― 0.20 t² (newtons) where t is in seconds. If the pulley starts from rest, what is the linear speed of a point on its rim 9.0 s later? Ignore friction and use the moment of inertia, calculated in Example 10–10.

Accepted Answer

Welcome back. Everyone in this problem. A wire is being un wounded from a spool initially at rest. As shown in the figure, if the tension T in the wire is given by T equals five T minus 0.4 T squared newtons, where T is the time in seconds. And the moment of inertia of this pool is 0.19 kg square meters. Determine the linear speed of a point on the outer edge of the spool after seven seconds, assume that friction is negligible for our answer choices. A says it's 11 m per second. B 29 m per second, C 41 m per second and D 110 m per second. Now, in this problem, we're trying to determine the linear speed of a point on the outer edge. Let's just first make note of all the information we have here. OK. So so far we know that, let's see, we know that the outer radius of the spool from our diagram are is 27 centimeters, which if we write that in si units, that's 0.27 m. We also know that the moment of inertia of the spool I, OK is 0.19 kg square meters. OK. We know that we have a function for the tension T, which is given by five T minus 0.4 T squared newtons. So in other words, our tension T is a function of time. OK. And we also know that we want to find the linear speed of a point on the outer edge after seven seconds. So the time that we're considered we're concerned with is when T equals seven seconds. OK. And we want to use this information here to help us figure out the linear speed. So that's what we don't know VV is what we're trying to figure out. Now, we're talking about rotational motion here. What do we know about linear speed and rotational motion? Well, recall, OK, that our speed V is equal to our angular velocity omega multiplied by I or radius in this problem. OK. So that means then that if we want to find V, we already know R now we need to figure out omega if we can find omega, we can substitute it to solve for V. So the question is, what do we know about omega? Think about the information we have here? Do we know anything that can relate? Well, we may not know omega directly. OK. But there are a few things we do know. We know that for a moment of inertia or tension T and our radios are OK. We know that our angular acceleration alpha is equal to RT divided by I, we also know that that angular acceleration is related to our angular velocity where our angular acceleration is the derivative of our angular velocity omega with respect to T. So in that case, since both of them are equal to the angular acceleration, this implies then that D or derivative of omega with respect to T is equal to RT divided by I, OK. And that means then that our change in omega, let me come over here a bit. That means I change in omega is going to be equal to RT divided by I with respect to TRDT. Now, here's the thing about it. Notice that as we said earlier, our tension T was expressed in as a function of our time T or our common T in seconds. So we could even rewrite this as R divided by I multiplied by TDT. OK. Because we're talking now about tea expressed in terms of time. So with that idea, then notice that here, if we want to figure out omega, we could integrate with respect to omega on the left hand side and on the right hand side, we could integrate our function T with respect to our time T. So that implies then that the integral of omega or change in omega eh between our initial omega, let's call that omega nut and omega OK is going to be equal to are divided by I multiplied by the integral, no R divided by I, those are constants, we can put them outside the integral multiplied by the integral of T. OK. With respect to T between the boundaries of zero and T. Now, what do we know about our initial angular velocity? Well, as we said, it was unwanted from a spool in spool initially at rest. So that means omega knot is actually going to be equal to zero. OK. So we can replace Omega knot here with zero. OK. And no, we can also substitute our function for T. So let me do that. OK? Because no, what we're saying is that the integral of DW between zero and sorry D omega between zero and omega is going to be equal to R divided by I. Yeah, are divided by I multiplied by the integral of T. OK. We know T is five T minus 0.4 T squared. OK. That's the integral of that function with respect to T. Now, if we were to actually integrate, let's go ahead and integrate here. And oh remember these are between the bonds zero and T. On our left hand side, the integral of D omega between zero and omega would just be equal to omega. OK. And on our right hand side, that's R divided by I multiplied by the integral of five T which would be five T squared divided by two. If I put a quick note here, remember that when we integrate a term race to a power. In other words, if we're integrating X squared with respect, sorry X to the N, with respect to X, we add one to the pore and divide by the new pore. And that's what we're doing here. We added one to the power of sorry, the power of T is one. So we added one to it. So it becomes two and we divide by our new power tool. Likewise, when we integrate 0.4 T squared, it's going to be 0.4 T cubed, that's two plus one divided by that new power, which is three. OK. Let me just write that properly. So that's 0.4 T cubed divided by three. And that's between T and zero. OK. Now, between T and zero here or initially, if we substitute T into our function, it's going to be the same expression. And if we substitute zero, all of that is going to be zero. So it's actually just this expression five T squared divided by two minus 0.4 T cubed divided by three. OK. So since we have an expression for Omega, now we can substitute it to solve for our linear velocity V. OK. Or our linear speed V because no, OK. When we substitute that into our expression, that means V which was equal to omega, I is now going to be equal to Omega which is R divided by I multiplied by this expression, five T squared divided by two minus 0.4 T cubed divided by three. OK. Multiplied by R which let me come over to the, let me come over to the, to the left here V is going to be equal to R squared divided by I because no, we multiply R by R multiplying that by R expression that we already have. Now we know our value of tea. Remember we said our time tea was we wanted to find a time after seven seconds. And we also know our moment of inertia and we know the radar of our sport. So now we can substitute all of those values because no, that means V is going to be equal to 0.27 m. OK. Squared. Divided by our moment of inertia, 0.19 kg square meters. OK? Multiplied by five T squared. That's five, multiplied by several seconds squared, divided by two minus 0.4 divided by three, multiplied by our time cubed, that's seven seconds cubed. OK? And that expression is in terms of Newton seconds, OK? So no, I think let me put this here. OK. Now we can go ahead and multiply through. OK. So you, you can, you, if we put this into our calculator here, then we should find that our linear speed V is going to be equal to 29.45 m per second. Remember that all of our answers were into two significant figures. So if we do the same, we end up with the being approximately equal to 29 m per second. Therefore, that tells us the linear speed of a point on the outer edge of the spool after seven seconds is 29 m per second. If we go back to our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Newton's Second Law of Motion

Moment of Inertia

Kinematics of Rotational Motion

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