(II) A softball player swings a bat, accelerating it from rest...

Question

(II) A softball player swings a bat, accelerating it from rest to 2.4 rev/s in a time of 0.20 s. Approximate the bat as a 0.90-kg uniform rod of length 0.95 m, and compute the torque the player applies to one end of it.

Accepted Answer

All right, everyone. Let's take a look at this practice problem dealing with torque. So in this problem, we have a 1.5 m long shaft and it's attached to a motor at one of its ends. The motor um accelerates the shaft from rest to a rotational speed of 3.2 revolutions per 2nd 0.5 seconds. And the question wants us to determine the torque exerted um at the end of the shaft by the motor. And we're considering this shaft as a um 0.82 kg uniform rod. We have four choices here um for our answers. Um A is 4.0 Newton meters. B is 16 Newton meters. D is 19 Newton meters and D is 47 Newton meters. So let's start off by drawing a quick little diagram of what we were given this problem. So here, I'm gonna draw the motor here as a circle. And here is our uniform rod which is going to be our shaft and this is going to rotate round um by the motor. So we were told several things we were told the length of the shaft which we can label as a lowercase L and that was 1.2 m that was given here at the beginning. Um We're also told the um mass of this shaft. So its mass was this 0.82 kg. We are also told its initial um angular speed, we're told that it starts from rest. So my initial angular speed which is omega I is going to be zero and our final um angular speed is going to be this 3.2 revolutions per second. Now revolutions per second is not an SI unit. We want to convert this into si units. And the si units that we need to have are radiance per second. So we can do that conversion really quickly because there are two pi gradients in one revolution. And this gives us an angular speed of 20.1 gradient per second. Now, the last bit of information we were given and the problem is the time it takes to go from rest to this final angular speed and that is 0.5 seconds. That's gonna be our time. Now to determine the torque, um we're actually going to use basically the equivalent of Newton's second law, but for rotational motion. And so that's one, tell us that our net torque is equal to our moment of inertia times our angular acceleration. Now, the only torque that's being exerted on the shaft is by the motor. So we'll just label that um torque as just tau. So we just have one torque in this problem. The moment of inertia is gonna be the moment of inertia of the shaft, which we're treating as a uniform rod and it's um rotation is about its end point. So that means its moment of Daria is given by one third times the mass times the length squared and the angular acceleration isn't given in problem. However, if we recall the definition of our angular acceleration, that's just going to be our change in our angular speed divided by our change in time. And we know both of our um initial and final um speeds and times. So we can actually just substitute that in or the angular acceleration. So our final angular um speed is gonna come first minus our initial angular speed divided by our final time minus zero, which is our initial time. So at this point, um we can plug in values for everything. So we'll have one third times our mass, which is the 0.82 kg times the length, which was the 1.2 m that gets squared and then, or angular acceleration on the numerator here, we're gonna have our final um angular speed, which is the 20.1 radiance per second divided by um are final time here which is 0.5 seconds. Now, on the numerator, my initial um angular speed is zero. So I don't have to worry about that. And my initial time is zero So I'm left with just this as the fraction. So we can um plug that into our calculator. And if I look at the answers, I only need to keep two significant figures and doing that, this comes out to be 16 Newton meters. And the units are correct here because I'll have kilogram meters squared per second squared, which is a Newton, which is a kilogram meter per second squared times meter. And if I look here at our um answers, this corresponds to answer B which is the 16 Newton meters. So in this problem, we um had to kind of piece together several different things. Um We had to use our, we had to recall and use our definition for angular acceleration. We had to figure out the um Vanessa or the rod. And so we had to piece together many different things. So be careful when you're doing these types of problems, you may have to reach back and recall some of the previous material that you have learned in order to deal with this type of problem. So I hope this helps and I'll see you in the next video.

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(II) A softball player swings a bat, accelerating it from rest to 2.4 rev/s in a time of 0.20 s. Approximate the bat as a 0.90-kg uniform rod of length 0.95 m, and compute the torque the player applies to one end of it.

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