The bunchberry flower has the fastest-moving parts ever observed in a plant. Initially, the stamens are held by the petals in a bent position, storing elastic energy like a coiled spring. When the petals release, the tips of the stamen act like medieval catapults, flipping through a 60° angle in just .30 ms to launch pollen from anther sacs at their ends. The human eye just sees a burst of pollen; only high-speed photography reveals the details. As FIGURE CP12.91 shows, we can model the stamen tip as a 1.0-mm-long, 10 μg rigid rod with a 10 μg anther sac at the end. Although oversimplifying, we'll assume a constant angular acceleration. b. What is the speed of the anther sac as it releases its pollen?

Video transcript

Hey, everyone. Let's go through this problem. A medieval throwing machine is used to launch stones. The machine has a throwing arm initially horizontal that pivots around a point P. The stones on the throwing arm are located four m from point P. As shown in the figure, the throwing arm takes 0.8 seconds to swing from its initial horizontal position to the 45 degree release angle. Find the speed of the stone as it leaves the machine. Assume that the throwing arm rotates at a constant angular acceleration. Option A 4.2 m per second. Option B 6.7 m per second, option C 7.8 m per second and option D 9.3 m per second. Fortunately, for us, this problem isn't nearly as scary as it might look because we're explicitly told that there is a constant angular acceleration, which means that we'll get to use our good old fashioned angular kinematics equations. In order to solve this problem, in order to find the speed of the stone as it leaves the machine, we first need to know what the, what the tangential acceleration of the pivot is at that point And in order to figure that out, we'll need to find what the final angular velocity or rather the angular acceleration is at that point. So let's go through the variables we're given, we're told that the pivot moves from some initial horizontal position. So its initial angle is zero to a 45 degree release angle. So the final position, the final angular position is 45 degrees or alternatively in radiance pi divided by four radiant, we're told that the motion happens in a period of 0. seconds. And the initial angular speed of the pivot is zero because it's not moving before the pivot begins accelerating the stone upwards. And the first thing we want to solve for is the angular acceleration. So that's unknown right now. And recall, we have a kinematics equation perfectly suited for this type of problem which states that the final angular position is equal to the initial angular position plus the initial angular speed multiplied by the time interval plus one half the angular acceleration multiplied by the square of the time interval. Now, we already discussed that our initial angle is zero and the initial speed is zero. So the, so this equation has already been simplified quite a bit to where the final angular position is equal to one half the angular acceleration multiplied by time squared. We want to solve this for angular acceleration. So let's do this by taking both sides of the equation. And dividing by one half T squared. And we find that the angular acceleration is equal to two, multiplied by the final angular position divided by the time interval squared. We let's plug that, let's now plug in the variables we were given in the problem. So for beta sub f, the final angle that's pi divided by four radians and T is given to us as 40.8 seconds. So we square that and put all this into a calculator and we find an angular acceleration of about 2.45 radiant per second squared. So that is the angular acceleration of the pivot. Next, we need to know the tangential acceleration of the pivot recall that when we're looking at rotational motion, an angular acceleration or or a tangential acceleration is equal to the lever arm multiplied by the angular acceleration. The lever arm R is the distance from the, the rotating point, the axle, the pivot point to where the stone is, which is given to us in the problem as four m. So four m multiplied by the angular acceleration. We just found if we put that into a calculator, we find a tangential acceleration of about 9.8 m per second squared. So now that we know the tangential or linear acceleration, let's now use our linear kinematics equations to find the final speed of the stone. So now let's consider the linear variables we're that we're dealing with. So you want to find the sub f the final linear speed of the stone, we know that V sub I is going to be zero since again, the stone is starting at rest and the time interval isn't going to change. And now we know the acceleration to be 9.8 m per second squared. So recall that one of our kinematics equations is again, pretty suited to this. It states that the final speed for a constant acceleration is equal to the initial speed plus the acceleration multiplied by the time interval. Once again, we discussed that the initial speed is zero. So we'll just have to take the acceleration, we determined 9.8 and multiply it by the time interval 0.8 seconds. If we put this into a calculator, then we find that the final speed of the stone is equal to about 7.8 m per second. And that is the answer to the problem. If we look at our multiple choice options, we can see that this agrees with option C. So option C is the correct answer to the problem. And that's all for this problem. I hope this video helped you out if it did. And if you need more practice, please check out some of our other videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye-bye.