(II) If a plant is allowed to grow from seed on a rotating pla...

Question

(II) If a plant is allowed to grow from seed on a rotating platform, it will grow at an angle, pointing inward. Calculate what this angle will be (put yourself in the rotating frame) in terms of g, r, and ω. Why does it grow inward rather than outward?

Accepted Answer

Welcome back. Everyone in this problem, consider a person seated on one of the platforms of a me own swing ride at an amusement park which has a radius R and rotates with an angular velocity Omega due to the centrifugal force, the person leans outward at the top of the ride. Calculate the angle phi at which the person deviates from the vertical direction. Considering Gr and Omega assume the ride rotates in a horizontal plane. A says the angle is the inverse tangent of R Omega divided by GB says it's the inverse tangent of R squared Omega squared divided by GC, the inverse tangent of Mr Omega squared divided by G and D, the inverse tangent of R omega squared divided by G. Now, if we're going to figure out the angle phi at which the person deviates from the vertical direction, assuming the ride rotates in a horizontal plane, let's try to sketch what's going on here to better understand our scenario. OK. So here we have our person. OK? That's deviating from our, our vertical direction. OK. And the person is deviating from that direction. Let me just draw over Vertica here, deviating from that direction by an angle of fact. Yeah. And now we assume that the person moves out in the direction of the local normal force. So here this would be FN OK, representing our normal force. We know that the person's weight is acting downwards as MG. And like we said, the red rotates in a horizontal plane, so the centrifugal force is going to be going in the horizontal direction. Now, here's the thing OK. In the rotating frame of the swing ride, there is an outward fictitious force of magnitude MVMV squared divided by R. But because we are concerned with our angular velocity here MV squared divided by R, we can rewrite that, OK. As Mr Omega squared, I know that we have that as Mr Omega squared, we can use what we know to help us find our angle phi from our free body diagram for the rotating frame of reference. Since the person is not accelerated in that frame of reference, the normal force must be the vector some of the other two forces. So we can resolve or we can write Newton's second law sorry in both the X and Y directions. And then we can eventually use this to help us solve for our angle phi so let's go ahead and do that considering upwards and rightwards as positive. First, let's start with Newton's law in the Y direction by Newton's law. We know that some of the forces in the Y direction equal zero. Now, if we think about it, the person's weight is acting downwards as MG while upwards we have the Y component of the normal force, which is going to be FN cos Phi because it is adjacent to our angle phi. So that means then that FN classify minus MG equals zero. OK. And now that tells us that FN classify equals MG and thus are normal force FN equals MG divided by class Phi. OK. Now what about our forces in the Y in the X direction? Well, some other forces in the X direction also going to be equal to zero. And no, that means if we think about it, we're going to have the horizontal component of FN which is going to be FN Sine Phi acting to the left. OK. And Mr Omega squared acting to the right. So that tells us then that FNS sorry, let me start with a positive one first. So that's Mr Omega squared positive minus FN sine Phi negative is gonna be equal to zero. Does FN signify equals Mr Omega squared? And thus our normal force equals Mr Omega squared divided by the sine of Phi. Now notice here we have two equations for our normal force, which means that both of our expressions must be equal. In other words, MG, divided by class OK, must be equal to Mr Omega squared divided by sin Phi. If we try to get sine and cosine on the same side that means and oh sorry. And we cancel M from both sides. Let me do that before I I transpose. Then that means Sine, Phi divided by the cosine of Phi is going to be equal to R omega squared divided by G. OK. Sine divided by cosine is the tangent. So that means the tangent of Phi equals R omega squared divided by G. And thus to solve for Phi, that means then that Phi is going to be equal to the inverse tangent of R omega squared divided by G. Thus, this is the expression for Phi in terms of R omega and G. Therefore, that tells us then that our correct choice is choice D. Thanks for watching everyone. I hope this video helped.

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(II) If a plant is allowed to grow from seed on a rotating platform, it will grow at an angle, pointing inward. Calculate what this angle will be (put yourself in the rotating frame) in terms of g, r, and ω. Why does it grow inward rather than outward?

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