A small bead of mass m is constrained to slide without frictio...

Question

A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about a vertical axis (Fig. 5–58) at a frequency f.

(a) Determine the angle θ where the bead will be in equilibrium within the hoop—that is, where it will have no tendency to move up or down along the hoop.

Accepted Answer

Welcome back everyone in this problem. Imagine a small ball positioned within a rotating drum which is open at the top. The drum rotates about a vertical axis with a frequency F. While the ball is restricted to traverse the inner frictionless circular path of radius R calculate the angle phi at which the ball reaches equilibrium inside the drum signifying that it does not ascend or descend along the circular trajectory. For our answer choices A says that Phi equals the inverse cosine of MG divided by two pi frb says it's the inverse cosine of MG divided by four pi squared. Fr C says it's the inverse cosine of G divided by four pi squared multiplied by F squared R. Sorry, all of that second expression is in the denominator. And finally, D says it's the inverse cosine of G divided by two pi squared F squared R squared. Now, if we're going to calculate the angle phi from our problem, and we want to find that angle at which the ball reaches equilibrium. That means we're thinking about forces. Let's try to draw a free body diagram to understand all of the forces acting on our ball So if we go to the right here, OK. And here's our circular path along the vertical frequence along the vertical axis. OK. And for our ball, yeah, we know that it traverses our circular path here, our inner frictionless circular path. No, in that case, since there's no friction between the ball and the inner circular surface, the surface will exert only a normal force on the ball. So we know the ball's weight is acting vertically downwards as MG and that normal force will be acting on the bar here that we can label as FN opposite. He at me draw that properly here. K so we have our first FN acting here, OK. Now, for the normal force that is exerted on the ball, it will have a vertical and horizontal component. OK. So if we have our angle, Phi here, then the horizontal component of our normal force is opposite to Phi, which means it's going to be FN Sine Phi acting in this direction while its vertical component is adjacent to Phi, so its vertical component is going to be FN class. OK? No, these are the forces that are acting on our wall, right? And since the ball moves in a horizontal circuit parallel to the floor, the central pedal force is horizontal and the net vertical force must be zero. So we can apply Newton's second law for both horizontal and vertical directions. Because recall that by Newton's law, it tells us that the force is directly proportional to the acceleration by the formula that says the sum of forces is equal to the mass multiplied by the acceleration. So we can resolve them vertically and then horizontally and then we'll see where we go from there. Now, first resolving them vertically, as we said, remember that the normal force has a vertical component that's acting against the weight of the ball. So that means F and classify minus MG is going to be equal to zero. Let me move that a bit over here. OK. Oops. Now, since that's the case that tells us that FN classify is it going to be equal to MG? Which means that our normal force FN is going to be equal to MG divided by cos Phi. OK. So here we have an expression for our normal force in terms of MG and Phi next, let's resolve the forces horizontally. Now, as we said, the horizontal force is our centripetal force. OK. So if we're looking at, if we're resolving it horizontally, we're really looking at the forces, sorry, we're really looking at the sum of the radial forces. So applying new to our second law that's going to be equal to our mass M multiplied by our acceleration in the radial or sorry, the radial acceleration. Now, what do we know about radial acceleration? Well, we know that the radial acceleration is going to be equal to the square of the speed divided by the radius, OK in the same direction. So in this case, it would be the X component of R here for our circular path. Now we know shaded in red is our radius R. So it's horizontal component Rx is going to be R sine Phi in the same way that the horizontal component of our normal force is FN sine phi. So we can replace Rx, OK with R sine phi. And since we already have an expression for uh or radial force, the X component of FN which is FN sin Phi. Using that idea, then the sum of the radial forces FN sine Phi are only force acting horizontally is going to be equal to V squared divided. Sorry the mass M because we're putting it into this formula. So the mass M multiplied by V squared divided by R sine Phi. So now we also have a way to get an expression for our normal force horizontally. But we have V squared here and V squared is not a known quantity. Recall that from our problem, we are given a frequency fa radius R, an angle phi and a mass M. To know we'll need to express our velocity in terms of one of those variables or more in order to eventually solve four P five. Now, what do we know about velocity that is related to frequency or radius or mass or anything like that? Well, recall, OK, recall that the velocity V is equal to two pi F multiplied by R sine phi. OK. And this is for our radial velocity as applied to our radial acceleration. So, in this case, no, we should be able to substitute the into our formula for FNSP five. We'll also be able to rearrange FN as MG divided by class Phi. And thus be able to get uh just be able to simplify and possibly get an expression for Phi in terms of MGF and R. So let's go ahead and do that. OK. And of course here, let me just fix those, those who have been capital MS. So in that case, OK, that means then in red here because we're just adapting our horizontal velocity formula here. That means MG divided by the cosine of Phi multiplied by the sine of Phi is going to be equal to M multiplied by V squared. So that's two pi fr sine phi all squared. And all of that is being divided by R sine phi. No, we can rearrange here. So we're going to have MG multiplied by sin divided by costs by being equal to M multiplied by two squared, which is four pi squared F squared R squared and sine five squared all divided by RS five. Now there's an opportunity to simplify here before we, we go any further. OK. We can simplify, we can cancel M on both sides. OK? We can cancel Sine Phi on both sides as well leaving us with just sine phi in the numerator on the right hand side. And again, we can cancel sine phi from the numerator and denominator. On the right hand side, we still can cancel more. We can cancel R from the numerator and denominator which instead of R squared, not just leaves us with R in the numerator on the right hand side and know that we've simplified to a point. This note tells us that what's left in the numerator on the left G divided by the cosine of Phi is going to be equal to four pi squared multiplied by F squared, multiplied by R no. Since that's the case, remember we're solving for five. So we can uh flip or basically find the inverse of our problem here. So if we rearrange this OK, then we're going to get the cosine of Phi. In other words, multiply both sides by the cosine of Phi and divide by what's on our right hand side here. So the cosine of phi, OK is going to be equal to G divided by four pi squared F squared R. In other words, we kind of just flip these here. OK. That's the math that we did. And now since we have an expression for the cosine of Phi to find Phi, we can find the inverse cosine. Therefore, Phi is the inverse cosine of G divided by or pi squared F squared R. So this would be the expression for our angle Phi at which the ball reaches equilibrium inside the drum. If we look back on our answer choices, that means c is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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