FIGURE P2.64 shows a fixed vertical disk of radius R. A thin, frictionless rod is attached to the bottom point of the disk and to a point on the edge, making angle Φ (Greek phi) with the vertical. Find an expression for the time it takes a bead to slide from the top end of the rod to the bottom.

Video transcript

Hi, everyone in this practice problem. We're being asked to actually find the time required for the box to slide from point A to point B. We will have a bar welded to a vertical hoop with a diameter D of 0.5 m between point A and point B located on the rim B is going to be the lowest point of the rim. And the bar makes an angle theta with the line joining the center of the hoop O and B A small box is placed on A or point A and sliding from rest on this frictional list bar. And we're being asked to determine the time required for the box to actually slide from point A to point B. The options are 0.32 seconds, 0. times cosine data seconds, 0.25 seconds and 0.25 times sine data seconds. So I am going to start us off with um actually drawing a better representation in our figure. So first, what we have is going to be the um acceleration. So the box is going to uh actually slide from rest on the frictionless bar under a constant acceleration of the gravitational uh created by the gravitational force. So the gravitational force is going uh vertical just like. So and in order for us to determine what the acceleration going in the direction of A and B of the bar A and B, we have to actually do the projection for the gravitational acceleration. So the angle between the bar and the gravitational acceleration here is going to also equals to theta. And therefore the gravitational acceleration in the direction of the bar between A and B can be calculated by multiplying G with cosine of theta just like. So, so essentially the acceleration acting A is going to equals two G multiplied by cosine of theta. We will model the box as a point like object particle sliding on this incline. And because of that, we will use the kinematic equation to actually calculate the time required for the box to slide from point A to point B. Also we will align the X axis with the bar itself. So the kinematic equation used is then going to be X B minus X A or essentially the displacement equals to half A T squared plus P knot X T. From the problem statement, we know that P knot is equals to zero m per seconds because it is starting from rest or pheno X essentially. So we can neglect the last part or the last term in our previous equation. Next, we also know the diameter given for the hoop is going to equal to 0.5 m. And the time B when the box arrives to point B is unknown and it is what's being asked for us to calculate the distance X B minus X A is still unknown. So that is what we're going to calculate next. We're going to utilize the fact that oh 0.0 A and B is going to form an isosceles triangle. So essentially O A is going to be the radius of the hoop and O B is also going to be the radius of the hoop. Therefore, those two are going to be equal and O A B is going to create an Isosceles triangle drawing a line straight from 0.0 to perpendicular to the bar A B. We will give that point to be point I. And essentially what we wanna do to calculate the distance of A B is to by calculating point B I and point A I and um summing those two up together. So the median O I is going to be perpendicular to the base. And therefore the way we wanna find B I B I is essentially just going to be the projection of O B onto the X axis or onto the A B bar. So B I right here is going to actually be, are multiplied by the cosine of the angle to get the projection cosine of theta. So B I is essentially R multiplied by cosine theta. Therefore, A I is also going to be R multiplied by cosine of data. And A B is going to equals to A I plus B I which will then equals to two R multiplied by cosine of theta. Or essentially A B is going to equals to the diameter D multiplied by cosine of theta. Now, we can actually start substituting all of our information or all of our known values into our kinematic equation. So what we have is X B minus X A or A B equals to oops, I'm gonna use the lower case A to stay consistent equals to half A T squared. X B minus X A is A B which will equals to D multiplied by cosine of data, which will also equals to half A T squared where D A is going to be G multiplied by cosine of data and multiplied by T squared just like. So therefore rearranging this, we can actually cross out the cosine of data and the T squared will then be two D divided by G, two D divided by G is going to be two multiplied by 0.5 m, which is the diameter divided by the G which is 9. m per second squared. And therefore, the T squared will then equals to actually one divided by 9.81 2nd squared. This will give us a T value by actually taking the square root of 1/9 0. 2nd squared, which will give us A T value of 0.32 seconds. So the time that it takes for the box to actually move from point A to point B is 0. seconds, which will correspond to option A in our answer choices. So answer A is going to be the answer to this problem statement where the time required for the box to slide from point A to point B is 0.32 seconds. And that will be all for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be all for this one. Thank you.