A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have at the top without flying off the road?

Video transcript

Hi, everyone in this practice problem, we will have a cyclist that is going to be going over a bridge on a bicycle racing track which is shaped like an arc of circle with a radius of 12 m. We're being asked to find the spit limit for the cyclist to actually avoid flying over the top of the bridge. And the options given are a 3.8 m per second. B 10.8 m per second. C 33.9 m Per second and D 37.6 m/s. So we will model the cyclist as a point like particle and on the bridge, the cyclist will actually undergo a circular motion. So I'm gonna draw our uh system as the full circle right here to make it easier for us to imagine our circular motion. We will have the two len on the left side and also on the right side of the bridge. And this is going to be our bridge where the cyclist will actually cross from. I'm going to represent our cyclist with a sphere or a circle here. And what we're interested in looking for is the speed of the cyclist to avoid flying over the top of the bridge. So I'm gonna put it, put the cyclist at the very top of the bridge. So at the very top of the bridge, the cyclist will experience or will be subjected to two different forces, we want to neglect any for shell forces. So the other two forces that the cyclists will experience are first, the gravity, shell force which is the weight pointing vertically downwards, which is going to be m multiplied by G. The mass is going to be the mass of the cyclists itself plus the mass of the bike. And the second one is going to be the normal force. So the normal force is going to be pointing upwards and which is going to be represented by this end right here. So at the top of the bridge, the cyclist will be subjected to two different forces, the weight pointing downward and the normal force pointing outward. I am going to complete our diagram right here and I'm going to add the radius Which is going to be R.R. is going to equals to 12 m. And also the speed of the cyclist co which is going to be represented by this fee right here. The V is the linear velocity. Awesome. So the way we want to tackle this problem is to by employing Newton's second law. So the Newton second law will allow us to actually substitute the velocity into itself in order for us to find the spit limit at the top of the bridge. So according to Newton's second law, the sigma of all the forces or the summation of all the forces in the radial direction will equals to M multiplied by A R or, or the radial acceleration. And the M here is going to be the mass of our cyclists and the bike or the mess of our system, the A R is going to be represented or we want to recall that A R can be substituted with V squared divided by R which is going to be the linear velocity squared divided by the radius of our circular trajectory. So we want to substitute debt into our new and second law in order for us to actually solve for the spit limit for the cyclists at the top of the bridge. So at the top of the bridge, there are two different forces acting upon our system. The first one is going to be the gravitational force and the second one is going to be the normal force. I'm going to uh create a convention for us. In order to make it easier for us to solve. For this problem, we wanna make a convention at which the positive direction in the vertical, um the positive direction in the vertical axis is going to be pointing downward. So therefore this is going to be the Sigma forces is going to be M G minus N equals to M multiplied by A R. So I'm gonna substitute the A R here. This will then equals to M G minus N equals to M multiplied by V squared divided by R. So to avoid the cyclist to not lose contact with the bridge, so to avoid the cyclist flying over the top of the bridge, we wanna set our normal force to equals to zero Newton. I'm gonna write down Newton here to make it not confusing for us. And at uh for the normal force to be zero, Newton, then the philo is going to be the limiting velocity at this condition. The cyclist will not lose contact with the bridge and still be going over the bridge. So uh we want to substitute this condition right here in order for us to be able to solve uh this problem into our Newton Second law expression. So in the Newton second law, we have M G minus N equals to M multiplied by V squared divided by R. We wanna imply the condition at which new uh the normal force equals to zero Newton and the velocity will then equals to uh V limit squared divided by R just like. So, and we wanna notice that uh we have M in both sides of our equation. So we want to cross that out so that our equation will then be G equals to V limit squared divided by R. We can then rearrange this equation in order for us to find the limit, which is going to then be the square root of G multiplied by R. So this is the equation that we're gonna use to find the spit limit for the cyclist to avoid flying over the top of the bridge. So we wanna substitute all the information given. So V limit will then be the square root of G multiplied by R which will equals to the square root of 9. m per second squared, which is the gravitational acceleration multiplied by the radius given in the problem statement which is 12 m. And that will actually give us a V limit value of 10.8 meters per second. and the velocity limit or the speed limit of 10.8 m/s is going to be the answer to this practice problem. So if the cyclist exceeds the speed of 10.8 m per second, then he will fly over the top of the bridge and the speed limit of 10.8 m per second will keep the cyclist to not lose contact with the bridge. So that will correspond to answer choice B in our uh option. And option B is then going to be the answer to this practice problem with a spit limit of 10.8 m/s. So if you guys still have any sort of confusion on this particular video, please make sure to check out our adolescent topics and that'll be all for this one. Thank you.