The normal force equals the magnitude of the gravitational force as a roller-coaster car crosses the top of a 40-m-diameter loop-the-loop. What is the car's speed at the top?

Video transcript

Hi, everyone in this practice problem. We're being asked to find the magnitude of the marble's velocity at the top where the marble will go around a loop to loop track by always being in contact with the loop, the radius of the vertical loop is going to be 12.5 centimeter. And at the top of the loop, the magnitude of the gravitational force is going to be equals to 1.25 times the magnitude of the normal force. We're being asked to find the magnitude of the marble's velocity at the top of the loop. So the options given are a 1.25 m per second. B 1.41 m per second. C 1.66 m per second and D 2. m per second. So at the top of the vertical loop, the marble is going to be subjected to two different forces which are going to be represented by this diagram right here. So first, I'm going to draw our loop right here and our marble is going to be located at the very top position. Next, the two different forces acting upon our marble are going to be first, the weight of the marble itself, which is represented by the M times G. And the second one is going to be the normal force, which is a result of the interaction between the weight and the circular loop itself. So both the normal force and the weight are going to be pointing vertically downward just like this red and blue arrow representing those two forces. So from the information given in the text, what we know is that MG or the gravitational force is going to equal to 1.25 times the normal force, which is going to be 1.25 times the N we wanna then rearrange our equation so that we have an equation for our normal force which will then be N equals to one divided by 1. of them multiplied by M times G. I'm gonna call this equation one just so that it's easier for us to refer back to this. And then what we wanna do, Dex is to employ Newton Second law. So according to Newton second law, all the forces acting upon our system or Sigma F R in the radial direction will equals to the mass of our system, which in this case is going to equal to the mass of the marble multiplied by A R or the radial acceleration. So the Sigma forces is then going to be M G and N but I'm gonna first define that the downward facing or the down vertical position, vertical direction is going to be our positive direction here. So there is the convention that we're gonna use. So the Sigma F R then going to be M G plus N equals to M multiplied by A R. We want to recall and we wanna connect that the radial acceleration is going to equals to the linear philosophy, which is what we are interested at here divided by the radius of the vertical circular motion, which in this case is the radius of the vertical loop. So we wanna employ that and substitute V squared divided by R into our Newton Second law expression. That will give us M G plus N equals to M multiplied by V squared divided by R. What we wanna do next is to substitute uh our first equation that we have here into our Newton Second law expression. So we want to substitute N equals to one divided by 1.25 M G into our Newton Second Law expression. And from there, we can then solve for our philosophy. All right. So I'm gonna start by doing that. So our Newton second expression will then be M G plus uh M G divided by 1. equals to M multiplied by V squared divided by R. Uh This will then give us a value of 1.8 M G equals to M multiplied by V squared divided by R D M can then be crossed out just so that we have an equation for our velocity. So V squared will then be equals to 1.8 multiplied by G multiplied by R. So our velocity will equals to the square root of 1.8 G R. Now we can actually start uh substituting all of our known values into this equation right here. So that our V Will then equals to the square root of 1.8 GR which will be the square root of 1.8 multiplied by G which is 9.81 m per second squared multiplied by the radius, which is going to be 12.5 centimeters which is 12.5 di multiplied by 10 to the power of minus two m. And that will give us a value for the velocity to be 1.4, 1 m per second around that. So the velocity value of 1. will actually represent the magnitude of the marble velocity at the top of the loop. And that will actually correspond to option B in our answer choices. So option B is going to be the answer to this particular practice problem. So if you guys still have any sort of confusion on this particular topic, please make sure to check out our other lesson videos and they'll be all for this one. Thank you.