I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.Delta2 said:The key to this problem is to "translate" the statement "It loses contact with the track" to a fairly simple equation. In order to be of more accurate help I will ask you "Which of the forces on the toy becomes zero when it loses contact with the track". Set this force to zero so you have one additional equation together with the equations of conservation of energy and the equations of centripetal force.
Also I find it easier instead of the height to calculate the angle that the radius to the point of lost of contact makes with the horizontal.
Witdouck K said:I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.
Yes that's right the normal force becomes zero. You got to write a detailed post giving all the details of what equations you made and how you worked with them and also a diagram with the track and the forces at the point of lost contact will be nice so I can pinpoint where you might have gone wrong.Witdouck K said:I think the normal force becomes zero when the toy loses contact. But when I use this and calculate the angle you specified, and then calculate the height h coresponding to this angle, I get the non sensical anwser of h = -1 / 3 R, which seems really strange. Maybe I made a mistake that I didn't notice, but I still don't know why I can't get the right answer.
On closer inspection, I indeed put the zeroth level of the potential energy at the starting point, I just didn't really realise it untill you pointed it out. When putting it at the center of the loop, I als got ##\frac{2}{3}R##, but I know this is just the same answer as ##\frac{-1}{3} R## because of the translation of the origin. Thanks a lot!Delta2 said:Yes that's right the normal force becomes zero. You got to write a detailed post giving all the details of what equations you made and how you worked with them and also a diagram with the track and the forces at the point of lost contact will be nice so I can pinpoint where you might have gone wrong.
If you cant type in Latex and make a diagram with some drawing software, I guess as a last resort you can post screenshots of your handwritten work of the problem.
By the way the answer I get for the angle is ##\sin\theta=\frac{2}{3}## . So the height is ##\frac{2}{3}R##. Come to think of it your height of -1/3R means that you might took as zero level of potential energy the level at the starting point. I took as zero level of potential energy the horizontal diameter of the loop.
P.S Actually no, its not where you put the zeroth level of potential energy but where you put your coordinate system. I put it centered at the center of the loop, it seems to me you put it centered at the starting point of the toy.
I think its not where you put the zeroth of PE but where you put the origin of coordinate system. Usually these two coincide, i.e the zeroth level of PE is the x-axis of the coordinate system but its not always the case.Witdouck K said:On closer inspection, I indeed put the zeroth level of the potential energy at the starting point, I just didn't really realise it untill you pointed it out. When putting it at the center of the loop, I als got ##\frac{2}{3}R##, but I know this is just the same answer as ##\frac{-1}{3} R## because of the translation of the origin. Thanks a lot!
The minimum speed required for the toy car to complete the loop can be calculated using the formula \( v = \sqrt{rg} \), where \( r \) is the radius of the loop and \( g \) is the acceleration due to gravity. This ensures that the car has enough centripetal force to stay on the track at the top of the loop.
The toy car falls off the track at the top of the loop if it does not have enough speed to generate the required centripetal force to counteract gravity. This means the car's velocity is too low to maintain contact with the track, causing it to fall.
Friction between the toy car's wheels and the track can both help and hinder its motion. While friction is necessary for the car to accelerate and maintain speed, excessive friction can slow the car down, reducing its speed below the necessary threshold to complete the loop.
The mass of the toy car does not directly affect the minimum speed required to complete the loop, as both gravitational force and centripetal force are proportional to mass. However, a heavier car may experience more frictional forces, which could slow it down more than a lighter car.
The shape and size of the loop are crucial factors in determining the car's motion. A larger loop requires a higher minimum speed to complete, while an uneven or irregularly shaped loop can introduce additional forces that disrupt the car's motion. Ensuring a smooth, circular loop helps maintain consistent centripetal force throughout the motion.