![]() ![]() I understand that at the top of the loop de loop, centripetal acceleration is attributed by gravity and normal force, but what EXACTLY IS PREVENTING the car from falling off? Gravity and normal force are pointed towards the centre of the loop de loop, and if centripetal acceleration needs to be equal to or greater than these combined forces, shouldn't the car just fall off the track because there is an overall net force towards the centre? I've had to revisit this topic because I realised that I didn't really understand it all! ![]() If the car did not go in a circular path, then it would not experience centripetal acceleration. That's literally the definition of centripetal acceleration. Remember that the car also has forward velocity, so the centripetal force affects the direction of the forward velocity in such a way that it makes the car go in a circle. The centripetal force always points towards the center, but that does not mean that the car will go towards the center. ![]() For a minimal speed, the only force that the car is experiencing (at the top) is the force of gravity, so by simply saying that the force of gravity is a centripetal force, the car has to stay in a circular motion. So if we want to make sure that the car does not fall, we have to maintain it's circular motion, which means that all the forces on the car have to sum up to a centripetal force. So if something is moving along a circle it's experiencing centripetal acceleration, or the other way around, if something is experiencing centripetal acceleration, then it has to move in a circular path. ![]() Every object that moves in a circular path, with a constant speed, experiences a centripetal acceleration. For the car to not fall, it needs to stay in a circular path. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |