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# Bowling ball in vertical loop

## Video transcript

imagine that in an effort to make bowling more exciting bowling alleys put a big loop de loop in the middle of the lane so you had to bold the ball really fast to get the ball up and around the loop and then only afterward it would go hit the bowling pins kind of like mini golf bowling or something like that well if you're going to build this you'd have to know at the top of the loop this structure is going to have to withstand a certain minimal amount of force you might want to know how strong you have to make this you can't have this thing breaking because it can't withstand the force of the bowling ball so let's ask ourselves that question how much force is this loop structure going to have to be able to exert while this bowling ball is going around in a circle and let's pick this point at the top to analyze we'll put some numbers in here let's say the ball was going eight meters per second at the top of the loop that's that's pretty darn fast so someone really hurled this thing through here now let's say the loop has a radius of 2 meters and the bowling ball has a mass of 4 kilograms which is around 8 or 9 pounds now that we have these numbers we can ask the question how much normal force is there going to be between the loop and the ball so in other words what is the size of that normal force the force between the two surfaces this is what we'd have to know in order to figure out if our structure is strong enough to contain this bowling ball as it goes around in a circle and it's also a classic centripetal force problem so let's do this what do we do first we should always draw a force diagram if we're looking for a force you draw a force diagram so what are the forces on this ball you're going to have a force of gravity downward and the magnitude of the force of gravity is always given by M times G where G represents the magnitude of the acceleration due to gravity and we're going to have a normal force as well now which way does this normal force point a common misconception people want to say that that normal force points up because in a lot of other situations the normal force points up if you're just standing on the ground over here the normal force on you is upward because it keeps you from falling through the ground but that's not what this loop structures doing up here the loop structure isn't keeping you up the loop structure is keeping you from flying out of the loop and that means this normal force is going to have to point downward so this is weird for a lot of people to think about but because the surface is above this ball the surface pushes down surfaces can only push if the surface is below you the surface has to push up if the surface was to the side of you the surface would have to push right and if the surface was to the right of you the surface would have to push left normal forces in other words always push so the force on the ball from the track is going to be downward but vice versa the force on the track from the ball is going to be upward so if this ball were going a little too fast and this we're made out of wood you might see this thing splinter because there's too much force pushing on the track this way but if we're analyzing the ball the force on the ball from the track is downward and after you draw a force diagram the next step is usually if you want to find a force to use Newton's second law and to keep the calculation simple we typically use Newton's second law for a single dimension at a time ie vertical horizontal centripetal and that's what we're going to use in this case because the normal force is pointing toward the center of the circular path and the normal force is the force we want to find we're going to use Newton's second law for the centripetal direction and remember centripetal is just a fancy word for pointing toward the center of the circle so let's do it let's write down that the centripetal acceleration should equal the net centripetal force divided by the mass that's going in the circle so if we choose this we know that the centripetal acceleration can always be re-written as the speed squared divided by the radius of the circular path that the object is taking and this should equal the net centripetal force divided by the mass of the object that's going in the circle and you got to remember how we deal with signs here because we put a positive sign over here because we have a positive sign for our centripetal acceleration and our centripetal acceleration points toward the center of the circle always then in toward the center of the circle is going to be our positive direction and that means for these forces we're going to plug in forces toward the center of the circle is positive so let's do that this is the part where most of the problems happen you got to be really careful here I'm just going to plug in what are the centripetal forces to figure that out we just look at our force diagram what forces do we have in our diagram we've got the normal force and the force of gravity let's start with gravity is the gravitational force going to be a centripetal force first of all that's the question you have to ask does it even get included in here at all and to figure that out you ask does it point centripetal e ie does it point toward the center of the circle and it does so we're going to include the force of gravity moreover because it points toward the center of the circle as opposed to radially away from the center of the circle we're going to include this as a positive centripetal force similarly for the normal force it also points toward the center of the circle so we include it in this calculation and it as well will be a positive centripetal force and now we can solve for the normal force if I solve algebraically I can multiply both sides by the mass and then I'd subtract mg from both sides and that would give me the mass times V squared over R minus the magnitude of the force of gravity which if we plug in numbers gives us four kilograms times eight meters per second squared you can't forget the square divided by a two meter radius minus the magnitude of the force of gravity which is four kilograms times G which if you multiply that out gives you eighty eight point eight Newtons this is how much downward force is exerted on the ball from the track but from Newton's third law we know that that is also how much force the ball exerts upward on the track so whatever you make this loop out of it better be able to withstand eighty eight point eight Newtons if people are going to be rolling this ball around the loop with eight meters per second now let me ask you this what if the ball makes it over to here right so the ball rolls around and now it's over at this point now how much normal force is there at this point is it going to be greater than less than or equal to eighty eight point eight Newtons well to figure it out we should draw a force diagram so there's going to be a force of gravity again it's going to point straight down and again it's going to be equal to at least the magnitude of it will be equal to the mass times the magnitude of the acceleration due to gravity and we also have a normal force but this time the normal force does not push down remember surfaces push outward and if this surfaces to the left of the ball the surface pushes to the right this time our normal force points to the right and let's assume this is a well oil track so there's really no friction to worry about in that case these would again be the only two forces so what the answer to our question will this normal force now be bigger less than or equal to what the normal force was at the top well I'm going to argue it's got to be bigger and I'm going to argue it's going to have to be much bigger because when you plug in over here into the centripetal forces you only plug in forces the point radially that is to say centripetal e-either into the circle which would be positive or radially out of the circle which would be negative if they neither point in tune or out of the circle you don't include them in this calculation at all because they aren't pointing in the direction of the centripetal acceleration in other words they're not causing this centripetal acceleration so for this case over here gravity is no longer a centripetal force because the force of gravity no longer points toward the center of the circle this force of gravity is tangential to the circle it's neither pointing into nor outer which means it doesn't factor into the centripetal motion at all it merely tries to speed the ball up at this point it does not change the ball's direction which means it doesn't contribute to making this ball go in a circle so we don't include it in this calculation so when we solve for the normal force we'd multiply both sides by M we would not have an mg anymore so we wouldn't be subtracting this term and that's going to make our normal force bigger moreover the speed of this ball is going to increase compared to what it was up here so as the ball falls down gravity's going to speed this ball up and now that it's speed is larger and we're not subtracting anything from it the normal force will be much greater at this point compared to what it was at the top of the loop so recapping when you want to solve a centripetal force problem always draw your force diagram first if you choose to analyze the forces in the centripetal direction in other words for the direction in toward the center of the circle make sure you only plug in forces that are into radially into the circle or radially out of the circle if they're radially end of the circle you make them positive if they were radially out of the circle you would make them negative and if they neither point radially inward toward the center of the circle or radially outward away from the center of the circle you just do not include those forces at all when using this centripetal direction