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## Physics library

# Bowling ball in vertical loop

In this video David explains how to find the normal force on a bowling ball rolling in a vertical loop. Created by David SantoPietro.

## Want to join the conversation?

- At7:53, David says that the speed of the ball in the outermost right position will increase compared to the speed in the top position. And in the case of vertical circular motion, it seems quite intuitive to me. But when we derived formulas for centripetal acceleration in earlier videos, we assumed that the speed is constant. Can you explain why we are still allowed to use these formulas in this case? Thank you!(25 votes)
- When we derive the formulas for centripetal acceleration, we are only looking at a single force at a time. The centripetal force will not increase the speed of the ball by itself; it will only change the
*direction*of the velocity just enough to make the ball travel in a circular path. If there are no other forces than the centripetal force, the speed will be constant. Therefore, we don't have to assume that the speed of the object is constant, only that the speed of the object is unaffected by the centripetal force.

In this example, however, we also have to consider the force of gravity. At any given moment, the centripetal force is equal to Fn + Fg ⋅ sin(x). When x = 180 degrees, the force of gravity does not contribute to the centripetal force. This does not mean that gravity doesn’t accelerate the ball, it only means that gravity doesn’t change the*direction*of the velocity. If it did, the ball couldn’t travel in a circular path anymore.(9 votes)

- What if the bowling ball was somewhere at an angle? Would gravity affect it?

(putting a clock for comparison) if the bowling ball was at 10 o'clock (which is between 12 o' clock and 9 o'clock) it would be at an angle...(6 votes)- Gravity will always be accelerating the ball by 9.8 m/s² downward no matter where the ball is. What changes is the normal force of the track pushing on the ball as it moves through the loop. Normal force will be greatest at the bottom of the loop, smallest at the top, and somewhere in between those two values based on the angle of the centrifugal force + gravitational force to the surface.(6 votes)

- Isn't the ball also exerting a force on the loop?(6 votes)
- Yes, that is the reaction force to the normal force, so it is equal to the normal force. The force diagram does not include the force that the
**ball is exerting**on the loop because the force diagram only shows what forces are**exerting on the ball**.(2 votes)

- I'm a bit confused; since at the top of the circle mg is acting downwards, why is there a normal force there in the first place? Should not (acc. to Newton's 3rd Law) the force exerted on the bowling track (Which is what we are looking for) be the pairing force with mg?(3 votes)
- It depends how fast the ball is going. If the ball is going very fast, mg will not be sufficient force to keep the ball on the circular path, so the track will have to do some pushing as well. The way we find the minimum required speed is to ask how slow can the ball go before the normal force becomes zero, which is when the ball is just losing contact with the track (i.e. "falling")(7 votes)

- How do you calculate the speed of the ball when it is at the outermost right or left side of the loop?(4 votes)
- In the video, you already know what the normal force is so you can equal it to mv¨2/r. Or, you could calculate v with energies, ET at the top= PE+KE, at the bottom ET= KE, and depending on how position changes, PE will change. Let's say PE decreases, KE will have to increase because energy is conserved around the loop (assuming it's a perfect circle)(3 votes)

- In case of rotating a string , is the normal force replaced by tension? Are their magnitude also equal?(3 votes)
- yes, the centripetal force is provided by tension instead of a normal force.

In other instances it might be provided by gravity, or by an electromagnetic force.(3 votes)

- Why do we need to include the normal force when calculating the centripetal force? Wouldn't the normal force get cancel out by the force exerted by the ball on the loop since it is an action-reaction pair?(4 votes)
- I hope I'm not misleading you now. It seems to me that 3rd law and action-reaction pairs are taken into account when dealing with a System of two or more bodies. Acc. to 2nd law, change in motion is due to External net force. Yes, the ball may communicate force upward at the top, i.e. on the loop de loop, but this force is not external, does not applies to the ball's motion. What loop experiences is of no interest; hopefully it's rigid.(1 vote)

- wait, isn't the normal force always perpendicular to the force of gravity? If all the forces are pointing toward the center of the circle (gravity, normal force) then why doesn't the ball just do that: go towards the center of the circle, maybe even fall towards the center?(1 vote)
- No, the normal force is perpendicular to the surface at the point of contact. For example it you have a ramp the normal force is only in the direction perpendicular to the ramp's surface, this only opposes a portion of the force of gravity and that is why things will tend to slid down a ramp if not stopped by friction.(4 votes)

- Let say if I want to calculate the centripetal forces at any given moment, am I plugging in force at that particular moment? In this case, the ball is speeding up when it is on the right hand side of the loop because its velocity is having the same direction with gravitational force. When I calculate centripetal force, am I plugging in just the initial velocity (8m/s in this case) of the ball or plugging in the sum of mg and the speed of the ball for v^2 ?(2 votes)
- For the sake of this video David simplified it to the initial velocity, i think. But in regular terms if you were solving for something in real life you would have to plug in the velocity at the current time, so in the case it would be larger than 8 meters.(1 vote)

- When the ball is in the second case where it is at the left of the track, do we include the velocity of it so as to calculate its force? Cuz I see there is only force of gravity pointing downward but there is no sign of velocity of the ball.(2 votes)
- Velocity is not a force, the only force was the gravity pointing downwards.(1 vote)

## Video transcript

- [Narrator] Imagine that in an effort to make bowling more exciting, bowling alleys put a big loop-the-loop in the middle of the lane,
so you had to bowl the ball really fast to get the
ball up and around the loop and then only afterward, it
would go hit the bowling pins kinda like mini golf bowling
or something like that. Well if you were gonna build this, you'd have to know at the top of the loop, this structure's gonna have to withstand a certain minimum amount of force. You might wanna know how strong
do 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 gonna 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 pretty darn fast
so someone really hurled this thing through here. Now let's say the loop
has a radius of two meters and the bowling ball has
a mass of four kilograms, which is around eight or nine pounds. Now that we have these numbers,
we can ask the question: How much normal force is there gonna 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 gonna 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 gonna have a
normal force as well. Now which way does this
normal force point? A common misconception,
people wanna 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
structure's doing up here. The loop structure isn't keeping you up. The loop structure's keeping
you from flying out of the loop and that means this normal force is gonna 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 gonna be downward but vice versa. The force on the track from
the ball is gonna be upward. So if this ball were
going a little too fast and this were 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 wanna 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 at time, i.e. vertical,
horizontal, centripetal. And that's what we're
gonna 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 wanna find, we're gonna 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 gotta 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 gonna plug in forces toward the center of the circle as positive. So let's do that. This is the part where most
of the problem is happening. You gotta be careful here. I'm just gonna 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 centripetally? I.e. does it point toward
the center of the circle? And it does so we're gonna
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 gonna 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 88.8 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 88.8 newtons if people are gonna 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 gonna be greater than, less than, or equal to 88.8 newtons. Well to figure it out, we
should draw a force diagram. So there's gonna be a force of gravity. Again, it's gonna point straight down, and again, it's gonna be equal to at least the magnitude
of it will be equal to the mass times the magnitude
of acceleration due to gravity. And then we also have a normal force, but this time, the normal
force does not push down. Remember, surfaces push outward and if this surface is
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 a well oiled track so there's really no
friction to worry about. In that case, these would
again be the only two forces. So what about 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 gonna argue it's gotta be bigger, and I'm gonna argue it's
gonna have to be much bigger because when you plug in over here, into the centripetal forces, you only plug in forces
that point radially. That is to say centripetally, either into the circle,
which would be positive, or radially out of the circle,
which would be negative. If they neither point into
nor 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 the 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 out of, 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 solved 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 gonna make
our normal force bigger. Moreover, the speed of
this ball's gonna 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 wanna solve the 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 into 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.