- Good morning.
Bo, please read the problem
and Billy please translate.
♫ Flipping Physics ♫
- A 453 gram toy car moving at 1.05 meters per second
is going over a semi-circular hill
with a radius of 1.8 meters.
- Please stop.
The mass of the car equals 453 grams,
which we multiply by one kilogram over 1,000 grams
to convert to 0.453 kilograms.
The speed of the car is 1.05 meters per second.
- [mr.p] Actually, Billy, that is the magnitude
of the tangent or velocity
of the car because the car is moving along a circle.
- Right, that makes sense.
Thanks.
And the radius of the circle is 1.8 meters.
Bo.
- When the car is at the top of the hill,
what is the magnitude of the force
from the ground on the car?
- The force from the ground on the car
is the force normal,
so force normal equals question mark.
- Great, Bobby, what do we do next?
- Well, we are solving for a force,
so do we draw a free body diagram
and sum the forces?
- Yes, but before we do,
let's review my five steps
to help solve any free body diagram problem.
Class, step one is?
- Draw the free body diagram.
- Step two?
- Break forces into components.
- Step three?
- Redraw the free body diagram.
- Step four?
- Sum the forces.
- And step five?
- Sum the forces.
- Again.
- Correct.
Remember, step five is to sum the forces
in a direction perpendicular to the direction in step four.
Booby, please begin.
- The force normal is up
and the force of gravity is down.
That's it.
So, there are no forces to break into components,
so we can skip doing that
and we can skip redrawing the free body diagram.
The net force in the y direction equals force normal
minus force of gravity
which equals mass times acceleration in the y direction
and the acceleration in the y direction equals zero,
so the force normal equals the force of gravity.
We know the equation for the force of gravity it equals.
- Hold up, Bobby.
I don't think what you're doing is right.
- Of course it is.
We're seen this free body diagram
an innumerable number of times
and the force normal always equals the force of gravity.
- But we have not done this with an object
moving in a circle before.
- [Billy and Bobby] Oh.
- Bo is correct.
Be careful.
The car is moving in a circle
which means the y direction is not the correct direction
to sum the forces in.
Billy, in what direction do we need to sum the forces in?
- Yeah, the in direction.
We need to sum the forces in the in direction. Nice.
- Correct.
Whenever an object is moving in a circle,
you need to identify where that circle is
so you can identify what direction
the in direction is.
In this case, the car is moving in a vertical circle
and therefore when the car is at the top of the hill,
the in direction is straight down.
Bobby, please sum the forces in the in direction.
- Okay, the net force in the in direction
equals force normal minus force of gravity
which equals mass times acceleration in the in direction.
- [mr.p] And what is the acceleration
in the in direction called?
- Centripetal acceleration.
- I thought in was positive and out was negative,
so the force of gravity should be positive
and the force normal should be negative, right?
- That is correct.
Students are so used to summing the forces
in the y direction that this is a common mistake.
Remember, we are summing the forces in the in direction
and inward is positive
and outward is negative.
Bo, in this example at the top of the hill,
what is the centripetal force acting on the car?
- Isn't that just the force inward,
so the force of gravity?
- No, the centripetal force is the net force
in the in direction.
- Right.
So, the centripetal force or the net force
in the in direction is the force of gravity
minus the force normal.
- [mr.p] Yes, remember, the centripetal force
is made up of other forces
because it is the sum of the forces in the in direction
and in this specific situation,
the centripetal force is the force of gravity
minus the force normal.
Bobby, please continue solving the problem.
- Okay.
Force of gravity or mass times acceleration
due to gravity minus force normal
equals mass times,
we can substitute tangential velocity squared over radius
for the centripetal acceleration.
We are solving for the force normal,
so subtract mass times acceleration
due to gravity from both sides
and now multiply the whole equation by negative one
to get force normal equals mass times acceleration
due to gravity minus mass
times tangential velocity squared over radius.
We can substitute in numbers to get 0.453
times 9.81 minus 0.453
times 1.05 squared
divided by 1.8
which is 4.1665
or 4.2 Newtons with two significant digits.
- Now, an important thing to realize here
is how the magnitude of the force normal
compares to the magnitude of the force of gravity.
The force of gravity acting on the car
is roughly 4.4 Newtons
which means the force normal
is less than the force of gravity.
- You know you can feel that, right?
- Hello, Kit.
- Who is that?
- Everybody, this is Kit from Gorilla Physics.
- Hi chaps.
- Yo, whatup?
- Hi, hi.
- Okay, Kit, now I know what you mean
but could you please explain yourself
a little bit more?
- Yeah, no bother.
Well, as you go over the crest of a hill,
you feel like you weigh less
because you feel less of a normal reaction force
from the ground
and the faster you go,
the less of the reaction force you'll feel.
You know that funny feeling like your stomach's rising
but it's fun.
It's like on a rollercoaster when you go over the top
of a crest.
The normal reaction force is less,
so you feel as if you weigh less.
- I love that feeling.
I love rollercoasters.
- Me too.
They are so much fun.
- Kit, thank you very much for your help.
- Yeah, my pleasure, thanks.
- Okay, this idea that the force normal decreases
the faster you drive the car
over the hill is one we are going to explore further
in our next lesson.
Until then, thank you very much for learning
with us today.
We enjoyed learning with you.