– Good morning. Bo, please read the problem,

and Billy, please translate. ♫ Flipping Physics ♫ – A 0.453-kilogram toy car

moving at 1.15 meters per second is going up a semicircular

hill with a radius of 0.89 meters. – Mass equals 0.453 kilograms. Tangential velocity equals

1.15 meters per second, and radius equals 0.89 meters. – When the hill makes

an angle of 32 degrees with the horizontal, what is the magnitude of the force normal on the car? – Theta equals 32 degrees and force normal equals question mark. – Mr. P, could you please

clarify that angle? I’m not really sure what it

means it says, “when the hill makes an angle of 32

degrees with the horizontal.” – Sure, Bobby. Let me clarify that angle. As the car goes up the

hill, the angle of the hill where the car is located increases. If we draw a line which

intersects the center of the two wheels, you can see how the

angle of the hill increases. – [Bobby] Sure. – [Mr. P] And if we pause

the car at the point we are talking about, you can see

how the hill makes an angle of 32 degrees with the horizontal. – [Billy] You know, that looks

just like the diagram we used when broke the force of

gravity into components on an incline and determined the equations for the force of gravity parallel and perpendicular to the incline. – [Mr. P] Yes, Billy, it does,

and we are going to use that fact in a few moments, thank you. – [Billy] You are welcome!

– [Bo] Nice. – Okay, Bobby, please start the problem. – Yeah, um. Let’s draw a free-body diagram. The force of gravity is straight

down and the force normal is up and perpendicular

to the hill. That’s it. So let’s break forces into components. The car is moving in a circle,

so we need to break forces into components in the in

direction and a direction which is perpendicular to the in

direction, which is, in this case, tangent to the circle or in

the tangential direction. The force normal is in the in

direction, so we don’t need to break that into components. That means we need to break

the force of gravity into components in the in direction

and the tangential direction. However, because theta is

just like an incline angle, we have our equations for

the force of gravity parallel and perpendicular, which use

the angle of the incline. We don’t have a numbers

dependency, so let’s not plug in numbers just yet, let’s just

redraw our free-body diagram. The force normal is still

there, but now we have the force of gravity perpendicular,

which is in the out direction, and the force of gravity

parallel, which is in the tangential direction and down. – Billy, please finish the problem. – Now we sum the forces

in the in direction. The net force in the in

direction equals force normal minus force of gravity

perpendicular, and the net force in the in direction equals

mass times centripetal acceleration, add force

of gravity perpendicular to both sides to solve for force normal, and now we can substitute in equations. Force of gravity perpendicular

equals mass times acceleration due to

gravity times cosine theta, and centripetal acceleration

equals tangential velocity squared, divided by radius. Next, let’s plug in numbers. So that’s 0.453 times 9.81

times the cosine of 32 degrees, plus 0.453 times 1.15 squared,

divided by 0.89, which equals 4.4418, or 4.4 newtons with

two significant digits. – Very nice. A couple of things to point out. First, the tangential velocity of the car, 1.15 meters per second, is

an instantaneous velocity at a specific point in time,

because the car slows down as it goes up the hill. Second, the magnitude of

the force normal decreases as the car goes up the hill. Both because the instantaneous

tangential velocity of the car is decreasing,

but also because the force of gravity perpendicular

decreases as theta, the angle of the hill, increases. Thank you very much for

learning with me today. I enjoyed learning with you.

β€οΈ

Thanks for the video! I'll never

*tire*myself out on a similar problem again.Yes BOY YOU ARE MY HERO I LOVE YOU

You deserve more subscribers. Keep up the good work π

your videos are so useful!!! thank you so much π

Thank you so much for all the videos you've uploaded; they've saved me in AP Physics this year! On a sidenote, do you know when you will have videos on AP Physics C E&M? If so, do you think that they may be out before the upcoming exam in May 2018?

Once again, thank you for all the effort you have put into this channel.

Why is force of gravity being added by centripetal force. Shouldn't it be subtracted or is this one of those situations where you have to use centrifugal force?

If there are agents for force and the force of gravity can't be seen, is the force of gravity a secret agent?

What a legend. I am from the UK and am so glad I found your videos. Keep up the superb work!!!!!

Thank you very very very very muchππππππππ

Superb

Should not we calculate cosinus for angle 90-32? I don't get it…

Hey Mr.P, quick question, is there no force of kinetic friction because the tangential velocity is instantaneous? Was just a little bit confused as to why there was no friction.