Introductory Centripetal Force Problem - Car over a Hill

[@yihongsun4536]

you got me a 5 in both AP Physics 1 and AP Physics C Mechanics, just came to say thank you.

[@FlippingPhysics]

That is awesome! You are welcome. Although, I will say, I may have helped you, however, _you_ got the 5s, not me. 👍

[@TheKiyoshiRyu]

Hey man, if you remember and don't mind me asking, how else did you study for the ap physics 1 exam? My test is coming up in May :(((

[@uah9031]

Good question. We don't have enough practice material either...

[@sosocute3078]

Very helpful, I'm studying for my finals and this video honestly made me solve lots of problems immediately. Thanks

[@FlippingPhysics]

So glad to help!

[@gh0stGD]

thank you this helped me better understand the hardest topic thus far in AP physics 1

[@FlippingPhysics]

That's great! Glad you understand it now.

[@lester8940]

You're for real a life saver

[@tinimunson1242]

Thank you for this video, but may I ask why do you feel you weigh less on the top of the hill? If the normal force is less, then that means the force of gravity is more than the normal force, so you should feel as if you weigh more because the force of gravity is larger, right? Please correct me if I am wrong. Anyway, your channel is amazing !!!!

[@FlippingPhysics]

1) You do not feel your weight (also known as the force of gravity). www.flippingphysics.com/apparent-weight.html 2) Your mass does not change in this situation, therefore, your weight also does not change. www.flippingphysics.com/weight-not-mass.html

[@2MinMaths]

Why don't you consider static friction in this problem? In the mint problem in your other video, you had kinetic friction in your force diagram, but it's not here-- why not/

[@carultch]

Static friction acts parallel to the surface of the road, and is perpendicular to the acceleration in this problem. Therefore, it isn't relevant to the force addition equation that matters.

[@2MinMaths]

@@carultch How would static friction be perpendicular to acceleration? Isn't acceleration of the car caused by static friction of the wheels?

[@carultch]

@@2MinMaths Acceleration can happens in any and all of the three axes. The static friction of the wheels can only cause acceleration parallel to the road surface. So static friction can only account for two out of three of the axes of acceleration. Gravity and normal force are the forces that need to account for any acceleration perpendicular to the road surface. In this particular case, we aren't interested in any changes in speed or steering, which are the kinds of acceleration that static friction would cause. The speed is already specified, and assumed to be constant, and car is assumed to be driving along a road that is straight as viewed from an aerial view, such that no steering is necessary.

[@carultch]

@@2MinMaths We could make a similar problem more interesting, and set it up so that static friction does play a role. Suppose the car is speeding up at a rate of 2 m/s^2, while passing over the crest of the hill with a 1.8 m radius hill crest, using its internal motor to drive the wheels. The car had a coefficient of static friction of 0.5 with the ground, and all four wheels are driven by its motor. What is the maximum speed it could have concurrently with this acceleration, so that the wheels don't skid?

[@2MinMaths]

@@carultch Thanks for your response-- the insight was really helpful. I took an attempt at your word problem and here are the steps I took: MY FBD looked like Fg pointing down, N pointing up, and fs pointing straight left, assuming the car was moving forward (in the direction of the acceleration). I know fs = MsN, but realized that wasn't useful so rewrote it as fs = ma, then substituted 2 for a and rewrote it as MsN = 2m. I didn't have mass, so I wrote another equation: Fg - N = mv^2/r, then rewrote N as m(g-v^2/r), and substituted that in for N in MsN = 2m. Now I had mass on both sides since the new equation is Ms(m(g-v^2/1.8)) = 2m. Canceled out the masses, then divided both sides by Ms where the equation now becomes g - v^2/1.8 = 2/Ms, and isolated v to get sqrt (1.8(g-2/Ms)), giving me velocity as 3.23 m/s. I'd love to know if my approach was correct or if there was an error in either my math or conceptually. Thanks!

[@jsivesh4323]

I have doubt in another kind of uniform circular motion of a car. If a car makes a curve on a level road which force will act as the centripetal force?

[@science1878]

force of friction between the tires and the road

[@dominickkrahnstoever7362]

7:02 you can tell who DOESNT like roller coasters.

[@physicsx5055]

Hope you will reach 100k subscriber and get silver play button. Best of luck☺️☺️😊😊👍💐

[@FlippingPhysics]

Someday...

[@Ndiedddd]

@@FlippingPhysics that day is soon!

[@gauravkumar-nn9mf]

Love from India

[@BanValsimot]

But if we are rotating on the planet Earth that means that the normal force and weight are never really the same? There is always a centripetal force due to a gravity that will make the difference between our weight and normal force?

[@FlippingPhysics]

This is essentially the same thing as the fact that we ignore air resistance, assume the acceleration due to gravity on Earth is constant, ignore buoyancy forces on most objects, etc. The radius of the Earth is roughly 6.375 x 10^6 meters and you know the planet rotates once every 24 hours, so you can figure out the angular velocity of the planet. Therefore, you can determine the centripetal acceleration of an object on the surface of the planet to determine how much error there is by calling that centripetal acceleration negligible.

[@BanValsimot]

@@FlippingPhysics That makes sense. Tnx a lot :D

[@griffinbrooks7705]

bro thx