What is the Maximum Speed of a Car at the Top of a Hill?

[@tanmayloshali8760]

I LOVE PRACTICAL PHYSICS! THANKS TO YOU THAT YOU PROVIDE US SUCH SITUATIONS TO MAKE US PHYSICS MORE COMFORTABLE. I NEVER SAW SUCH A TEACHER LIKE YOU WHO ILLUSTRATES EACH CONCEPT SO CLEARLY WITH LOTS OF EXAMPLES. LOVE YOU FLIPPING PHYSICS!

[@FlippingPhysics]

Thank you for the love!

[@tanmayloshali8760]

@@FlippingPhysics YOURS WELCOME!!

[@MegaMisch]

This channel is so awesome. Thank you for all the great content. :)

[@FlippingPhysics]

You are welcome. Thanks for the kudos!

[@kainwalker7287]

Totally agree with you

[@adjierexafantasy]

This explained what i want to know easily, while i having a hard time looking for knowing how much is the car max linear velocity without being lifted off the ground

[@phenomenalphysics3548]

2:02 why does normal force decrease with increase in velocity?

[@FlippingPhysics]

Net force in the in-direction equals force of gravity minus force normal equals mass times centripetal acceleration or mass times tangential velocity squared. As tangential velocity gets larger, the net force increases. Force of gravity will not change, so force normal must decrease because the net force increases.

[@phenomenalphysics3548]

@@FlippingPhysics ohhh!! Thank you

[@sosocute3078]

I swear this is awesome! Great

[@FlippingPhysics]

Thanks again!

[@nilluth]

great lesson! but i have one confusion: if Fn=mg-m.Vt^2 /r when Vt is greater than its max value that makes Fn zero, shouldn't Fn be towards" in the ground "?

[@FlippingPhysics]

The point here is that a surface cannot pull on an object, therefore the force normal is always a push. In this example the surface can only push upward, therefore the force normal can only be up. Getting a result that the force normal is down means the force normal no longer exists because the car is no longer in contact with the ground. Hope that helps!

[@eaglekraft687]

@@FlippingPhysics Oh that makes more sense

[@rainbow_doglover8301]

Does this mean that if the radius is larger the net force is smaller? If net force = mass times v^2/r, a larger r would have to mean a smaller net force in? It doesn't feel right that going over a taller hill would mean you have LESS net force in. Does anyone know? Thanks.

[@dannydevito400]

thanks, this was exactly what I was looking for :)

[@FlippingPhysics]

Glad I could help!

[@learningisecstatic9348]

Sir can you please analyse the situation at a point between the top and the starting point. It will be a great learning experience.

[@FlippingPhysics]

Sorry, however, I am not going to take the time to create that video. I do have this in-class lecture which should be helpful though. kzbin.info/www/bejne/oHXZlZl_jtGHaMk The lecture notes are on page 6 of this document. www.flippingphysics.com/uploads/2/1/1/0/21103672/ap_physics_c_flipped_lecture_notes_chapter_05-06_-_all_of_them.pdf

[@learningisecstatic9348]

@@FlippingPhysics sir thank you for your quick reply as usual. But the provided links are not much helpful in this regard to me. Only a free body diagram for the mentioned case would be helpful enough. Sir there is no urgency. During your leisure kindly share a free body diagram mentioning all the forces acting on the car for the case that is when the car is halfway to the top of hill.

[@FlippingPhysics]

Does this video not help? www.flippingphysics.com/car-hill-force-normal.html

[@learningisecstatic9348]

@@FlippingPhysics yes sir it helped a bit. But I wanted the the situation where the car runs on a convex hill like this one and applies force by its engine and don't just run due to the inertia of motion. And friction also appears in the scenario. But anyway thank you sir. By the way sir I am not an examination aspirant, I am just passionate about physics and sir I am glad to say that your videos have answered me a number of questions that were so far tormenting me. One such is why is direction of angular velocity thus defined? And it is due to your video I got the satisfying answer. Thank you so much sir. Learning from you is a real joy.

[@carultch]

@@learningisecstatic9348 When a force is applied by a car's engine, it really is applied by the force of traction (i.e. static friction) from the road pushing forward on the car. The car engine causes the wheels to push backward on the road, and since the road is ultimately attached to Earth, the road acts as a "work mirror" and reflects the force and power onto the car. The road applies its N's 3rd law pair force to push forward on the car. The work can't be done on the road, since the Earth's mass is so large and its movement due to this interaction is negligible. So the work is applied onto the car instead. Construct the free body diagram of the normal force, the gravitational force, and the force of traction, to show the forces that enable a car to climb a hill. This also enables you to determine the maximum acceleration from the traction force limits. Only the powered wheels can participate in supplying the force for speeding up the car, so if the car is not all wheel drive, you'd need to determine how normal force is distributed among the wheels to get the traction limit. It won't be the same as the normal force distribution when the car is parked.

[@goranjohnson4093]

BEST PHSYICS VID OF ALL TIME

[@FlippingPhysics]

THANKS

[@listen_2_d_melody]

2:33 why does Fn equal to 0?

[@FlippingPhysics]

The same reason the force of tension is zero in this example: www.flippingphysics.com/water-bucket-minimum-speed.html

[@kainwalker7287]

Shouldn't the maximum velocity be lesser than what you found? If Fn is zero, it means that there is no contact between the tires of car & road :/

[@FlippingPhysics]

This is really a rhetorical question. There would be no contact, however, there also would be an immeasurable distance between the tires and the ground. So, are the tires touching the ground? Does it really matter?

[@ren5124]

What about the minimum speed required to go over a hill?

[@softlocked9586]

you can solve that problem with potential and kinetic energy. Find the gravitational potential energy at the top of the hill using the bottom as height 0. Then set that potential energy equal to kinetic energy and solve for velocity using the kinetic energy equation. That should be the minimum velocity it takes to hit the very top of the hill.

[@carultch]

If the car is powered at every point along the way, there is no minimum speed. Minimum speed would apply if you were coasting up the hill first.

[@Phoenix-ly5yz]

Isn't a force normal of 0 theoretically impossible? How does max tangential velocity exist if the Force Normal is 0 and the object is tunneling towards the bottom of the Earth?

[@softlocked9586]

It's not tunneling towards the bottom of the earth. The force of gravity in the inward direction minus the force normal is the centripetal acceleration which increases as the tangential velocity increases. That means if the tangential velocity gets bigger and bigger, the force normal must get smaller and smaller (the force of gravity cannot change because the acceleration due to gravity on earth at sea level is always 9.8 meters per second, and the toy car has a constant mass throughout the entire problem). Eventually, the tangential velocity and, by extension, the centripetal acceleration will be so great that the normal force hits zero. The tangential velocity can increase, but the normal force cannot. This means that if the car goes any faster, it will launch itself in a tangent when attempting to traverse the hill. The normal force is not 0 because it doesn't exist and the car is plummeting to the bottom of the earth, the normal force is 0 because the car is traveling so fast over the hill that once the car is going downhill, the car is not pushing down on the hill and the hill is not pushing up on the car. It's at that perfect balance where the car is gliding just on top of the hill.

[@RoudhaAlhamli]

@@softlocked9586 yeah but there is still contact between the car and the hill

[@softlocked9586]

@@RoudhaAlhamli there’s contact but there’s no force going down. It’s just brushing against the hill at that point.

[@RoudhaAlhamli]

@@softlocked9586 thank you so much! you explain better than my physics teacher

[@mybabyalulu]

@@softlocked9586 you are amazing for this. thank you for explaining it so well !!