Physics 9 Conservation of Energy (10 of 11) Skiing Down A Hill

[@MichelvanBiezen]

You may want to look at this playlist: PHYSICS 8 WORK, ENERGY, AND POWER

[@AZèro026]

This channel is amazing ! It is the best physics channel out there! Keep up the good work !

[@ahmedal-ebrashy3691]

5:08 was very omtersting because I never thought of it this way. Brilliant work.

[@binoanil6954]

Same here bro

[@rubegoldberger]

It blows my mind that regardless of gravity, mass, or radius, the angle will always be 48.2 degrees.

[@MichelvanBiezen]

Physics is filled with these interesting discoveries.

[@emilianorodriguez807]

What solves this problem is setting PE final equal to 0, the position h = 0, relative to the moment the skier leaves the dome; thus, an equation for the Final Velocity, which we plug into the equation for centripetall force. In other words, by finding the Final Velocity that the centripetal force will have, we can find the angle. Had to think about that, realized that because of energy conservation, the amount of mass and the length of radius doesn't matter, they simply cancel out. Thank you Professor!

[@lologomez7093]

You made it look so simple thank you

[@louiscorpuz3761]

isn't it that the sign of the mgcostheta that you used in centripetal force is negative because it lies on the negative y axis?

[@MichelvanBiezen]

Only if you were to express it as a vector, (and then you would need both the x and y component). But not as a magnitude as was done in the video.

[@paulcon1038]

Excellent explanation!

[@MichelvanBiezen]

Thank you. Glad you found our videos. 🙂

[@syedhayat7275]

God bless you, your a life saver

[@faicornelius2601]

Thank you so much for the lecture, Michel.

[@MichelvanBiezen]

You're most welcome!

[@flawns]

It's a shame you stopped right here.....There's more lectures to do!

[@alexmorfi8321]

When doing the centripetal force, what happens to the normal force? Isn't it going to be N-mgcos(theta) = mv^2/R ?

[@MichelvanBiezen]

alex morfi Technically, the centripetal force is the perpendicular component of the weight = mg cos(theta) When the velocity reaches some point, mg cos(theta) will no longer be enough to keep the skier on the slope.

[@mokarrommolla5820]

Hello sir. I get a question for you. If the mass is changing, are we allowed to use the Conservation of Energy method? or it's a good idea to use the principle of impulse? thanks

[@ernestbeckham2921]

Brilliant work professor. 💕💕💕

[@MichelvanBiezen]

Thank you! 😃

[@vaibhavbangia]

Sir, in your videos on circular motion, the concept that is used is that, if the CeintriFugal force is greater then the mg component then the object/skier will leave the circular path . can that concept be used here?

[@MichelvanBiezen]

Yes, that would work great in this example.

[@vaibhavbangia]

Thank you sir ☺

[@الاستاذحسينعلييعقوب]

Please sir what is the difference between this example (vertical motion )and that with body of (5kg) rotating also vertically when you said that the velocity will remain constant ( 6. m/sec). - - -thank you very much sir

[@MichelvanBiezen]

We are sorry, but we didn't understand your question. What specifically would you like to know?

[@الاستاذحسينعلييعقوب]

Michel van Biezen thank you very much . . . . You have another vedio with an object its mass(5kg)rotating by arod its velocity in the top is (6 m/s) and you said it will remain same in bottom .why it is remain constantbut in this example the velocity is not constant .thank you again

[@veertjuh7]

Hey there it's me again with another question. I don't really understand the part where you mgcos(θ) = (mv^2)/R. Why do you equal these 2?

[@MichelvanBiezen]

When mgcos(theta) becomes smaller than mv^2/R the skier would leave the ground (go airborne). mgcos(theta) represents the weight component perpendicular to the surface. mv^2/R represents the centripetal force.

[@AzNDuM]

Why is V presented as a constant in this problem? If I am correct, shouldn't you only be able to substitute the centripetal force equation in if it's constant? In this case, isn't V not constant? The initial velocity to the drop off point.

[@MichelvanBiezen]

William, v is not constant in this problem. In this problem you are trying to find the location where the skier will leave the hill and become air borne. That happens when the centripetal force is equal to mgcos(theta)

[@AzNDuM]

Michel van Biezen Hi Michel, thanks for your prompt response. I think I might have just been confused as to how you were able to quickly deduce that you were able to set the centripetal force equal to mgcos(theta). I had to use force analysis to establish that mgcos(theta)-Fn = mv^2/r. Can i assume that centripetal force will always equal the projection of gravity force? Isn't it necessary to assume that Fn = 0? This is my work through force analysis. mgcos(theta)-Fn = ma mgcost(theta)-Fn = mv^2/r

[@MichelvanBiezen]

William Zhou William. Now you are thinking about it in the correct way. You have to consider what will be providing the centripetal force. And the answer in this case is the component of the gravitational force that is normal to the surface.

[@varunnithin5326]

What is the distance travelled by boy before leaving from mountain??

[@MichelvanBiezen]

Once you know the angle, at the point the boy leaves the mountain, then you use distance = radius x angle (in radians)

[@ibrahimnazemqader9153]

دەستەکانت خۆش بێت Thank you so much .....

[@hailunkara]

Where is the 11th video of the playlist ?🤔

[@ajaygrewal5354]

sir here you took center seeking force=mgcos(theta).sir in this type of circular motion wont be any role of reactionary force? i want to know how this reaction force will act in this kind of problem or not act?

[@dimascarlos74]

Very helpful thanks a lot

[@danieliussipovic4704]

It may sound stupid, but don't understand how R-Rcos could be replaced by 1-cos

[@MichelvanBiezen]

No, not stupid. Sometimes we just don't see something. (Happens to me all the time). I just factored out R. R - R cos(theta) = R ( 1 - cos (theta))

[@danieliussipovic4704]

Totally didn't see that. Thought about identities which didn't make sense at all. Thank you!

[@abdulraheem2046]

please, can anyone tell how to calculate horizontal distance travelled after losing contact?

[@pipertripp]

sure. You know the velocity vector of the skier when they leave the hill. You can then break that vector down into its horizontal and vertical components. Just like you would in any 2-D projectile motion problem. You also know the height at which the skier leaves the hill, Rcos(48.1deg). Knowing these things, you should be able to come up with the horizontal distance. It would have to be in terms of R because we don't have any actual value for the height of the dome.

[@kindjupiter]

So perfect.merci bcp

[@tsoojbaterdeneharvard3187]

Will you teach me how to solve arccos 2/3=theta

[@MichelvanBiezen]

We have several playlists on trigonometry where we show you how to do that.

[@stefmartens7588]

What to do when the skiër has a starting velocity

[@MichelvanBiezen]

Then you have an additional energy term ( 1/2 m v^2) on the left side and the skier will leave the "hill" sooner (at a smaller angle)

[@alimohammadzade5906]

Thank you a lot! سپاسگزارم

[@orchoose]

why not calculate landing distance, since this is frequent problem on exams:D

[@iamdanish99havocx79]

hi, wont it be N-mgcosx=mv^2/r?

[@MichelvanBiezen]

Try working the problem that way and see what happens.

[@Yashpandey467]

and that was something awesome)