Amazing explanation! Simple, clear, and straight to the point. Thumbs up! ...Just wondering, do you think it can be solved using calculus and optimization?
You are supposed to take into account the energy of rotation of the ball in the final energy. Making the assumption that the ball is pure rolling in all of it's travel, the tangential velocity (Vt) must be equal to the velocity of the center of mass (Vc) which we know is Vc = sqrt(g*R). Then the angular velocity is Vt = Rω ; Vc = Rω ; (Vt=Vc) ω = Vc/R = sqrt(g*R)/R (Vc=sqrt(g*R). Then the rotational energy is: Kr = 1/2*ω^2*I ; Kr= 1/2* sqrt(g*R)/R* I. Remenber that "I" is the moment of inertia of a ball (sphere) or a disk if you simplify the problem (If you need to substitute it search for it (should be in Wikipedia or any proper physics book about classical mechanics)).
Ei=Ef according to conservation of energy. I believe if no external net forces act on a system, then energy is conserved. So the potential energy initially equals the potential energy at the end, plus the kinetic energy at the end. Also, just because Ef is potential + kinetic, doesn’t make Ei < Ef. Gravitational potential energy is greater at the beginning than the top of the loop. The kinetic energy at the top of the loop adds with that potential energy to equal Ei.
Ettoyea z no: this problem is viewed from an energy standpoint, not a dynamical one where variables such as displacement, acceleration or acceleration along track are considered (so not using suvat equations) Considering conservation of mechanical energy incredibly simplifies the physics, and tedious maths that would be required if a dynamical approach is taken.
The normal force (and centripetal force) will be greater. This would (in real life) result in higher friction with the surface of the loop (so, lower efficiency, but higher speed).
first give both sides a common denominator by multiplying both the numerator and denominators of the fractions on both sides (in this case I multiplied the left side by 10 and the right side by 20). the reason they are still equal to the original values is because they are equivalent; try dividing the values below (ie 10/20 should give 0.5, or 1/2, and 200/20 should give 10, which are both the original values). 10/20R + 200/20R now just add them: 210/20R which can be reduced to 21/2R.
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amazing content
Amazing explanation! Simple, clear, and straight to the point. Thumbs up! ...Just wondering, do you think it can be solved using calculus and optimization?
very good, very clear, nice clear writing, pleasant vocal presentation. thanks! edit: make sure your v's look like v's and not u's!
You saved my tail, thank you!
You are supposed to take into account the energy of rotation of the ball in the final energy. Making the assumption that the ball is pure rolling in all of it's travel, the tangential velocity (Vt) must be equal to the velocity of the center of mass (Vc) which we know is Vc = sqrt(g*R). Then the angular velocity is Vt = Rω ;
Vc = Rω ; (Vt=Vc)
ω = Vc/R = sqrt(g*R)/R (Vc=sqrt(g*R).
Then the rotational energy is:
Kr = 1/2*ω^2*I ;
Kr= 1/2* sqrt(g*R)/R* I.
Remenber that "I" is the moment of inertia of a ball (sphere) or a disk if you simplify the problem (If you need to substitute it search for it (should be in Wikipedia or any proper physics book about classical mechanics)).
How passive aggressive would it be to send this to my teacher?
Thanks
very nice
thanks - exactly what my textbook had!
May I know which text book it is?
thank you!
Hi patrickJMT,
Can you explain why Ei has to equal Ef? Shouldn't Ei
Ei=Ef according to conservation of energy. I believe if no external net forces act on a system, then energy is conserved. So the potential energy initially equals the potential energy at the end, plus the kinetic energy at the end. Also, just because Ef is potential + kinetic, doesn’t make Ei < Ef. Gravitational potential energy is greater at the beginning than the top of the loop. The kinetic energy at the top of the loop adds with that potential energy to equal Ei.
@@joshuasteiner5568 Okay I see, thank you.
4:51 i got preety far while solving this but i forgot about the normal force
thx
Can u make a video about ' Intermediate Axis Theorm'?...
Btw nice video!
This is so so cool
Why is this recommended to me?
Can you make one with friction loop to find the minimum speed to go around the loop and you know the height
did you ever find out a way to calculate that
5:37 doesn't this step assume uniform circular motion?
right, because the object is going around a loop, which implies "circular" motion
Smallest possible ratio of dislikes to likes?
Why didn't we take the distance from the launch point into consideration???
Because we assumed there's no friction, so it doesn't matter how far away in the horizontal axis the start point is
Ettoyea z no: this problem is viewed from an energy standpoint, not a dynamical one where variables such as displacement, acceleration or acceleration along track are considered (so not using suvat equations) Considering conservation of mechanical energy incredibly simplifies the physics, and tedious maths that would be required if a dynamical approach is taken.
@@donegal79 Ok, got it
Coefficient of Friction is taken to be 0
@@donegal79 Exactly.
You never included centripetal force, wtf?
It is from which text book?
Hello. Thank you for the video. I had a query. what will happen when the vehicle have more speed than it needs?
go faster
The normal force (and centripetal force) will be greater. This would (in real life) result in higher friction with the surface of the loop (so, lower efficiency, but higher speed).
it will travel further along the loop before loosing contact with it
Whats 1/2R + 10R?
first give both sides a common denominator by multiplying both the numerator and denominators of the fractions on both sides (in this case I multiplied the left side by 10 and the right side by 20). the reason they are still equal to the original values is because they are equivalent; try dividing the values below (ie 10/20 should give 0.5, or 1/2, and 200/20 should give 10, which are both the original values).
10/20R + 200/20R
now just add them: 210/20R
which can be reduced to 21/2R.
@@hummus9118 thanks dude
@@abbbyggrim3288 no problem :)
Hi, great video. I just had one doubt. Why is the normal force equal to 0?
Since v is the minimum velocity, it barely touches the top of the loop so the loop isn’t exerting force back on it
which syllabus is this ?
thank you so much
H= .5R+2R becomes h=5/2 ???????
2/2 =1
5/2 = 2.5
You transform 2R into 4R/2 to be able to add it with 1R/2, thus getting to 5R/2.
2 + .5 = 2.5