Centripetal Acceleration & Force - Circular Motion, Banked Curves, Static Friction, Physics Problems
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- Опубліковано 5 жов 2024
- This physics video tutorial explains the concept of centripetal force and acceleration in uniform circular motion. This video also covers the law of universal gravitation, weightlessness, banked curves with friction, kepler's third law of planetary motion and other stuff. This video contains plenty of examples and practice problems.
Circular Motion - Formula Sheet:
bit.ly/4gIUL3r
________________________________
Centripetal Acceleration:
• Introduction to Centri...
What Is a Centripetal Force?
• Physics - What Is a Ce...
Centripetal Force Physics Problems:
• Centripetal Force Phys...
Tetherball Physics Problems:
• Tetherball Physics Pro...
Tension Force - Vertical Circle:
• Uniform Circular Motio...
_______________________________
Static Friction & Centripetal Force:
• Static Friction, Centr...
Normal Force on a Hill:
• Normal Force on a Hill...
Banked Turn Physics Problems:
• Banked turn Physics Pr...
Non-Uniform Circular Motion:
• Non-Uniform Circular M...
Gravitational Force:
• Gravity, Universal Gra...
______________________________
Universal Law of Gravitation:
• Gravity and the Univer...
Gravitational Acceleration:
• Gravitational Accelera...
Gravitational Field Strength:
• Gravitational Field St...
Satellite Speed:
• Speed of a Satellite i...
Kepler's Third Law of Motion:
• Kepler's Third Law of ...
____________________________________
Kinetic Energy and Potential Energy:
• Kinetic Energy and Pot...
Work, Energy, and Power:
• Work, Energy, and Powe...
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Physics PDF Worksheets:
www.video-tuto...
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Circular Motion - Formula Sheet: bit.ly/4gIUL3r
Physics PDF Worksheets: www.video-tutor.net/physics-basic-introduction.html
Full-Length Videos & Exams: www.patreon.com/MathScienceTutor/collections
thanks so much
Timestamps for this lesson:
Note: He sometimes teaches concepts(Kepler's third law, tangential acceleration...)during the practice problems so make sure to look out for those
0:00 - Intro
0:15 - Centripetal Force
14:31 - Centripetal Acceleration
15:22 - Period and Frequency
16:32 - Tangential Velocity
32:40 - Vertical Circle Problem(Varying centripetal acceleration)
1:10:59 - Gravity and Newton's Law of Gravitation
Problems:
1. 18:50
2. 23:42
3. 26:27
4. 35:26
5. 38:05
6. 40:17
7. 46:31
8. 50:15
9. 58:12
10. 1:01:41
11. 1:07:31
12. 1:15:51
13. 1:22:29
14. 1:24:35
15. 1:31:22
16. 1:36:53
17. 1:43:24
18. 1:47:45
19. 1:50:34
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I know it’s too much too ask, but timestamps in these kinds of long videos would be extremely helpful. Other than that, very well explained and you’re a without a doubt a lifesaver!
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I'm just curious why the problem at 45:00 does not use formula FT = sqrt(FTx^2 + FTy^2), taking into account the effects of gravity
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Slight correction for the teatherball problem, for T x all you use is mv^2/r and T y is mg so you get 29.6 and 2.45.
By pathagorean T = 29.7N right?
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Great video, but one question. At about 46:28 did you just not count the tension force for the Y component when calculating the overall tension force because the Y component tension is almost negligible? I counted the Y component and got 29.7 N instead of 29.6 N.
The circular motion and, the effect of magnitude of radius on the time period of an object was
investigated in this practical. According to this the relationship between the radius and the period
as follows,
T
2 = r
According to the collected data it was able to prove that above statement.
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Is anyone else watching this the day before their big physics test over centripetal force, acceleration, and gravitation?
yup
35:21 I got a question. Shouldn't the forced acting on the ball at point A be T-Fg because only T is pointing towards the circle. Point C should be T+Fg because both T and Fg are providing the centripetal for, so they should both be positive
Awesome Teacher Ever found in Life
The force causing the vertical acceleration comes from the centrifugal force. This causes the
swings to move at a faster speed which then ultimately leads to the swings moving outwards and,
in the direction, going up.
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Hey I think you made a mistake at around 35:00, the tension at the top is equal to the sum of the weight force and the centripetal force, and the difference of them at the bottom.
Archer i’m thinking the same thing too but thinking about it it kind of makes sense because at the top the tension would be less than the tension at the bottom since gravity is helping the fc at the top and relieving tension, and at the bottom fc is trying to go up wards and gravity is pulling the opposite direction, increasing the tension. so the tensions there should be T at point A>C which proves his equation but i also get what you mean so i’m very confused lmao
Nope..actually it'a correct...he said that centripetal forces are positive when the direction is seekin to the center of the circle..He wrote it as Fc=T+Mg at the top since both Mg and T are directed through the center..and Fc=T-Mg at the bottom since T is directed through the center and Mg is away from the center :)
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8:49 in this example, I would think that the normal force is balanced by a force of the magnitude of mgcosΘ and in an opposite direction; while, the acceleration along the direction of the velocity is due to a net force equal to the (mgsinΘ-friction) force. At the initial point when the moving object is no longer in touch with the ground i.e. when the gravitational force is not balanced by the normal force, the object accelerates downwards by gravity.
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1:01:30 the answer for net acceleration is done right but due to intermediate rounding the answer became off by 0.4 In this scenario, 0.4 is a slightly large difference since the answer itself is less than 10. Rather if a direct input of the v^2/r in the Pythagorean equation then it would yield about 5.26. You taught the right way but it was just off due to rounding. But still you helped me learn how to approach these problems
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1:07:10
I want to understand. you said that if the speed exceeds the value that makes the normal force equal to zero it's going to fly off the road. but if the normal force was already 0 the car will lose contact so the normal force needs to be little more than zero, not exactly zero. isn't that?
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it has been 5 years that u made this video. I am not sure whether u r gonna answer my question or not. But still, (56:39) u solved Normal Force equals to mg/cos. What makes it different from the other questions that the normal force was usually mg x cos? why makes the normal force to be mg/cos? Also, Thank you for making these videos. I like how u explain the general concept first and then u solved questions that are related to the concepts that u just taught.
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One thing that really helped me understand the free-body diagrams is that remember that fNET= Fac all the time. So in the roller coster example, at the bottom, Fnet= Fac= FN-mg and then at the top it is Fnet = Fac = mg - FN
arham siddiqui what do you mean by fac? I guess i have studied too much and am starting to mix things up
@@ro1882 i meant force of centripetal acceleration
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force is linear with R, because V itself is a function of R. and when you square V you get an R^2 and R^2 divided by R is R.
so R is linear with force.
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As always, great video. Perfect for my physics class.
I think there's an error; the value of the speed is 251.3m/s because the period T is equal to 0.25 sec not 0.2
For the frictionless bank problem. I had to research the Normal force being greater than the weight force. Still would like more explanation cause is interesting
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At 36:43 isn't it supposed to be T+mg=Fcg for the top and T-mg for the bottom?
yes, because mg points downward along with Tension at the top.
yes it is, be he put the equations in terms of Fcg. "Fcg=..."
S. R. Oooooo, makes sense now. Thxs
35:05 Somebody could help me? Why they are both negative if they are pointing to the center (positive)?
Nope..actually it'a correct...he said that centripetal forces are positive when the direction is seekin to the center of the circle..He wrote it as Fc=T+Mg at the top since both Mg and T are directed through the center..and Fc=T-Mg at the bottom since T is directed through the center and Mg is away from the center :)
for that tetherball example i just equated fc = the tension force and took 0.25(9.4)^2/1/2=29.4n
Thanks for helping me understand this subject!!
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Very well explained and excellent presentation. A fantastic work!