Centripetal Acceleration & Force - Circular Motion, Banked Curves, Static Friction, Physics Problems

[@TheOrganicChemistryTutor]

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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'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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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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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

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[@roberto3196]

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.

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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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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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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

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arham siddiqui what do you mean by fac? I guess i have studied too much and am starting to mix things up

[@arhamsiddiqui7083]

@@ro1882 i meant force of centripetal acceleration

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[@greatpharoh0303]

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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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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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.

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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

[@ralphanthonymallillin2712]

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 :)

[@kabouch]

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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1:07:10 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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[@tatiana1310]

35:05 Somebody could help me? Why they are both negative if they are pointing to the center (positive)?

[@ralphanthonymallillin2712]

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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[@kal7708]

At 36:43 isn't it supposed to be T+mg=Fcg for the top and T-mg for the bottom?

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yes, because mg points downward along with Tension at the top.

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yes it is, be he put the equations in terms of Fcg. "Fcg=..."

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