lucky, im in 6th grade and my dad could have explained it fastly but hes making me watch videos that i cannot unerstand (at all)

lol thats like every single comment here

same fact for to bro

Should be, the profs are not professional at teaching

glad to hear! please tell your friends :)

stfu

@@roastfire7242 Chill dude :(

@@roastfire7242 you stfu you peice of shit no one speaks to Jesus like that

@@salahshayah9864 Lol. *Peace

if you subscribe to the channel youtube will notify you every time i post! that way you won't miss any of the sweetness!

I almost commented that 😳

Hehe Speaking of circles..

haha reply go brrrrrrr

i have a few on electricity coming next week!

stfu

@@roastfire7242 you stfu

@@roastfire7242 stfu

why are you here mary

with how much of my grades he's saved, he pretty much is

No

What's wrong with Khan Academy?

Hey vaishnavi where are you from????

Hey vaishnavi where are you from?????

You are not from India I guess...

Right I don't get it I did the same thing and it's 6.25

You are confused about this, because you are losing track of what value you are holding constant during this time. The exact term you hold constant, will affect your answer, of what a change in radius does to the centripetal acceleration. For instance, if you hold angular speed constant, then tangential speed would be proportional to radius. Accounting for this fact will allow us to conclude that centripetal acceleration will decrease with a decrease in radius. The equation v^2/r becomes omega^2*r instead. Hold omega constant, and centripetal acceleration is proportional to radius. You would see this if the object were something that we place on a rigid body with a constant rotating speed, and we are choosing where to put it on this rotating body. Like placing a coin on a turntable, and where you put the coin. By contrast, if you hold tangential speed constant, and change the radial position (or change the center position so that the radius also changes), then decreasing radius will increase centripetal acceleration in an inverse proportionality. An example of where you see this, is if you take two turns in a car at the same speed, but different turning radii. The speed is given to be the same, and the radius changes because the center of the circle relative to the car changes position. You have to steer tighter to navigate the curve with less radius of curvature. You will also notice that if you turn too tightly, your wheels will skid and you won't be able to turn. This is why you have to slow down for tight turns.

Rate of change in momentum = net force When the rate of change in momentum is entirely perpendicular to the velocity, that rate of change in momentum is the force that causes centripetal acceleration. Centrifugal force is a fictitious force, that is a consequence of looking at this problem from within a reference frame in circular motion, rather than from the inertial reference frame. What it really is, is a body's tendency to proceed in straight line, reacting to a force that attempts to accelerate it inward, and assuming (incorrectly) that it is due to a force that attempts to pull it outward. That apparent force attempting to pull it outward, is the centrifugal psuedoforce.

If you fix the tangential speed between examples, closer to the center means a greater centripetal acceleration. It is inversely proportional to the radius, a=v^2/r. If you fix the period of rotation between examples, which is equivalent to fixing the angular speed, closer to the center means less centripetal acceleration. It is proportional to the radius, a=omega^2*r.

Due to the fact that the wheels rotate, the point of contact between the tires and the road becomes an instantaneous center of zero velocity for each wheel. This is a point where the wheel that is in both translation and rotation, is locally at rest. The force of static friction prevents the tire from skidding, so that the car only moves according to the direction set up by the various axes of rotation of the wheels. Static friction enables the force of traction, that allows the drive wheels to propel the car forward, and allows the steering wheels to change the direction of the car on a flat road. It also works with the braking of the car. The kinetic friction occurs between the brake drum and shoe, so that it doesn't have to occur between the tire and road. If your brakes lock up, your car will skid, and steering won't let you control it. This is why we have ABS brakes, so that the wheel-road remain in static friction, and the kinetic friction is kept inside the brakes. In pre-2000 vehicles that didn't have ABS brakes, you would have to manually pump your brake pedal to get the same result.

Perhaps tangential acceleration is stronger than centripetal acceleration, and the object should follow the stronger force more

@@user-hv6ef9ie1g there is no tangential acceleration if there is velocity so the tangential acceleration is zero

how do you mean?

Wouldn't the radius equal 1m, if the string is 2m long?

but you don't hold the middle of the string, you hold the end of the string.

The length of the string is the radius.

@@ProfessorDaveExplains Ah, silly mistake from me there! Thanks for the reply

There is a component of acceleration that is parallel to the velocity. There is a component of the acceleration that is perpendicular to the velocity. The component that is parallel to the velocity causes a change in speed. The component perpendicular to velocity causes a change in direction. If it is not directed toward the center, it will not move in a path around that particular circle of curvature.

Nice explanation, considering it actually helped me with a certain problem. Might I add the piece I noticed that helped me out with it. The only change in tangential velocity for uniform circular motion is in the direction, while we know the magnitude is constant. As you say, the vertical component of the acceleration controls causes change in direction, while the horizontal component causes change in magnitude. No change in magnitude means there’s zero horizontal acceleration, which means the total centripetal acceleration must just be the vertical acceleration.

Vignesh Nithyanandam non inertial frame of reference do not follow Newton's laws of motion are u from SSC board

Ajinkya Deshpande i am from cbse mam

@@ajinkyadeshpande6271 It isn't that non-inertial frames of reference "do not follow Newton's laws of motion", they still do. It is just that you have the acceleration of the reference frame to account for, something that is not immediately obvious, which will feel like it is a fictitious force.

The period is a kinematics quantity. It is a mathematical description of the motion. Factors like forces, mass, energy, and momentum are not considered when we consider the definition of period for uniform circular motion. If you want to know whether mass affects the period of a condition of uniform circular motion, you need to define what happens to the other kinetics quantities. If you make constant the magnitude of the tension in string that holds it in circular motion, mass will affect the period of the circular motion. If the kind of force that holds it in circular motion comes in proportion to the mass, then the period is independent of the mass (for instance, circular orbits around an astronomical body of significantly greater mass).

what does that even mean lol