Some doubts about centripetal acceleration in high 1 physics

Definition: The acceleration of a particle pointing to the instantaneous center of curvature when it moves in a curvilinear motion. The formula is v2 r, where v is the tangential velocity of the mass motion and r is the radius of curvature of the motion path.

Direction: Pointing to the center of the circle. It can be understood as the component of the acceleration of a body moving in a circle in the direction of pointing to the center of the circle.

Formula: a=r 2=v 2 r=4 2r t 2 All objects that move in a curvilinear motion have centripetal acceleration, and centripetal acceleration reflects the speed of the change in the direction of velocity. Centripetal acceleration is also known as normal acceleration, which means acceleration in the normal direction of the curve.

When the velocity of the object also changes, there is also acceleration along the tangent direction of the trajectory, which is called tangential acceleration. The direction of centripetal acceleration is always perpendicular to the direction of velocity.

v can only represent the magnitude of the change in the amount of velocity, a=v 2 r is only used for uniform circular motion, in which the magnitude of velocity does not change, the direction always changes, so the centripetal acceleration can be found through the magnitude of velocity and the radius of motion, in addition, the force f is a vector, m mass is a scalar, acceleration a is a vector, high school math will learn, scalar multiplication vector is equal to vector, so you can use this formula, to be honest, I don't know you very well, you can ask me again.

A acceleration represents the speed of the change of velocity, which is equal to the change of velocity divided by the time a v t and the velocity is a vector, with magnitude and direction, so v In linear motion, it is mainly the change of velocity magnitude, and the resulting acceleration can also be called linear acceleration.

In curvilinear motion, the direction of velocity also changes, and in uniform circular motion, the magnitude of velocity does not change. And the direction of speed is changing all the time. v represents the magnitude of the change in the direction of velocity, v t represents the speed of the change in the direction of velocity, and the resulting acceleration is called

Centripetal acceleration a=v 2 r

Newton's second law must be true whether it is a straight line or a curve: f ma so f=ma=mv 2 r

Who said that f=ma is a change in velocity? It is derived from a=f m, which means the rate at which the velocity of m changes when f acts on m. Moreover, when the circumference is moving, it is not only the direction that changes, but also the speed.

You have to look at it in the direction of the force, not in terms of the total velocity. v depends on the definition, you can only represent the size, in the kinetic energy theorem commonly used e1 = mv1 2 2, e2 = mv2 2 2, v = root number (2e2 m) - root number (2e2 m). It can also be said that the momentum theorem is commonly used, i= p= vm has a direction.

v and a are both vectors, and they have magnitude and direction and are vectors.

Scalar quantities, on the other hand, only represent values and have no direction.

f=ma Newton's second law macroscopic is applicable at all times.

v and a are both vector quantities, and there are quantities of magnitude and direction.

f=ma is the law of the ox, and it is the computing power, not the change of speed.

Weight drop 1

m, the instantaneous velocity is.

v==m/s=2

m s Obviously, the linear velocity of each point on the edge of the pulley is also 2m s, so the angular velocity of the pulley rotation, i.e., the angular velocity of rotation at each point on the edge of the pulley, is =rad s=100

The centripetal acceleration of rad s is a=

2r=100

2×m/s2

m/s2.

AC plots and then provides the centripetal force based on the resultant force of gravity and tension. According to the angle is 45°, so the gravitational force is equal to the centripetal force, so the centripetal acceleration is g The tensile force is the hypotenuse of the triangle of the tensile force with the gravity force and the centripetal force vector so it is the root number 2*mg

First of all, it is known from the kinetic energy theorem mgh=mv 2.

When the balls are at different heights, the velocity of the descent to the lowest point is different v = 2gr again, by the centripetal force formula f (to) = mv r

And f(to) = f(pressure)-mg

f(pressure) = 3mg So the pressure of the ball on the two orbits is the same, and the centripetal acceleration a=v r=2g

So the centripetal acceleration of the ball is the same.

I choose AC. The derivation is as follows:

From the law of conservation of mechanical energy: 1 2*m*v 2=m*g*h, we get: v 2=2*g*r, and the gravitational acceleration a=v 2 r, so a=2*g, which has nothing to do with the radius, and f-m*g=m*a=3*m*g, which has nothing to do with the radius, so choose ac

The direction of acceleration remains the same, and the direction of velocity must also remain the same. is wrong. If the velocity is 100 m s and the acceleration is -10 m s, then the velocity will be 0 in the tenth second, and the direction of the velocity will be reversed.

Draw a V-T diagram for yourself, take a look at it, study it, and study it.

Relationship between the direction of acceleration and the direction of velocity: The direction of acceleration is the same as the direction of the change in velocity δv.

In linear motion, if the velocity increases, the direction of acceleration is the same as velocity, and if the velocity decreases, the direction of acceleration is opposite to velocity.

Be sure to remember! Acceleration is not speed! Acceleration is just a variable ......There are positives and negatives!

1.Correct, the direction of acceleration determines the positive or negative of a, so a does not change, and u does not change! Negative!

3.The object has a constant velocity, a is 0! When A is negative, it is the opposite!

Wrong. In a uniform circular motion, the centripetal acceleration a direction is always directed towards the center of the circle, and the direction of velocity is tangent to the point and constantly changing. In one-dimensional linear motion, if the direction of initial velocity is positive, the direction of acceleration during deceleration motion is opposite to the initial velocity.

In two-dimensional curvilinear motion, the direction of acceleration is the direction of the amount of velocity change, and has nothing to do with the direction of velocity (of course, in linear motion), by the way, you are in a high school, hehe, I am also in a high school...

The direction of acceleration has nothing to do with the direction of velocity, and the angle between the two directions can be between 0-180.

The direction of acceleration is the same as the direction of velocity change

What is centripetal acceleration equal to? a=v squared r right? Because this is a belt conveyor, the linear velocity of point A is equal to the linear velocity of point B, and because the radius of point A is greater than the radius of point B, so AB is greater than AA, do you understand?

Then because the centripetal acceleration is equal to w [angular velocity] square r because point A and point C rotate in the same park, so the angular velocity is equal, and because the radius of point A is greater than the radius of point C, so AA is greater than AC, and the above AB is greater than AA is greater than AC, does the landlord understand? If you don't understand, please ask, if you understand, can you adopt it?

As shown in Figure A, a particle moves in a uniform circular motion around point O, and the tangent line from point A to point B, that is, the linear velocity VA and VB, are equal in magnitude. Then the centripetal acceleration a is the unit change vector from vb to va linear velocity. Method:

As shown in Figure B, the vector va is translated so that its starting point coincides with point b, then the vector v = vector vb vector va (that is, the change of linear velocity when turning a certain radian), and the angle between vector va and vb is the angle of rotation of the particle in a uniform circular motion (expressed by the radian system).

Another example is Figure Ding (part of the circle O, i.e. fan, Oq = Op = R, and there are chord pq and arc pq at the same time), let the number of radians of the angle between oq and op (in fact, it is the ratio of the arc length corresponding to this angle to the radius of the circle mathematically, that is, the arc pq: the value of the radius r, such as a radian, then we know x·y x=y, then the length of the arc pq can be expressed as "radius r·arc pq radius r", that is, arc length = radius corresponding to radians. When the angle is very small, it can be approximated that the arc pq = chord pq, that is, the length of the curved arc is almost the same as the length of the straight line segment, which provides a basis for the subsequent finding of v.

Back to Figure B, as shown in the figure, when the angle between ob and oa (equal to the angle between vb and va) is very small, then the corresponding v is very small, and with b as the vertex, the arc v swept from point a to point b in the fan with b length va (or vb) can be approximately equal to chord v, that is, according to the introduction of figure d, if the radius r in figure d is regarded as the linear velocity va (or vb), the arc length = radius corresponding to the radian (that is, the previous v = ·r) is used in figure b is the arc v = v =Linear velocity (regarded as radius r) Radian (ratio of arc v to linear velocity va or vb which can be regarded as radius r of a circle) And when the quantity v is small to a unit (i.e. the amount of v in one second), then this v is what we call centripetal acceleration a, centripetal acceleration a= v t, and arc v = chord v, so centripetal acceleration a = arc v t.

First of all, the radian is the number of radians of the angle of the particle turning in a circular motion at a certain time (t), then the angular velocity = t, which represents the number of radians that have been turned in one second, that is, the radian = · t, and v=arc v=centripetal acceleration a t. Then according to the arc length = radius corresponds to the radian, the arc v = v = linear velocity v radian (as shown in Figure C, when it is small to a certain extent, the arc v = v, and there is such a relationship when it is small to the unit radian) and then according to the two formulas, the centripetal acceleration a t = linear velocity v (the magnitude of this vector is always the same) angular velocity · t, and at the same time remove the t left and right of the equation, so the final simplification is: centripetal acceleration a = linear velocity v angular velocity, that is, a(n) = ·v, and a(n) = 2·r, a(n) = v2 r and so on It's all based on this equation and v= ·r.

f=mv²/r

Because it's doing a circular motion.

So v=2 r t (t is the period).

So v = (2 r t).

Bring in f=mv r.

f=(m4π²r²/t²)/r

m4π²r²/t²*r

m4π²r/t²

m(2π/t)²*r

Because 2 t = angular velocity w (beating Miga).

So f=mw r

Because 2 t = 2 * 1 t

And because 1 t = f or 1 t = n (f frequency, n speed) 2 t = 2 f or 2 n

So f=m(2 f) r or f=m(2 n) r, so f=mv r=mrw =mr(2 f) =mr(2 n) a=f m

First, the direction of acceleration is determined by the direction of the resultant force experienced by the object and is therefore independent of the initial velocity.

Secondly, I think the meaning of this question is unclear, if it is understood that A is doing a uniform motion, then the acceleration of A is zero.

If, at a certain moment, the acceleration of A's motion is unknown, d is fine.

Solution: Your problem lies in the choice of coordinate system.

The direction can be arbitrarily selected, however, in the same problem, only one coordinate system can be selected, it is impossible to select two coordinates, and the vector of the same direction will not have both positive and negative coordinates.

In a coordinate system, the direction of acceleration is determined for the direction of velocity, since the direction of velocity is positive or negative, then, the direction of acceleration is also determined, not both positive and negative.

Therefore, d is wrong.

In the same problem, all objects must be in the same positive direction.

Therefore, if the initial velocity of the two objects is not the same, one must be positive, and the initial velocity of the other is negative.

Therefore, in the same topic, the opposite sign must be in the opposite direction.

A negative acceleration means that the acceleration is opposite to the selected positive direction, not to the initial velocity of the object.

This is a problem of scalars and vectors in physics, acceleration is a vector like velocity, vectors have directions, and positive is the positive direction of the selected coordinates, and negative is the opposite direction. The meaning of the selection here is that the selected direction is the positive direction, and the velocity, acceleration, and force under the reference frame after selection are subject to this direction, like the direction of acceleration and velocity in this question is only related to the selected reference frame, and has no relationship with each other.

Acceleration is a vector quantity, plus or minus only represents the direction, so the acceleration magnitude only compares the numerical magnitude, and the acceleration direction has nothing to do with the muzzle velocity, only with the force, as long as you look at the plus or minus.

You are right, the answer is wrong, the positive and negative before acceleration represents whether it is acceleration or deceleration, regardless of direction.

Acceleration should default to a positive direction, so negative acceleration must be reversed.

In the same topic, an exhibition department has been set up.