A physics problem about centripetal force.

A false centripetal force is a force of the same nature as gravity, elastic force, and frictional force.

b False can also be a variable speed curve movement.

c Right. d False can also be a variable speed curve motion.

A is false, because the centripetal force is only a resultant force, which is produced by the joint action of multiple forces.

dFalse, the object is affected by the centripetal force, and it can also do variable speed circular motion, such as a pendulum ball that does circular motion in a vertical plane.

The centripetal force formula is f-direction = mrw 2

The centripetal force of A is f1 = 2mrw 2

The centripetal force of B f1 = mrw 2

The centripetal force of C f1 = 4mrw 2

According to Newton's second law, it can be known that the centripetal acceleration of C is the largest, so A is paired.

When turning, friction provides the centripetal force of the person. From the above analysis, we can know that the friction of B is the smallest, so B is right.

The centripetal force required to maintain the circular motion of C is the largest, followed by A and the least required by B. If the centripetal force is not strong enough, the person will fly out. But the friction force is limited, so as far as A and B are concerned, B reaches the maximum value of friction a little faster, so it will slide before A, so C is right.

The same is true for D.

From the centripetal force formula, it can be seen that f is towards m* 2*

r The masses of the two objects are equal, and they have the same angular velocity when there is no relative disc sliding, so the object farther from the center of the circle requires a greater centripetal force. Considering that the friction factor between them and the disc is equal, i.e., the maximum static friction between them and the disc is equal in value, therefore, when the angular velocity of the disc gradually increases, the object farther from the center of the circle begins to slide first.

When an object starts to slide (it must be the one farther from the center of the circle), there is an f-direction mg m* 2*(2r).

The angular velocity obtained is the root number[

g（2r）]

The maximum static friction provided is UMG

When the angular velocity is w.

The centripetal force required by the object is oh mw square r

and MW side 2R

So the outside slides first.

The angular velocity when sliding w = below the root sign (ug r).

Summary. For example, there is a circle that rotates around the center of the circle.

Then the angular velocities of the points on this circle are equal.

The linear velocity of the points above this circle that are equal to the center of the circle is equal.

If the angular velocity of this circle is a (unit: radians, seconds).

If the length of a point on a circle is l, then the linear velocity v of this point

al about the physics of centripetal force.

Hello, please wait.

The gravitational force of an object mg when it slides to the highest point of an orbit its pressure on the orbit is 3 4mg ask it what is the centripetal force at this time.

Why is the angular velocity of A the same as B, the radius of B is larger than A, so B slides first than A, and this is why.

For example, there is a circle that rotates around the center of the circle. Then the angular velocities of the points on this circle are equal. The linear velocity of the points above this circle that are equal to the center of the circle is equal.

Addendum: If the conveyor belt is connected to two flywheels, the radius of flywheel 1 is r1 and the radius of flywheel 2 is r2; The linear velocity of each point at the outer edge of the flywheel 1 is a, and the linear velocity of each point at the outer edge of the flywheel 2 is b; The angular velocity of flywheel 1 is a, and the angular velocity of flywheel 2 is b, then the linear velocity: a=b; Angular velocity:

a/b=r2/r1

I'm asking why B moves first than A, and according to the formula f mw2r shouldn't be B's son, so he needs more force than B, so why does B move first?

Because the radius of b is larger than a, the position of b is more rounded.

Why is it sleek, and the contact surface is different?

Yes, the radius is different, and the contact surface is also different.

There are no units for a given data: g ?The mass of the earth ?

Centripetal force f=mv 2 r=gmm r 2(=gravitational force) gm=rv 2=r(2 r t) 2=r 3*(2 t) 2r 3=gmt 2 (4 2).

r=v=2πr/t=2π*

The calculation results of the linear velocity seem to be too far out of the cosmic velocity?!

The instantaneous speed cannot be changed, a is wrong.

It can be seen from a=v 2 r.

Since r becomes smaller, the acceleration suddenly becomes larger, and b is correct.

The centripetal force is the trillion resultant force.

It can be seen from f=ma.

A becomes larger, centripetal force becomes larger, C is correct.

It can be seen from f=t-mg.

f becomes larger, and Li Shan's rent mg remains the same, so t becomes larger, and d is correct.

Choose BCD

BCD is correct, the answer is yes.

The above answerer's explanation is quite correct, the key is that the velocity is constant when it hits the nail, and the reason is due to inertia. Another example of a corresponding is a ball tied under a non-extendable rope, and when it falls vertically from a position to the lowest point, the velocity at the lowest point will suddenly change to zero.

A=V2 R, because R decreases when it hits a nail, but V does not change, so A increases, and Option A is correct.

The centripetal force is the combined external force, f direction = ma, so the f direction increases, and c correctly answers auspicious slippery.

F = F-mg, so F increases, D is correct.

There are also more questions, asking how the angular velocity changes at the lowest point? From v=wr and r, w increases.

The slope of gravity teased and laughed.

GCOSA provides this centripetal force.

So as not to "fly fingers out".

He fully provides the centripetal force.

The centripetal force f can be found from the velocity and radius

Both are equal.

The large chaotic slip force of the tensile force is equal to the gravitational force of the ball 10N, and the centripetal force is provided by the combined force of the tensile force and the static friction force, so the value range of the early heart force is 4n to 16N

According to f mr 2

Get r minimum

r max

The energy balance can solve the final velocity equal, and I will help you analyze the convex and concave situation.

Force analysis, you can get a component of the concave gravity to offset the frictional force, and at the same time provide as a driving force, (there is a component and centripetal force in one direction, this force is useless for solving the problem), this component provides power, you can judge that the car is doing acceleration movement when it is descending, the same force analysis, and then the uniform deceleration process, and then the uniform speed movement on the horizontal road.

Similarly, the force analysis is first uniform on the convex.

Then reduce evenly. Then add evenly.

Finally homogeneous. The above is the force relation for preparation.

Let's talk about the similarities and differences between the two above. (The level pavement is the same, not to mention).

The distance of the concave and convex is the same, and the friction force is just the opposite, one is the maximum positive direction to the maximum negative direction, and the other is the maximum negative direction to the maximum positive direction, and the direction is self-determined, so it is not added, which is the maximum positive direction to the maximum negative direction. The judgeable average frictional force is the same, and the work done by f*s=frictional force is the same. (The energy balance yields the same velocity).

Let's talk about the biggest difference between the two to reach the midpoint of the time is not the same, this is also to solve the problem to give the difference, the initial velocity and the end velocity are the same, but in the concave is the first acceleration and then deceleration, the same analysis of the convex is the first deceleration and then acceleration, you can know that the average speed on the concave is concave, simply put, with the initial velocity as a reference, the velocity in the concave is greater than the initial velocity, and the velocity in the convex is less than the initial velocity, because the final velocity is the same, you can know that the speed of the two cars when they reach the end of the concave and convex point reaches the same.

In this process, the distance is the same, the average speed is concave, so the time = distance speed, you can know that the car walking on the concave road takes less time, so that the concave car first arrives in city B.

Let the mass of the earth be m, the orbital velocity of the space station is v, the acceleration of the space station is a, the gravitational force of the earth on the space station is f, the mass of the space station is m, the gravitational acceleration of the earth's surface is g 10 m s 2, and the height of the space station above the earth is h 400 km 4*10 5 meters.

To the space station there is a lead ma

i.e. gmm r h) 2 m* v 2 r h) and at the ground, gm r 2 g where r is the radius of the earth.

Therefore, the orbital velocity of the space station is v root number [ gm r h ] r * root number [ g r h ] ].

The radius of the Earth is r 6400 km.

Get the v root number [ 10 ] m s

The acceleration of the space station is a v 2 r h) (2 m s 2

The gravitational pull of the Earth on the space station is f-gravitational ma (123*1000)* N.

f = maf citation = gm1m2 r squared.

The centripetal force formula is f=(mv2)r and f=mrw2

You should connect the angular velocity and the linear velocity together to solve the solution.

Speaking of ideas, the time for a water droplet to make a flat throw motion with the initial velocity (linear velocity) of wr from the edge of the board is (2h g).

Flat throwing distance l=wr(2h g).

Since the water droplets are thrown in the tangential direction, perpendicular to the radius of the small circle, the radius of the large circle r=(r 2+l 2).

Note 2 denotes square and denotes open squared.

The horizontal distance of the water droplet from the tangent direction of the round plate is wr and the height is h.

It is denoted as s, and the radius of the large circle is under the root number (s 2+r 2), which you can understand when you think about the top view.

Hello, when the water droplets at the edge fly away from the layout, the velocity extends the tangential direction of the disc, the size v=rwAfter leaving the disc, the water droplets do a free fall motion in the vertical direction1 2gt 2=h t=root number(2h g).

During this time, the distance of movement in the tangent direction s=rw The root sign (2h g) draws a right-angled triangle, with the center of the disk as the center of the circle, the radius of the disk and the tangent distance s are the right-angled edges, and the hypotenuse is the radius sought r=root number (2r 2w 2h g + r 2).

The math symbols can't be typed, so I can barely take a look at it......

v=wr h=1/2gt^2

Find ts=tv

This is the distance of the tangent.

Draw to find the distance of the perpendicular tangent s (Pythagorean theorem).

Finally, r=r+s

Analysis: To at least keep the rubber block from falling, there is only to balance the gravity of the rubber block, and to provide the upward force of the rubber block is only friction, which is only generated by the rubber block pressing the cylinder wall, and this pressure is the centripetal force of the rubber block.

Since the rubber block rotates with the cylinder, the pressure of the cylinder wall on the rubber block provides the centripetal force n=f according to the centripetal force formula f=mrw 2

Friction f=un

Equilibrium gravity f=mg

To sum up, the total formula: mg=umrw 2

The solution is w=under the root number (g ru).