What is the edge speed of the rolling wheel edge speed Rolling speed? ） 10

The motion state of the scroll needs to be broken down into two motions.

One is the horizontal translation of the rolling axis, and the speed v is set, that is, the forward speed of the wheel is v.

The other is that the edge of the wheel rotates around its own axis, and the angular velocity is set to w.

Study the point where the ferry comes into contact with the ground. His actual velocity is the combination of the translational velocity of the axis and the velocity about the axis.

Axis translational velocity, horizontal to the right v.

Relative axial velocity v'=wr, horizontally to the left.

So the actual speed is V-WR

Because there is no slippage in the roll, the actual velocity of the contact point (relative to the ground) is 0

i.e. v-wr=0

So w=v r

The angular velocity around the axis at any point on the wheel is w, so the relative velocity is wr=v, but the direction is different.

Through the vector synthesis of the relative velocity v and the axial velocity v, the velocity of any point relative to the ground can be obtained.

For example, the highest point is 2V, and the speed of the left and right points is 2V

I hope it helps you, if you have any questions, please ask o( o haha

Are you referring to the vector velocity, or the minute velocity parallel to the rolling surface?

Vector velocity is linear velocity.

The average velocity parallel to the rolling surface: for example, if you roll on a horizontal plane, the distance s is equal to the circumference c is 2 r (where r is the radius of the wheel), and the time is equal to the period t is 2 v (where v is the linear velocity), so the horizontal minute velocity (or horizontal average velocity) is s t=2 r (2 v) = r v = the square of r).

However, I understand that "edge velocity" refers to linear velocity, which is vector velocity.

It is the linear velocity of the edge, which can be calculated using angular velocity and radius.

The acceleration at the edge point is indeed the square of v divided by r, and the direction is towards the center of the circle. Here's why: The acceleration of a point at the edge can be broken down into:

The acceleration relative to the center of the circle and the acceleration and vector sum of the center of the circle to the ground. Since the uniform speed of pure rolling, the center of the circle is moving in a straight line at a uniform velocity, and the acceleration to the ground is zero, so the acceleration to the ground at a point on the edge is equal to the acceleration relative to the center of the circle.

The velocity of a point at the edge relative to the center of the circle is equal to the velocity of the center of the circle, so the acceleration is equal to the square of v divided by r.

The magnitude of the acceleration does not change, but the direction changes at all times, so it is a variable acceleration motion.

The two points A and B are the points on the edge of the collinear transmission wheel, and the linear velocity is equal, because the radius of the large wheel is 3 times the semi-lead diameter of the small wheel, according to V=R, know A

b3:1, and the centripetal acceleration orange degree is a=v, and the round beard is equal to the linear velocity then aaab a

b3:1, so the answer is: 3:1 3:1

The speed of the wind turbine is around 10 rpm at full load, and the maximum speed is 25 rpm. The most common impeller diameters are 77 meters, 82 meters, etc. Therefore, the linear velocity of the blade edge is roughly 40 m/s to 108 m/s.

Since the belt is not sleepy and slippery, the arc length of the two wheels passing through the same time is equal, and the linear velocity can be seen to be equal A and B are wrong, and the ruler is C correct

According to the formula =v

r big wheel radius is large, angular velocity is small, so d is correct

Therefore, the first acre is selected: cd

The two points of cab are collinear relations, the linear velocity is the same, which can be seen from the period t=2 r v, the radius hail is different with the oak, the period is different, and the angular velocity formula w=v r can know that the velocity next to the angular source is different, and b is wrong; a=<>

It can be seen that the acceleration is different, d is wrong; Collapse.

A. Since the belt does not slip, the arc length of the two wheels passing through the same time is equal, and the linear velocity can be seen to be equal a, b is wrong, c is correct

d. According to the formula v= r, the radius of the big wheel is large, and the angular velocity is small, so d is wrong

Therefore, c

<> the speed direction of point A is the tangent direction of the point, if the small wheel is the driving wheel, then the velocity direction of point A is as shown in the figure, if the small wheel is the driven wheel, then the velocity direction of point A is the opposite direction of the figure, and the acceleration direction of point B always points to the center of the circle;

This question is kind of interesting. But it's a bit over-the-top, and I use a little college knowledge.

First of all, have you ever thought about why you roll?

Even on a stationary inclined plane, why do cylinders of the same mass roll and blocks don't? Don't tell me because the surface of the cylinder is smooth!

Knock off one of the strings of the cylinder, and you put it on and roll!

Is it the same as the square force analysis (the support force is along the tangent direction)? In fact, the reason for the cylindrical rolling is the "force couple", and the total force in the vertical bevel direction is 0, but don't forget that the friction in the horizontal direction is not the center of mass! The components of friction and gravity on parallel inclined planes form a force couple, causing the cylinder to rotate and roll faster because there is no other reverse couple to equilibrate!

And these two forces are not necessarily equal, one is mgsin (theta) and the other is the static friction with a maximum of umgcos (theta) and a minimum of 0. That is to say, in addition to the rolling due to the coupling, there will also be an acceleration change in the center of mass by the bull two law.

Let the magnitude of the force couple x*r, let the moment of inertia of the cylinder be i, its angular acceleration be b, and there is b=x*r iThe angular velocity of the edge point is bt (similar to velocity, the angle of rotation is 1 2bt 2 (in radians), and the distance in the parallel direction in t time is 1 2bt 2*r, that is, the acceleration of the center of mass in the parallel inclined direction is 1 2br, and the sum of the component forces is m*1 2br=mgsin(theta)-x (where x=friction = force of the couple).

x=mgsin(theta) [1+1 2*mr 2 i].

It can be seen that for the cylindrical case, there is always x but! If it is an inclined plane of accelerated motion, which exceeds the resulting acceleration 1 2br, the cylinder will eventually roll upwards!

The principle of the block is the same, the difference is that the support force of the block is distributed on a plane, that is, the support force is not equal everywhere on the surface! Moreover, the way the block rolls is more complex, and calculating the centroid line of the block rolling involves more complex equations. I won't analyze them one by one here...

In other words, this process can be solved exactly, but to solve it well, it may require knowledge such as calculus.