How to calculate the friction when the disc is placed on the tab

1. The disc rotates around a fixed axis, and the radius of rotation is one half of the root of the radius r of the disc, because the disc can be regarded as numerous sectors with uniform mass. Draw an arc at one half of the root of the center of the circle, and the area of the two sides of the arc is equal, which is half of the area of the sector. Therefore, the arc on the whole disc is connected into a circle with one half of the root of the radius R, This circle is the center of mass of the circular motion of the disk

2. It depends on whether you are talking about static friction or dynamic friction. Dynamic friction is also divided into sliding friction and rotational friction. Only when the topic is clear can we discuss it

3. DF = u * D (mg) integral
DM = P * 2 π R * Dr P is density
m = P* π R ^ 2
friction has nothing to do with angular velocity

4. 1. Using the double integral of angle and distance
2, first calculate each ring with thickness of DR, and then integrate r from 0 to R

5. Take the ring with radius R and width Dr as the infinitesimal element, and the mass of the infinitesimal element DM = (m)/ π L²) two π rdr=(2mr/L²) Dr
friction of micro element: DF= μ( dm)g=(2 μ mgr/L²) Dr
friction torque of micro element: DM = - RDF = - (2 μ mgr²/ L²) Dr
friction torque of the whole disc:
m = ∫ DM = - (2 μ mg/L²) ∫r² Dr
by substituting the lower limit of integral 0 and the upper limit of integral L, M = - 2 can be obtained μ MGL / 3
so, you are right and the teacher is wrong

6.

Principle of moment of momentum

m △ t = I ω-ω 0)

calculate the moment of momentum M = ∫ R μ g(M/ δπ R^2) δ two π rdr=(2/3) μ gMR

(0---> R)

look up table moment of inertia I = (1 / 2) Mr ^ 2

(2 / 3) μ gMR△t=-((1/2)MR^2) ω 0

transit time △ t = - 3R ω 0/4 μ The g-second disk will be stationary

negative sign because the moment of momentum m and ω 0 in the opposite direction

let R be the radius of the disk. Find the moment of momentum M = ∫ R μ g(M/ δπ R^2) δ two π rdr=2 μ Gmrdr, and the moment of momentum is 2 / 3mgr μ Then we can use his method

extended data:

the moment of momentum of a particle with mass m, velocity V and radius r to the origin of R is L = R × mv The moment of momentum is a vector, and its projection on an axis is the moment of momentum on that axis. The moment of momentum on the axis is a scalar. The moment of momentum of a particle system or rigid body to a point (or axis) is equal to the vector sum (or algebraic sum) of the moments of momentum of all particles to the point (or axis)

Because the velocity of each point of a translational rigid body is the same (see translational motion of rigid body), its moment of momentum to a point is equal to the vector proct of the vector diameter of the center of mass of the rigid body taking the point as the origin and the momentum of the rigid body. If the angular velocity of a particle in uniform circular motion with radius r around the center O is, then the angular momentum of the particle to o is, where I is the moment of inertia of the particle to the center. The moment of momentum of a rigid body rotating around a fixed axis is the angular momentum of the rigid body, where I is the moment of inertia of the rigid body, ω Is the angular velocity of the rigid body around the axis

7. The question of the landlord is not complete. This question is very old. I saw it several years ago. At that time, the question was right. The friction factor between plate and cloth, plate and table is U2. Right uniform speed (or no uniform speed? Forget) pull out the cloth and the tray just stops at the table. I forgot whether to ask for the speed or the pull. This question is very difficult. It took a lot of time at that time. If the question is really what I said. Let's understand this question. To make the tray just stop at the edge, the tray must first break away from the cloth when accelerating to 1 / 4 of the table length. As long as you know that, whether it's speed or pull. You can establish a relationship through this length. You can establish three equations. In fact, two are enough. You can't blame me. You didn't make the topic clear.

8. Determination of friction force:
when determining the size of friction force, special attention should be paid to whether the friction force between objects is static friction or sliding friction, because the change of the size of the two is different. The size of sliding friction force is proportional to the pressure FN, and has nothing to do with the size of the external force causing the sliding friction force; The magnitude of static friction is independent of the pressure FN, which is determined by the external force causing the friction, but the magnitude of the maximum static friction is related to the pressure FN. Therefore, when the magnitude of friction is determined, the magnitude of static friction should be determined by the external force causing the static friction instead of F= μ FN calculation.
the common formula of sliding friction is f= μ The value of static friction is usually determined by the equilibrium condition.
1. It is determined by the equilibrium condition.
[example 1] when a wood block with mass m slides on the board placed on the table, and the board is still, its mass m = 3M. The sliding friction coefficient between the board and the wood block or between the board and the table is known to be 0 μ. The friction force of the board on the table is ()
a μ mg B.2 μ mg C.3 μ mg D.4 μ Mg.
[answer] A.
2. Determine according to Newton's second law.
[example 2] put a block of wood m on the horizontal disc, and the block will rotate at an angular velocity with the disc ω When the object rotates at a constant speed, the distance between the object and the rotating shaft is R. what is the friction force on the object? What's the direction
[analysis] when an object rotates with a disk, it needs a centripetal force, while the vertical object is balanced by Mg and FN, so it is impossible to provide a centripetal force. When the object rotates around the axis, it tends to be centrifugal, so it is subject to the friction force pointing to the center of the circle. The centripetal force comes from the friction force between the disk and the object.
[solution] the friction force acts as a centripetal force. According to Newton's second law, f = f = M ω^ 2R
extension: if the dynamic friction coefficient between the object and the disc is known to be μ, How far is the object from the axis to make it stationary relative to the disk
analysis: the farther away the object is from the axis, the greater the centripetal force is required. When the centripetal force required is greater than the maximum static friction, the object will slide centrifugally relative to the disk.
analysis: when the centripetal force required is greater than the maximum static friction, the object will slide centrifugally relative to the disk μ mg=m ω^ 2R R= μ g／ ω^ two

9. According to the meaning of the question, the rolling friction is zero, that is, the last disc W is constant
the condition of pure rolling is that the translational velocity of the center of mass is equal to the linear velocity of the edge of the disc
at the beginning, the translational velocity of the center of mass is zero. Acceleration: a = UMG / M = UG
for disk rotation: angular acceleration: a = UMG / M = UG ε= Umgr / J, j is the moment of inertia of the disk. J = Mr ^ 2 / 2
then: ε= UG / 2R
at the end of sliding: VC = w1r
VC = at
W1 = w- ε T
has: at = (W)- ε t)R，t=WR/(a+ ε R)
for the sliding distance, only the displacement of the center of mass can be calculated< br />L=at^2/2=ug(WR/(ug+ ε R) ) ^ 2
that is, the sliding distance is: UG (WR / (UG)+ ε R) ) ^ 2
in fact, the disc also rolls when it slides.

10. Determination of friction force:
when determining the size of friction force, we should pay special attention to whether the friction force between objects is static friction or sliding friction, because the size changes of the two are different. The sliding friction is proportional to the pressure FN, but not to the external force; The value of static friction has nothing to do with the pressure FN, which is determined by the external force causing the friction, but the value of maximum static friction is related to the pressure FN. Therefore, when determining the magnitude of friction, the magnitude of static friction should be determined by the magnitude of the external force causing static friction, instead of F= μ FN calculation
the common formula of sliding friction f= μ FN, and the static friction force is usually determined according to the equilibrium condition
1. Determined by equilibrium conditions
[example 1] when a wood block with mass m slides on the board placed on the table, the board is still, and its mass m = 3M. It is known that the sliding friction coefficients between the boards and the boards, and between the boards and the table top are all 0 μ The friction force of the board on the table is ()
a μ mg B. 2 μ mg C. 3 μ mg D. 4 μ mg
[answer] a
2. According to Newton's second law
[example 2] a wooden block m is placed on the horizontal disc, and the wooden block moves with the disc at an angular velocity ω The distance between the object and the axis of rotation is r. What is the friction force on the object? What's the direction
[analysis] when the object rotates with the disk, it needs centripetal force, while when the vertical object is balanced by Mg and FN, it cannot provide centripetal force. When the object rotates around the axis, it tends to be centrifugal, so it is subject to the friction force pointing to the center of the circle. The centripetal force comes from the friction force of the disc on the object
[answer] friction acts as a centripetal force. According to Newton's second law, f = f direction = M ω^ 2R
extension: if the dynamic friction coefficient between the object and the disc is known to be μ， How far is the object from the axis to make it stationary relative to the disk
analysis: the farther away the object is from the axis, the greater the centripetal force required. When the required centripetal force is greater than the maximum static friction, the object slides centrifugally relative to the disk
answer: μ mg=m ω^ 2R R= μ g／ ω^ 2

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