What is the momentum of the homogeneous disc around the central axis not angular momentum and what

Originally, it was inappropriate to discuss the momentum of a rotating disk, so angular momentum was quoted.

momentum).

Momentum is the concept of translational motion. If it is rotated on a fixed axis, the momentum is 0.

Momentum moment is the concept of rotation and is not 0 and is related to the mass speed radius.

1. Newton's second law f ma in translational motion, the resultant external force mass line acceleration. In turning, it becomes m i; Resultant external moment, moment of inertia, angular acceleration.

2. In translation, the momentum expression of Newton's second law is: the rate of change of the momentum of the line of external force; Linear momentum, mass, velocity. In rotation, the angular momentum of Newton's second law is expressed: the rate of change of the angular momentum of the resultant external moment; angular momentum, moment of inertia, angular velocity.

3. The kinetic energy in translation, ek mv mass, and the square of the linear velocity. Kinetic energy in rotation ek mv moment of inertia squared of angular rate.

Momentum is equal to mass multiplied by velocity, which can be done in a micro-element method, for example, a rod (with mass m) moving in a uniform circular motion, taking a section of symmetry on both sides, with equal mass, and the velocity of the two adds up to exactly twice the velocity at the midpoint, so that the cumulative momentum is m multiplied by half of v, and the same is true of the disk, and the momentum moment is half the radius.

Are moment of inertia and momentum the same thing.

Angular momentum =Moment of inertiaAngular velocity =1 2mr 2 w.

The amount that describes the state of rotation of an object. Also known as momentum moment. If the mass of the particle is m, the velocity is v, and its sagittal diameter with respect to point o is r, then the angular momentum of the particle to point o is l r·mv.

Angular momentum is a vector quantity, and its projection on an axis through point o is the angular momentum (scalar) of the particle on that axis.

The angular momentum of a mass system or rigid body to a point (or axis) is equal to the vector (or algebraic) sum of the moments of the momentum of each particle to that point (or axis). A mass m of mass motions in a uniform circular motion around point o with radius r.

The angular velocity of rotation is , then the angular momentum of the particle to the point o l r·mv r·mr mr2 i0 , where i0 is the moment of inertia of the particle to the center of the circle o.

A rigid body rotating around a fixed axis z with angular velocity, in which each point moves in a uniform circular motion on each plane perpendicular to the z-axis, and their center is the intersection point of each plane and the z-axis.

Angular momentum describes the mechanical quantity of an object that rotates around a fixed point or axis. Select a reference point, and if the object's motion has no components tangent to the radial direction, then the object's angular momentum relative to that point is zero. Quantitatively, angular momentum l is equal to the vector diameter (i.e., the vector length of the reference point pointing at the object) cross-product momentum.

For example, if an object is moving in an accelerated straight line, the angular momentum is zero if the reference point is taken on the moving line. If the reference point is not in a line of motion, there will be rotation at that point, and there will be angular momentum.

<> surface density = m ( r 2 ) , micro ring stool area da = 2, moment of inertia of the microring.

Disc moment of inertia Blind Zen Li io= substitution integral grinding delay (0-->r) disc moment of inertia io=((m ( r 2))2 (r 4) 4=

The moment of inertia of the ring is the diameter of the pin.

Seek the law. Take the microelement dm= (m2)d

Then the moment of inertia of the ring on the straight equilibrium trail: J=(mr 2 ) sin d is substituted into the upper limit of the integral block 2 and the lower limit is 0 points

j=mr²/2

The moment of inertia of the disc around the central axis is 1 2mr 2.

For example, a disk with radius r and mass m rotates around the axis of mass perpendicular to the plane of the disk, and the moment of inertia j. is obtained

Solution: Divide the mass element as a torus, and the radius of the torus is r and the width is dr, then the mass of the torus: dm=dm=m (pi*r 2)* 2pi*rdr, and then substitute j= r 2dm from 0 to r to integrate, and get j=1 2mr 2.

Note: The magnitude of moment of inertia depends on the shape of the object, the mass distribution, and the position of the shaft. The moment of inertia of the rigid circle carrying the ulnar body has important physical significance, and it is also an important parameter in scientific experiments, engineering technology, aerospace, electric power, machinery, instrumentation and other industrial fields.

For an object with a uniform mass distribution and an uncomplicated shape, the moment of inertia relative to a definite rotating shaft can be calculated by a formula from the mass distribution of its dimensions.

For a rigid body with a simple geometry and uniform mass distribution, its moment of inertia relative to a definite rotating shaft can be calculated directly with a formula. However, for objects with complex shapes and uneven distribution of orange mass, the moment of inertia of the object can only be accurately measured by experimental methods, so the experimental method is more important. <>

<> this has to be the moment of inertia from a thin circle around the central axis.

Counting, linear density = m (2 r) , j = r >2 ) = moment of inertia of the disc around the central axis.

The areal density m ( r 2 ) and the mass of the microring dm = (2 r) drj = r 2 dm=∫σr^2(2πr)dr=σ.2π(r^4)(0-->2π)=

Calculus can be used to derive the process, and the specific process is as follows:

Angular momentum l= mr 2 2, also known as "momentum bai moment".

A definite integral can be used to prove:

du takes the distance from the center of the disk du is the circle of r to r + dr, then the mass of the ring is: m * 2 * pi * r * dr) (pi * r * r);

The moment of inertia is: 2m*r 3 r 2dr

So the moment of inertia of the disc is 2m*r 3 r 2 r from 0 to r 2m*r 3 r 2dr = 1 2(mrr).

Angular momentum l= mr 2 2, also known as "moment of momentum".

1.Look at it as two semicircles of the Land Information Bureau.

2.Ask my sister where the center of mass of the semicircle is (not the midpoint of the radius, calculate it with calculus, (find the area of the two parts equal after the semicircle is divided), I won't do this, I forgot the formula).

3.Calculate what is the linear velocity of this center of mass.

is the linear velocity, m is the mass of the semicircle (everyone knows this)5Don't add up the two sides and say that this is momentum (only energy can be calculated)6His external momentum is still 0.