College Physics, Torque Work Related, up to 40 points

It means that there is an angular velocity w0 and the radius of the ball r0 rotates in a circle, and then the rope is pulled so that the radius becomes r0 2, and the work of the new angular velocity and tensile force is found.

This problem is actually a simple application of the conservation of angular momentum.

Since the ball is only subjected to radial tension, the resultant moment experienced by the ball is 0, so the angular momentum is conserved (towards that center).

mw0r0^2=mw1(r0/2)^2

w1=4w0

Since the pulling force only increases the velocity of the ball, the work done by the pulling force is equal to the increase in the kinetic energy of the ball.

w=3/2m(w0r0)^2

Tragedy.,Example problems in physics books.。。

I wouldn't have done it!!

To2 floor.

Patronizing and sighing, I didn't pay attention.

There is obviously no moment here that belongs to kinematics.

oksolution:

1) as the angle between f and r is 0 at any time, the moment: m=0

according to the formula: =dl/dt, the angular momentum l=l0(constant)

becauce l=mwr^2, l0=mw0r0^2=mw(r0/2)^2,hence, the new angular speed: w=4w0

2)the work done by the pull force(w) equals the change of the kinetic energy(et-e0): w=et-e0

e=1/2mv^2, v=wr,so,e=1/2mw^2*r^2

hence,w=et-e0=1/2mw^2*r^2-1/2mw0^2*r0^2

1/2m(w^2*r^2-w0^2-r0^2)

1/2m[(4w0)^2*(r0/2)^2-w0^2*r0^2]

3/2mw0^2*r0^2

Need Chinese?

Are you sure it's the University Moment?。。 I think it looks like a high school question.

And according to the topic, it is impossible to have a relationship with doing work. . . At best, it's balance ...

Maybe I don't understand???

It is recommended that you take a good look at the book.

There are two commonly used definitions of moment. One is the force multiplied by the force arm, and the other is the force fork multiplied by the position vector, and the position vector is a vector line segment from the point o to the point of application of the force, and the horizontal position is directly multiplied by the force arm.

The first question is that the moment of the system is equal to the moment of the two balls plus the moment of the rod. But the moment has a direction. The direction of the moment of the ball and the half-section to the left of the o-point is outward (the trend is counterclockwise), while the direction of the moment of the ball and the half-section to the right of the o-point is inward (clockwise).

So subtract it. It's clear that the end result is clockwise, so subtract the left from the right.

The m bar is the moment of the whole rod, and the moment of the right half is subtracted from the moment of the left half, which is the formula you have above.

3/4mg）((3/4)*.1 2) L)-(1 4mg)((1 4)(1 2)L)=1 4Mgl, where the gravity arm should take the distance from the center of the half-section rod to the O point.

The moment of the ball on the right is positive and negative on the left, and the algorithm is the same.

Who made this answer, so wordy! You draw a force diagram and write it out directly! You're all college students, and you're still writing questions like junior high school students, how can you do it!!

Three forces: the gravity of the left and right two balls, the gravity of the rod!

1) The resultant external moment is m=mg*l 4+mg*3l 4-mg*l 4=3mgl 4

2） m(l/4)^2+m(3l/4)^2+ml^2/12+m(l/4)^2=34ml^2/48

3) (3mgl/4)/(34ml^2/48)=(3mgl/4)*(48/34ml^2)=36g/34l

Question 24 is the one that asks for the moment.

Analysis: Since the current i2 in the AB wire is perpendicular to the direction of the magnetic field generated by the long straight wire, the direction of the ampere force F experienced by AB is perpendicular to AB.

The ampere force of the AB wire can be obtained by integrating the moment at point O.

Torque m b * i2 * dr , b 0 * i1 ( 2 r ).

The integration interval of r is from r0 to (r0+m. (I changed the distance from point O to end A to r0, so that it is not easy to be confused, please forgive the landlord).

i.e. m [0 * i1 ( 2 r )]i2 * dr

μ0 * i1* i2 / ( 2π )1/ r）* dr

μ0 * i1* i2 / ( 2π )lnr

Substituting the integral interval of r from r0 to (r0 to get the moment that is sought.

m＝[μ0 * i1* i2 / ( 2π )ln(r0＋

μ0 * i1* i2 / ( 2π )ln[(r0＋ r0]

Take the whole:

fx=0 nax-fcosθ+ndx=0 --1)∑fy=0 nay-fsinθ+ndy=0 --2)∑ma=0 m'-fsinθ.l-m+ -3).

fx=0 --4)

fy=0 ndy+ -5)∑md=0 -m+ -6)

Substituting the known data, the above 6 equations can be solved to solve 6 unknown constraints. If the result is negative, the binding force is in the opposite direction to the one set.

The direction of the moment is the direction of the moment, and the direction of the angular momentum is the direction of the angular momentum. There is no necessary relationship between them.

Similar to the relationship between the direction of force and the direction of momentum, there is no necessarily perpendicular or parallel relationship between them.

The moment determines the direction of angular acceleration.

When you push open a door with your hand, the direction of the moment is perpendicular to the ground. (Different from the direction of the force, defined by the right-hand rule).

The angular momentum gained by the door is also perpendicular to the ground (again defined by the right-hand rule).

Here the angular momentum is conserved, strictly speaking, the angular momentum component is conserved, as can be seen from the diagram, within the angular velocity vector of the system.

Vertically upward, so the capacitive angular momentum l (vector) = j*w (angular velocity vector), vertically upward. The moment r g of gravity on the axis of rotation is in the horizontal direction, so the component of this moment in the vertical direction is zero, so the external moment of gravity does not affect the angular momentum in the vertical direction.

By the way, this is also the case with a person on a horizontal turntable, moving along the radius: the component form of conservation of angular momentum, i.e., although the external moment as a whole is not zero, but the component of the external moment in a certain direction is zero, then in that direction, the angular momentum is conserved.

Orthogonal decomposition of f into: fcos and fsin set the moment counterclockwise as"+", clockwise as "-".

The FSIN crosses point A and the moment towards A is zero.

The result is the solution in the problem.

This problem is a moment problem, as long as it is pulled at the beginning, the force that needs to be applied later is getting smaller and smaller. According to F*L1=Mgl2, and the resistance arm and the power arm are perpendicular to the direction of their respective forces, the power arm (L1) is 1m. Whereas the resistance arm (L2) is.

So there is: f*l1=mg*l2 , g= , so f=98n

If e represents the kinetic energy of the slider and s represents the distance, then the amount of change in the kinetic energy in the whole process is equal to the amount of work done by friction

f=μn=μmv²/r=2μe/r

de=-fds=-2μeds/r=-2μerdθ/r=-2μedθde/e=-2μdθ

e=(mv0²/2)e^(-2μθ)

The work done by friction is e( )e(0):(mv0 2)(e(-2)1).