College Physics Rigid Body Rotation Problem, College Physics, A Problem of Rigid Body Rotation

The problem with this question is in the title:

At a certain moment, in"Moment"Under the action of the angular velocity and angular acceleration can be 0 is such that under this moment, the angular acceleration cannot be 0, and the angular velocity may be 0; It is like adding a force to a stationary object instantaneously, the velocity is 0 but the acceleration must not be 0;

At a certain moment, the object is under the action of moments, it is not said how many moments there are, if there are two moments, and they cancel each other, then both angular acceleration and angular velocity can be equal to 0

For example, for a stick with a fixed center, a clockwise moment perpendicular to the stick is applied to the left end, and a counterclockwise one is added to the right end. It won't move. But can you say that at this moment, the object did not receive a moment action? Apparently not.

It seems that the angular velocity should be okay, for example, the object is stationary, and then the moment is given instantly, and the angular velocity is 0 A.

Angular acceleration doesn't seem to be.

However, your question is not very exact, because the moment, the angular acceleration is all relative to a certain point. For example, if a disc has a fixed center, the angular acceleration at the center is 0., regardless of the momentTherefore, the proposition is not precise enough, and it is more difficult to judge it.

It is better to state what the moment of one point is for another point, and find the angular acceleration relative to a certain point.

Stick turn to kinetic inertia:

Source i1 = (l m1) 12

The bullet turns. Inertia: i2=m2(l 2) =(l m2) 4 Conservation of rotational momentum: (i1+i2)* =i2*v0 (l 2) obtain: =(6m2*v0) (lm1+3lm2).

Solution: (BAI1)MF=j ( adds du to the angle

velocity), zhi = -5rad s, and the negative sign indicates that the dao is a deceleration motion.

In the linear velocity v=30m s, then the angular velocity is w=v r=60rad s by 0-w= t, and the tolerance t=12s

2) 0-w = 2, angular displacement = 360rad the number of turns of the wheel rotation n = 2 = 180

then the distance is s=n2 r=180m

Mechanical energy is conserved, initial potential energy = final potential energy.

mgl(1-θ)2mgl(1-cosθ'/2)cosθ'=2cosθ=1/2

cosθ=1/4

Calculate the angular velocity at the beginning first, and then calculate the second angular velocity through the conservation of angular momentum, and after this velocity is calculated, you get the centripetal acceleration, I didn't bring the number in to calculate it for you, do you calculate it yourself, do you understand? Ask me again if you don't understand something.

Good question! I've thought about this issue too :

But which direction do you mean by "the direction in which the rigid body turns"? Specialized in rigid bodies.

The object has different directions of rotation in different directions around the circle, and it is difficult to describe the rotation;

How can it be described in a counterclockwise way? If it is clockwise from the front, then it is counterclockwise from the back, and I can also think of the front as the tail, and it cannot describe the rotation;

So with the right-hand rule to regulate, so that no matter how you look at it, well, the direction is always the same;

And as long as the angular velocity vector is used, the radius vector (cross product) = the velocity vector of a point in the rigid body, that isrv

Good question! I've thought about this issue too :

But which direction do you mean by "the direction in which the rigid body turns"?

Taking the rigid body as the object, the rotation direction has different directions in different directions around the circle, and it is difficult to describe the rotation.

How can it be described in a counterclockwise way? If it is clockwise from the front, then it is counterclockwise from the back, and I can also think of the front as the tail, and it cannot describe the rotation;

So with the right-hand rule to regulate, so that no matter how you look at it, well, the direction is always the same;

And as long as the angular velocity vector is used, the radius vector (cross product) = the velocity vector of a point in the rigid body, that isrv

Good question! I've thought about this issue too :

But which direction do you mean by "the direction in which the rigid body turns"?

Taking the rigid body as the object, the rotation direction has different directions in different directions around the circle, and it is difficult to describe the rotation.

How can it be described in a counterclockwise way? If it is clockwise from the front, then it is counterclockwise from the back, and I can also think of the front as the tail, and it cannot describe the rotation;

So with the right-hand rule to regulate, so that no matter how you look at it, well, the direction is always the same;

And as long as the angular velocity vector is used, the radius vector (cross product) = the velocity vector of a point in the rigid body, that isrv

Good question! I've thought about this issue too :

But which direction do you mean by "the direction in which the rigid body turns"?

Taking the rigid body as the object, the rotation direction has different directions in different directions around the circle, and it is difficult to describe the rotation.

How can it be described in a counterclockwise way? If it is clockwise from the front, then it is counterclockwise from the back, and I can also think of the front as the tail, and it cannot describe the rotation;

So with the right-hand rule to regulate, so that no matter how you look at it, well, the direction is always the same;

And as long as the angular velocity vector is used, the radius vector (cross product) = the velocity vector of a point in the rigid body, that isrv

The left question is divided into 3 processes:

Conservation of mechanical energy in clay fall h: final velocity v= (2gh) ;

Conservation of angular momentum of clay-disc collision (excluding clay gravity): , jo=mr 2 2+m(r 2) 2

> the end of the collision, the angular velocity of the disc =

The velocity of the clay v'=ω.r=

After the collision is completed, v the law of rotation: angular acceleration = , tangential acceleration of clay at=r, normal acceleration an=v'^2/r 。

Right, conservation of angular momentum: > = tangential acceleration of the bullet at=0, normal acceleration an=rω^2=r.(v/(2r))^2=v^2/(4r^2)