Rigid body collision problems, rigid body collision problems

1) Obviously, when the moment of inertia of the ball is equal to the moment of inertia of the straight rod around the O point, the ball is just stationary after the elastic collision between the ball and the rod.

The moment of inertia of the small ball: J1 = ML 2, the moment of inertia of the straight rod: J2 = ML 2 3

Then there are: ml 2 = ml 2 3, l = 3l 3

That is, when the length of the rope is 3l 3, the ball is just stationary after the elastic collision between the ball and the rod.

2) The binding force on the fulcrum is actually the centrifugal force of the ball and the straight rod. The least binding force is the least centrifugal force.

This is a bit of a problem, because the binding force is not calculated to be related to the length of the wire, unless l=0

There is a formula: the centrifugal force of the fine wire: f=mg+2mg(1-cos) (when l=0, there is no 2mg(1-cos), the binding force of point o is equal to mg).

The binding force of the straight rod to the O point is related to its angular velocity, and the angular kinetic energy of the straight rod is proportional to the square of the angular velocity, then the smaller the angular kinetic energy, the smaller the binding force, and when the elastic collision occurs, the angular kinetic energy is provided by the small ball, then l=0 is the smallest (the straight rod does not obtain the angular velocity. The binding force of the point o is equal to mg).

Maybe it's the most binding force. When l=l, there is a maximum.

Because when l=l, the potential energy released by the ball: ep=mgl(1-cos) is maximum. The angular velocity obtained by the straight rod is the largest, the centrifugal force of the straight rod on the O point is the largest, and the binding force of the O point is also the largest.

l=l 2.

That is, when the ball hits the center of gravity of the rod.

The ball is just stationary.

The second time to get the minimum (small) value is more complicated.

The answer to the first question is l 2 and the second question is 0. Explain in detail that you will give it tomorrow and rest today.

The formula for a fully elastic collision is v1'=[m1-m2）v1+2m2v2]/（m1+m2），v2'=[m2-m1）v2+2m1v1]/（m1+m2）。

According to whether the kinetic energy of the collision process is conserved or not, it is divided into:

Fully elastic collision: the kinetic energy of the system is conserved before and after the collision; Non-fully elastic collision: the pure training of the kinetic energy of the system before and after the collision is not conserved; Completely inelastic collision: The system moves at the same speed after the collision.

Fully elastic collisions are conserved according to energy and momentum.

If two objects with mass m1 and m2 collide with initial velocity v1 and v2, the only thing that happens is that the velocity after the collision is v1',v2'。

Then according to: shirt and pants m1v1 + m2v2 = m1v1'+m2v2'。1/2 m1v1^2 + 1/2 m2v2^2 = 1/2 m1v1'^2+ 1/2m2v2'^2。

E-Proof: v1' =m1-m2)v10 + 2m2v20] /m1+m2)，v2' =m2-m1)v20 + 2m1v10] /m1+m2)。

(1) At the moment when the bullet is fired into the disc, the velocity of the bullet can be regarded as the linear velocity of rotation around the center point of the disc.

Then: angular velocity 1=v0 r (1).

The resultant moment of the system is zero, which is consistent with the conservation of angular momentum.

Let the angular velocity of the system after injection be: 2

The moment of inertia of the bullet with respect to the disc is: i1=mr 2, and the moment of inertia of the disc is: i2=mr 2 2

Conservation by angular momentum:

i1ω1=（i1+i2)ω2 (2)

1) formula is substituted into (2) formula.

2=i1v0/r（i1+i2)

Then: The total kinetic energy of the system is:

ek=（i1+i2)ω2^2/2

i1+i2)(i1v0/r（i1+i2))^2/2=(i1v0)^2/(2（i1+i2))

mv0)^2/(2m+m)

2) When the bullet is fired, it can be seen as a disc rotating around the lower fulcrum.

Then there is: the moment of inertia of the bullet is: i1 = m(2r) 2, and the moment of inertia of the disc is: i2 = 3 2 mr 2

The angular velocity of the bullet 1=v0 (2r )3) is conserved by angular momentum:

i1ω1=（i1+i2)ω2 (4)

3) formula is substituted into (4) formula.

2=i1v0/（2r（i1+i2)）（5）ek=（i1+i2)ω2^2/2

Bring in (5)ek=(i1v0) 2 (8r 2(i1+i2))=16(mv0) 2 (8m+3m).

The solution upstairs is the solution of high school, which is an overly simplistic model, and the answer is wrong.

This problem should be solved using the momentum theorem and the angular momentum theorem.

The first question is that the whole process m, the net moment of the m system to the center of the disc is zero. Angular momentum is conserved.

The second question is that in addition to the conservation of angular momentum, momentum is also conserved. It should be noted that the disc has translation and rotation. After injection, the m-velocity should be the translational motion of the center of mass of the disc plus the rotation around the center of mass.

I don't know the specifics. If you still have questions, please feel free to ask.

If it helps you,.

Let the obtained common velocity be v1 and the total kinetic energy of the system is ek

From the law of conservation of momentum: mv0=(m+m)v1 1) from the kinetic energy expression: ek=1 2*(m+m)*(v1) 2 2) synthesis (1)(2) get:

ek=1/2*[(m^2)/(m+m)]*v0^2

1. The collider has mechanical energy.

energy loss, rigid body is no mechanical energy loss.

2. To create a collision, you must add rigidbodies and colliders to the GameObject, which allow the object to move under the influence of physics. A Collider is a type of Physics Component that is added to a GameObject along with Rigid Nucleoslip in order to trigger a collision. If two rigid bodies collide with each other, the physics engine will only calculate the collision if the two objects have colliders, and in the physical modification flushing simulation, the rigid bodies without colliders will pass through each other.

<>1. The collider has mechanical energy loss, and the rigid body has no mechanical energy loss.

2. To create a collision, you must add a rigid body (including a rigidbody) and a collider to the game object, which allows the object to move under the influence of physics. A collider is a type of physics component that is added to the first round of the play object along with the rigid body in order to trigger a collision. If two rigid bodies collide with each other, the physics engine will not calculate the collision unless the two objects have colliders, and in physics simulations, rigid bodies without colliders will pass through each other.

1 All m=m? The two are considered to be equal.

Let the velocity at the end of a be V', the velocity at the end of b is v''

Conservation of momentum mv = mv'+3mv'' ①

If it is a fully elastic collision, it is also satisfied.

1/2mv^2=1/2mv'^2+1/2*3mv''2 Simultaneous solution v'=-v/2v，v''=v 2 can be rewritten as if it is a completely inelastic collision.

mv=(m+3m)v''

v''=v/4

The value of the final velocity of b should be between the above two values, i.e., v 4<=v''<=v 2, so the answer is b

If the speed of the B ball is v2, there are three possible scenarios.

Elastic collision: m * v = m * v1 + 3m * v2 (1 2) *m * v 2 = (1 2) *m * v1 2 + 1 2) *3m * v2 2

v = v1 + 3*v2

v^2 = v1^2 + 3*v2^2

v1 =v2 =

Completely inelastic:

m * v = (m + 3m) *v2v2 = (1/4)*v

Generally inelastic:

1/4)v < v2 < 1/2)v

There is no energy loss in elastic collision, which is the critical value of the maximum velocity of b, and the maximum energy loss of inelastic collision is the maximum value of the velocity of b, so v is located between to.

So only (2) the condition is satisfied.

It is stipulated that elastic collisions are the conservation of momentum and energy, which is the ideal state. If you look at the concept of elastic collisions: kinetic energy is conserved only when there is a fully elastic collision, and there is a loss of energy in all other cases (in this case the energy is only kinetic energy.

For the sake of simplicity, allow me to set the initial velocity of m2 to zero.

By the law of conservation of momentum m1*v1=m1*v1'+m2*v2'Can be solved v2'=(m1*v1-m1*v1')/m2

The total kinetic energy after the collision is e=m1*v1'*v1'/2+m2*v2'*v2'2 Substituting the results above.

e=m1*v1'*v1'/2+(m1*v1-m1*v1')*(m1*v1-m1*v1')/(2*m2)

(m1+m2)*v1'*v1'-2*m1*v1*v1'+m1*v1*v1]*m1/(2*m2)

m1+m2)*v1'*v1'-2*m1*v1*v1'+m1*v1*v1 is based on v1'is a quadratic function of the variable, the image is a parabola with an opening upward, and it can be mathematically known that the parabola y=a*x*x+b*x+c takes -b 2a as the axis of symmetry, that is, in.

When x=-b 2a, y has a minimum value. then in v1'=m1*v1 (m1+m2), e has a minimum value. This velocity is the velocity of m1 and m2 at the same velocity, which indicates that the kinetic energy gained by the completely inelastic collision is the smallest, and the mechanical energy loss is the largest.

From the above equation, it can be obtained that when u3=u4, the mechanical energy after the collision is the smallest, that is, the mechanical energy loss is the largest.

Let the mass of the light ball be m, then the mass of the heavy ball is 3m

The total kinetic energy of the two balls before the collision eo = (1 2) (3m) v 2 + (1 2) mv 2 = 2mv 2

After the collision, the total kinetic energy of the two balls e=0+(1 2)mv 2=2mv 2e=eo has no energy loss, and they have an elastic collision.

The mass of the light ball is m and the velocity is v

That is, according to the conservation of momentum: before the collision: mv-3mv, after the collision is -2mv

If the two equations are equal, it is an elastic collision.

This is the ideal state prescribed by man, and the elastic collision is the ideal state where momentum and energy are conserved.

Take a look at the concept of elastic collision: under ideal circumstances, after the collision of objects, the deformation can be restored, without heat, sound, and kinetic energy loss, this kind of collision is called elastic collision, also known as full elastic collision. True elastic collisions occur only between molecules, atoms, and smaller particles.

In life, when a hardwood or steel ball collides, the loss of kinetic energy is small and negligible, and their collision is usually regarded as an elastic collision. Momentum is conserved at the time of collision. When two objects have the same mass, the velocity is exchanged.

In fact, it is all ideals that produce conservation.

Kinetic energy is conserved only in the case of a fully elastic collision, and in other cases there is a loss of energy (in this case, energy is stored only in the form of kinetic energy);

This problem feels a bit backwards, we call collisions when kinetic energy is conserved as fully elastic collisions, and momentum conservation does not require the condition of elastic collisions (the above discussion is based on the collision of objects in ideal space); The certificate should be on the book, right?

If a system is not subjected to an external force or the vector sum of the external forces is zero, then the total momentum of the system remains the same, and this conclusion is called the law of conservation of momentum. The law of conservation of momentum is one of the most important and universal conservation laws in nature, which applies to both macroscopic objects and microscopic particles; It is an experimental law that can be deduced from Newton's third law and the momentum theorem.

The velocity does not change, and the kinetic energy is conserved.

Mechanical energy is conserved.

In fact, it should be reversed, and people define the collision in the ideal state where the momentum and kinetic energy of the system are conserved as an elastic collision. So fundamentally, you understand it the other way around.