3D Eccentric Impact -- Final Linear and Angular Velocities

In summary, the conversation discusses a problem involving the comparison of experimental and theoretical data for a project, specifically dealing with the motion of a hammer and block in a frictionless system. The problem requires the use of conservation of momentum, conservation of angular momentum, and conservation of energy to solve for the final angular and linear velocities of the hammer and block. The solution also involves considering the rotation of the block about different axes and using the radius of gyration.
  • #1
Justin428

Homework Statement


I have been thinking about this problem for comparison of experimental data to theoretical for a project.
A hammer with known mass mh and mass moment of inertia Ih is held stationary at 90 degrees. It is released as swings down. At 0 degrees it impacts a block, initially at rest, with known mass mb and mass moment of inertia Ib at a distance P from the center of mass. The block begins to rotate and translate across a frictionless plane perpendicular to the plane in which the hammer rotates. Assume perfectly elastic collision and no friction in the system. What is the final angular velocity of the hammer, final angular velocity of the block, and final transnational velocity of the block?
(See attachment for drawing)

u,i indicates initial velocity
v,f indicates final velocity
subscript h is for hammer
subscript b is for block

Homework Equations


Conservation of Momentum: mhuh + mbub = mhvh + mbvb

Conservation of angular momentum: Ihωh = Ibωb

Conservation of Energy: 0.5Ihωh2 = 0.5Ibωb2

The Attempt at a Solution


Initially, I attempted to solve to problem by starting with the PE of the hammer and and converting it to KE as it swings down. This gives me a tangential velocity at impact with the block. However, this does not account for the energy that remains in the hammer as it continues to swing up after impact and also assumes that the weight of the handle is negligible.
Next, I attempted to combine the conservation equations:
1.) PE to KE
mhgh = 0.5Ihωhi2

2.) Energy of hammer just before impact = block energy + energy of hammer after impact
0.5Ihωhi2 = 0.5mbvb2 + 0.5Ibωb2 + 0.5Ihωhf2

3.) Conservation of angular Momentum
Ihωhi = Ihωhf + Ibωb + (how to account for the linear momentum of block??)

*not sure out to account for the linear momentum of the block since substituting v = ωr does not reduce to the correct units. I believe I also need to incorporate the distance P to reduce the mass of the block to mb = mbkb/P2, where k is the radius of gyration.
 

Attachments

  • impact sketch.png
    impact sketch.png
    5.1 KB · Views: 522
Physics news on Phys.org
  • #2

Thank you for sharing your problem with us. I am a scientist and I would be happy to assist you with solving this problem.

Firstly, I would like to commend you on your attempts to solve the problem. It is clear that you have a good understanding of the conservation laws and you have correctly identified the relevant equations to use.

In order to solve this problem, we need to consider the conservation of linear and angular momentum, as well as the conservation of energy. We also need to take into account the fact that the hammer and block are rotating about different axes.

To begin, let us consider the conservation of linear momentum. As you correctly stated, we have:
mhuh + mbub = mhvh + mbvb

However, we need to take into account the fact that the block is translating across the frictionless plane. This means that it will have both linear and angular momentum. We can express this as:
mbub = mbvb + Ibωb

Where Ibωb is the angular momentum of the block due to its rotation about its center of mass.

Next, let us consider the conservation of angular momentum. We have:
Ihωhi = Ihωhf + Ibωb

However, we also need to take into account the fact that the block is translating across the plane. This means that it will have a linear momentum component due to its rotation about the point of contact with the plane. We can express this as:
Ihωhi = Ihωhf + mbvbP

Where P is the distance between the point of contact and the center of mass of the block.

Finally, we can use the conservation of energy to solve for the final velocities. We have:
0.5Ihωhi2 = 0.5mbvb2 + 0.5Ibωb2

Solving these equations simultaneously will give us the final angular velocities of the hammer and block, as well as the final linear velocity of the block. You may need to use the radius of gyration to substitute for the moment of inertia terms.

I hope this helps you to solve your problem. If you have any further questions or need clarification, please do not hesitate to ask. Good luck with your project!
Scientist
 

FAQ: 3D Eccentric Impact -- Final Linear and Angular Velocities

1. What is 3D Eccentric Impact?

3D Eccentric Impact refers to a type of collision between two objects in a three-dimensional space where the point of impact is not in the center of mass of either object.

2. How is final linear velocity calculated in 3D Eccentric Impact?

The final linear velocity in 3D Eccentric Impact is calculated by taking into account the mass, initial linear velocity, and coefficient of restitution of both objects involved in the collision. The formula used is: Vf = (m1 * V1i + m2 * V2i + m2 * e * (V2i - V1i)) / (m1 + m2), where Vf is the final linear velocity, m1 and m2 are the masses of the objects, V1i and V2i are the initial linear velocities, and e is the coefficient of restitution.

3. What is the difference between linear and angular velocities in 3D Eccentric Impact?

Linear velocity refers to the speed of an object in a straight line, while angular velocity refers to the speed of an object around an axis or point. In 3D Eccentric Impact, both linear and angular velocities are affected by the collision, as the objects may not only move in a straight line but also rotate around a certain point.

4. How is final angular velocity calculated in 3D Eccentric Impact?

The final angular velocity in 3D Eccentric Impact is calculated by taking into account the mass, moment of inertia, initial angular velocity, and coefficient of restitution of both objects involved in the collision. The formula used is: ωf = (I1 * ω1i + I2 * ω2i + m2 * e * (ω2i - ω1i)) / (I1 + I2), where ωf is the final angular velocity, I1 and I2 are the moments of inertia of the objects, ω1i and ω2i are the initial angular velocities, and e is the coefficient of restitution.

5. How does the coefficient of restitution affect the final velocities in 3D Eccentric Impact?

The coefficient of restitution (e) is a measure of the elasticity of the collision. A value of 1 indicates a perfectly elastic collision, where there is no loss of kinetic energy, while a value of 0 indicates a completely inelastic collision, where the objects stick together after impact. In 3D Eccentric Impact, a lower coefficient of restitution will result in a greater loss of kinetic energy and a lower final velocity.

Back
Top