Q: "I read that if I were to drop a steel ball bearing and a rubber ball at the exact same height, I would find out that the steel ball would rebound higher than the rubber ball. This was only if the balls were dropped on a very hard surface. The steel ball would rebound higher because of its elasticity. I was wondering how I could calculate elasticity and also what kind of surface would be appropriate to demonstrate this."
A: This is a fun experiment. To demonstrate it, the best for you would be to buy a set of balls called "happy balls" and "sad balls" from any science catalog. They are both the same size, the same shade of black color, but are made of different materials: one of them is elastic and bounces back to about the same height where it started, while the other ball will absorb the collision energy and will not bounce at all (will just remain dead down in contact with the floor.) Viewers will be very surprised to see this. Any floor, such as linoleum, wood, cement, etc. would be good enough for the experiment. The idea is that the elastic ball will recover completely after the collision with the floor, although the region at the surface of contact will be temporarily compressed. In the non-elastic (or better to say less elastic) ball, after compression at contact, the energy of collision will be dissipated inside the ball, and therefore less energy will be left available as kinetic energy for the ball to re-bounce.
Try this experiment. It is guaranteed successful.
If you still want to try your steel and rubber balls, the best floor that I can think of would be steel of the same quality as your ball. I suggest that you demonstrate instead how a steel ball suspended as a pendulum would bounce back after collision with a vertical steel wall or with another steel pendulum.
Q: What kind of deformation does a Jell-O cube exhibit when it jiggles?
the Jell-O cube is well prepared (the mass of Jell-O is not running and
the base of the cube is fixed to the plate beneath the cube,) then
jiggling means vibrating of the mass of Jell-O around and equilibrium
position. In this case the deformation of the mass of Jell-O is
mainly of "bending" type.
Q: How do the elastic properties of a solid-sole (one without air) tennis shoe help cushion the foot?
A: When the weight of the tennis player is applied on the shoe, the force of weight is distributed over the surface of the sole, and the sole is being squeezed. The elastic material of the sole, then, allows the squeezed sole to recover to the original unsqueezed shape. Squeezing of the sole (similar to squeezing of an air bag) takes time. The longer this squeezing time, the longer the stopping time of the moving body of the tennis player. The longer this time, the smaller the "shock" force on the stopping body will be. The mathematical correlation between the parameters involved in this problem is F (t) = m(v), where F = force, m = mass, v = change of velocity when the moving mass stops, and t = time interval during which the moving mass comes to rest. From this relationship, the force experienced by the moving body while stopping is F = m (v/t). The larger the time of stopping t, the smaller the shock force (F) will be. Any shock absorber has the purpose of increasing the time interval t during which a moving object comes to rest. For details about the physics principles involved in this example, please review the definitions of linear momentum (mv) and impulse (Ft).
[Top] [Previously Asked Questions] [References]
Strain: A fractional change in length or
deformation of a material body under the action of applied forces.
Stress: Force per unit area. A force exerted when one body or body part presses on, pulls on, pushes against, or tends to compress or twist another body or body part.
Modulus: The ratio of the stress in a body to the corresponding strain.
Young's Modulus: Measures the resistance of a solid to a change in its length.
Shear Modulus: Measures the resistance to motion of the planes of a solid sliding past each other.
Bulk Modulus: Measures the resistance that solids or liquids offer to the changes in their volume.
|Elastic Moduli||stress = modulus X strain|
|Tension and Compression||F/A = E (L/L)||E is Young's Modulus|
|Shearing Stress||F/A = G (x/L)||G is the shear modulus|
|Hydraulic Stress||p = B (V/V)||B is the bulk modulus|
[Top] [Previously Asked Questions] [References]
|Measurements||Newton's Laws||Potential Energy and Conservation of Energy||
|Vectors||Forces and Fields||Linear Momentum||Angular Momentum||
|Motion of Point-Mass Objects in One Dimension||The Gravitational Field||Collisions||Torque||Mechanical Waves|
|Motion of Point-Mass Objects in Two and Three Dimensions||
|Circular Motion of Point-Mass Objects||Equilibrium||Sound|