Wednesday, November 4, 1998 Chapter 8: Angular Momentum Chapter 9: Density, Stress, Strain, Young’s Modulus, shear modulus
We can construct a completely analogous argument to define angular momentum Where L is the angular momentum
A solid cylinder with moment of inertia I=MR 2 /2 is given an angular velocity of 1 rad/s. If the cylinder has radius 0.5 m and mass 2 kg, what is its angular momentum?
Solids & Fluids We’ve spent a lot of time looking at systems of point masses, objects with no physical extent. Let’s now look at more realistic representations: objects that flow, stretch, and compress.
Let’s start by characterizing a solid mass of uniform composition. What quantities can we directly measure? If our block was made of copper, and we doubled its size, what would happen to its mass?
Remember our scale arguments (way back in the first week of the course)!!! L 2L V 0 = L 3 V = (2L) 3 V = 8L 3 = 8V 0 M 8M
In fact, if we made a plot of the mass of our copper block versus its volume, we’d find m V This line has a slope that characterizes the type of material from which the block is made. We define the slope of this line as the density ( ) of the material. slope = How would the slope of the line for a block of styrofoam compare to that for a block of lead?
The unit of mass, the gram, was chosen to be the mass of 1 cm 3 of liquid water. So water has a density of 1 g/cm 3 or 1 kg/L or 1000 kg/m 3. The term specific gravity refers to the ratio of the density of a given substance to that of water. Objects with a specific gravity less than 1 will float in water; those greater than 1 will sink.
Note: density is often written in grams per cubic centimeter or g/cc. There is a factor of 1000 difference between the two sets of units.
Stress on an object results in strain. You guys are looking pretty strained…I mean drained right now! A stress results when a force is applied across a surface of a given object. Stresses result in deformations of the object known as strains.
We push on a piece of copper with a force F. If we apply the force over the entire cross- sectional area of the end of the tube, the stress is given by F F compressive stress
F F What happens to real objects when we exert stress such as applied in this figure? In this case, the copper rod will be compressed. The fractional change in its length is known as the strain l lflf UNITLESS QUANTITY!
When stresses are applied, objects undergo strain. The ratio of the stress to the strain turns out to be a characteristic of the material from which the object is made. We call the property of a material related to the way it strains under stress
Notice that Young’s modulus will have the same units as stress: N / m 2 We’ve looked at compressive stress. What happens when we try to stretch an object instead of trying to compress it?
F F lflf l tensile stress We name the stress that tries to pull an object apart the “tensile stress.” Again, the ratio of the stress to the strain is still a characteristic of the material from which the object is made. In fact, Young’s modulus describes this type of stress as well as compressive stress.
A 500-kg load is hung from a 3-m steel wire ( Y = 2 X N/m 2 ) with a cross-sectional area of 0.15 cm 2. By how much does the wire stretch? Well, the force exerted by a 500-kg load is
A 500-kg load is hung from a 3-m steel wire ( Y = 2 X N/m 2 ) with a cross-sectional area of 0.15 cm 2. By how much does the wire stretch?
We can conceive of yet another way to try to deform our copper tube... F F l shear stress This type of stress tries to move one end of the tube in one direction and the other end of the tube in the opposite direction! Unfortunately, it is NOT characterized by Young’s modulus...
Let’s look at a side view of what happens to our copper cylinder under the influence of shear stress. FsFs FsFs xx l Shear stresses produce shear strains “shear force” where A is the area of the top of the cylinder.
FsFs FsFs xx l Note: the units of shear stress and shear strain are the same as our tensile and compressive forms of stress and strain.