First, recall that angular speed = w = Dq/Dt, a = Dw/Dt, vt = wr, at = ar, and torque = t = Fd.
Consider a mass, m, fixed to a central point by a string of length r, and moving about that point in a circle on a frictionless surface. If a tangential force is applied to the mass, it will experience a tangential acceleration given by Ft = mat. Multiply both sides of this expression by the radius, so that the left hand side is the torque, to get Ftr = t = mrat. Now use the relation at = ra and the expression becomes t = mr2a. This relation says that the torque is proportional to (or produces) the angular acceleration, times a factor of mr2. The factor mr2 is called the moment of inertia.
What if the rotating object is not a small mass at the end of a string, but a larger object, like a wheel, a cylinder, or a sphere?
In this case, consider the extended object as being composed of lots of little pieces, each piece feeling a torque.
Then the total torque is the sum of the torques on each piece:
ttot = St = S(mr2a).
Since the angular acceleration, a, is the same for all the pieces of the object, it can be factored out of the sum in the last expression to yield:
ttot = (Smr2) a.
We call I = Smr2 the moment of inertia of the object, and then write ttot = Ia, the rotational analog of F = ma.
The moment of inertia is an important part of the above result. It plays the same role as mass in Newton's second law, but has some special properties. While an object has only one mass, it can have different moments of inertia which depend on the axis of rotation, and the distribution of mass (applicable for people and object with moving parts). Table 8.1 lists the moments of inertia for a number of shapes and axes of rotation.
If an object is (purely) rotating, then each particle of the object has kinetic energy. The sum of all the kinetic energies is the kinetic energy of rotation, KErot = S(½mivi2) = ½(Smiri2)w2 = ½Iw2. Notice the similarity of the expressions KE = ½mv2 and KErot = ½Iw2.
Rotational kinetic energy is a useful concept, especially when an object is both rotating and translating, like a wheel rolling down an incline. The total kinetic energy of the wheel is the sum of the kinetic energies of each particle in the wheel. Due to the combined rotation and translation, the speed of each particle depends on its angle on the wheel, and the computation becomes quite complex. But we can consider the translational and rotational motions separately.
The total kinetic energy of a rolling object is the sum of the translational kinetic energy (mass of the object, speed of the axis of rotation), and the rotational kinetic energy:
KEtot = KEtrans + KErot = ½mv2cg + ½Iw2.
If we ignored the rotational kinetic energy, then we would predict that both will move down an inclined plane at the same speed, since after moving downward a vertical distance h, the kinetic energies would be ½mv2 = mgh, or v = Sqrt{2gh}.
Since these objects are rolling, we must include the rotational kinetic energy in the conservation of energy equation.
Then we find that:
½mv2 + ½Iw2 = mgh.
The moment of inertia of a solid disk is Idisk = ½mr2, and the moment of inertia of a ring is Iring = mr2.
Inserting these into the conservation of energy expression, yields:
½mv2 + ¼mr2w2 = mgh for the disk, and
½mv2 + ½mr2w2 = mgh for the ring.
Divide through by the mass, and substitute v = wr, resulting in:
½v2 + ¼v2 = gh for the disk, and
½v2 + ½v2 = gh for the ring.
Thus we find that the disk has a speed of vdisk = Sqrt{4gh/3}, and the ring has a speed of vring = Sqrt{gh}.
The disk is moving faster by a factor of Sqrt{4/3} for the same distance traveled.
Demonstration.
The disk has a smaller moment of inertia than the ring. Therefore a smaller fraction of its kinetic energy goes to rotational kinetic energy, and a larger fraction goes to translational kinetic energy, and its the translational kinetic energy that gets the object down the incline.
Continuing to draw parallels between rotational and translational motion, we can define an angular momentum. Linear momentum was defined such that F = Dp/Dt. Similarly, we can write t = Ia = IDw/Dt = D(Iw)/Dt. Now it is easy to see that the rotational analog of p=mv is angular momentum = L = Iw.
We can write t = DL/Dt, which states that "the torque acting on an object is equal to the time rate of change of the object's angular momentum."
If the net torque on an object is zero, then the angular momentum is conserved.
Li = Lf or Iiwi = Ifwf.
We applied the principle of conservation of linear momentum to collisions involving two bodies. Since angular momentum depends on the moment of inertia and angular speed, there are a number of interesting applications that involve only one object, but an object whose moment of inertia can change. For instance, a figure skater will begin a spin with her arms extended, then bring her arms in close to her body, seemingly spinning faster as she does so. To a reasonable approximation, the ice is frictionless, and her angular momentum is conserved. The moment of inertia of the skater with her arms extended is larger than with her arms close to her body. To conserve angular momentum, her angular speed must increase when she brings her arms in, so she doesn't seem to spin faster, she does spin faster!
Demonstration with rotating turntable.
The Physics 2130 instructor is standing on a turntable holding two 3.0 kg masses. When his arms are extended, the masses are 1.0 m from the axis of rotation, and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the instructor and turntable is 3.0 kg m² and is assumed to be constant. the instructor then pulls the masses to 0.30 m from the axis of rotation. (a) Find the new angular speed of the instructor. (b) Find the kinetic energy of the instructor before and after the masses are pulled in.
The initial moment of inertia of the two masses is Imi = 2mr2 = 2(3.0 kg)(1.0 m)2 = 6.0 kg m². The final moment of inertia of the two masses is Imf = 2(3.0 kg)(0.30 m)2 = 0.54 kg m². Including the moment of inertia of the instructor and turntable, the initial moment of inertia is Ii = 9.0 kg m², and the final moment of inertia is If = 3.5 kg m².
(a) Using conservation of momentum, Iiwi = Ifwf, or
wf = w iIi/If = (0.75 rad/s)(9.0 kg m²)/(3.5 kg m²) = 1.9 rad/s.
(b) All kinetic energies are purely rotational.
KEi = ½Iiwi2 = ½(9.0 kg m²)(0.75 rad/s)2 = 2.5 J
KEf = ½Ifwf2 = ½(3.5 kg m²)(1.9 rad/s)2 = 6.3 J
Notice that the kinetic energy increased! Work (negative) was done by the centripetal force when the masses were pulled in, resulting in the increase in kinetic energy.
I will be covering primarily sections 9.3 to 9.6.
You are all familiar with the 3 common (on earth) states of matter: solid, liquid, and gas. Physicists recognize that there is a fourth, and just recently, a fifth state of matter. The fourth is called a plasma, basically a collection of highly ionized atoms, as found in the sun and other stars. The fifth is a Bose-Einstein condensate, a state of matter only recently created in laboratories by cooling atoms to ultra cold temperatures using laser beams. For our discussions, we will concentrate on solids, liquids, and gases.
We can think of a solid as a material where the position of one atom is basically fixed with respect to another. I say basically, because if we compress or stretch a solid, the atoms will move slightly with respect to one another, that is, we say that solids are elastic. (Remember Hooke's law for springs.)
The molecules in liquids and gases are basically free to move about. From a physics perspective, the difference between a liquid and a gas is primarily the space between molecules: the space is about 10 times larger in a gas than in a liquid.
"The density of a substance of uniform composition is defined as its mass per unit volume."
Density = r = M/V, and has units of kg/m³ in SI units.
The specific gravity of a substance is the ratio of its density to the density of water at 4°C, which is 1.0x103 kg/m³
A fluid exerts a force on any object submerged in it. This force tends to compress the object, and is perpendicular to the surface of the object. We define the pressure exerted by the fluid as the force divided by the area that it acts on, P = F/A. The units of pressure are newtons per square meter (N/m²) also called pascals (Pa).
To determine pressure, we need to know the force, and the area that this force is applied to. This idea explains the sharp edges of knives, and the sharp point of an ice pick or a push pin (to concentrate force and produce a large pressure) and the idea behind snow shoes (to spread out force and produce a low pressure). The pressure inside a tire produces a force on the road equal to the weight supported by the tire.
How much pressure is applied to snow by a 700 N man standing in boots with an area of 0.060 m² (2 boots of 10 by 30 cm), or using snowshoes with an area of 0.20 m² (2 shoes of 20 by 50 cm).
In boots the pressure is P=F/A = 700 N/0.060 m² = 12,000 Pa, and in snow shoes P = 700 N/0.20 m² = 3,500 Pa.