physics I exam 4
units of heat
Joules (after all it is energy!) J
temperature must be expressed in
Kelvin (K) NOT F or C
A diver tucks her body in midflight, decreasing her moment of inertia by a factor of two. By what factor does her angular speed change?
L = Iω Iω is constant they are inversely proportional *ωf/ωi = 2*
chapter 11
Torque
If the object is uniform, and we are rotating it about an edge, the center of gravity is the center of mass. If we have multiple objects, we can calculate the center of gravity like we calculate the center of mass:
Xcg = (W1x1 + W2x2 +....) / (W1 + W2 + ....)
Positive heat when
energy is transferred TO the system -obtains heat = +
system
ex: heating a room = room is the system
Example: Inside most computers is a hard disk drive. In a certain computer, the drive is made of a disk with a radius of 4.0 cm and a mass of 100 g. There is also a metal arm that hovers just over the disk to read the magnetic data on the disk. A crash occurs when the arm crashes onto the disk, quickly stopping the disk. If the drive is spinning at 7200 rpm at the moment the arm crashes, what is the work done by the arm during the crash? If the disk rotates 0.5 rad during the crash, what is the magnitude of the torque of the arm?
m = 100g ω = 7200 rev/min θ = 0.5 rad r = 4cm τ = ? ω = ? spinning = use moment of inertia instead of mass --> I = 1/2mr^2 --> I = (0.1kg)(4cm)^2 --> I = 8.0 x 10^-5 kg•m2 ω = 7200 rev/min x 1min/60sec x 2pi --> 753 rad/s W = ΔKE = KEf - KEi --> W = 0 - 1/2Iω^2 --> 1/2(8.0 x 10^-5)(753)^2 --> *W = 22.8 J* W = τθ --> τ = W/θ --> 22.8/0.5 --> τ = *45.5 N•m*
The units of torque are
newton•meters (N•m)
Avagadro's number
used for large numbers NA = 6.022137 x 10^23 mol^-1 -1 mole of any substance has the same number of molecules per mole
example: An 8.00 m ladder with a weight of 355 N leans against a smooth vertical wall leaning at an angle of 50º with the floor. A 875 N firefighter stands l1 = 6.30 m up the ladder. If the ladder is in equilibrium what forces do the wall and the ground exert on the ladder. (picture on slide 12)
∑Fx = 0, ∑Fy = 0, ∑τ = 0 (not net force and no rotation) Fx - Fw = 0 --> Fx = Fw Fy - (Wf + WL) = 0 --> Fy = (875 + 355) --> *Fy = 1230 N* τ = Fd --> τ = -WL(L/2)cosθ - Wf(li)cosθ + Fw(L)sinθ --> 0 = -355(4)cos50 - 875(6.3)cos50 + Fw(8)sin50 --> *Fw = 727 N* Fx = Fw --> *Fx = 727 N* -WL is negative because it is clockwise -chosen axis of rotation is Fy and Fx intersection (Fy/Fx = 0)
Iω is the rotational momentum of the object, so we can rewrite Newton's 2nd Law:
∑τ = IΔω/Δt = ΔL/Δt
Application: Bimetallic strip
-Two metal strips with differing values of α -Strip bends since one metal expands more than the other
To determine the location of her center of mass, a physics student lies on a lightweight plank supported by two scales 2.50 m apart, as indicated in the figure (Figure 1). Assume the left scale reads 300 N , and the right scale reads 132 N 1.) Find the student's mass. 2.) Find the distance from the student's head to her center of mass.
1.) F1 + F2 = mg --> 300 + 132 = m(9.8) --> *m = 44.1 kg* 2.) ∑τ lefthand = x2F2 - xcgF --> x2F2 - xcgF = 0 --> (2.5)(132) - xcg (44.1 x 9.81) = 0 --> *xcg = 0.764 m*
0 C = ___F
32
A person slowly lowers a 3.9 kg crab trap over the side of a dock, as shown in the figure (0.70 m radius) What torque does the trap exert about the person's shoulder?
F = ma --> F = 3.9 * 9.8 --> F = 38.22 N τ = rF --> τ = (0.70)(38.22) --> τ = 26.754 N⋅m
Instead of talking about the number of particles, we will often talk about the number of moles:
Number of moles n = N/NA Number of moles = total number of particles / how many particles there are per mole
Thermodynamic variables:
Pressure P Volume V Temperature T Number of molecules N
Chapter 17:
Some preliminary notions
What about holes? Say we have a 10 m by 10 m metal sheet with a 50 cm diameter hole in the middle. If the entire sheet of metal is heated, will the hole get bigger or smaller?
The entire sheet expands, so the hole must also get bigger! -hole also expands with entire structure
if work done is negative, then KE (friction)
decreases
Historically important unit: calorie 1 calorie =
energy required to raise the temperature of 1 g of water by 1 ºC= 4.186 J 1 calorie = 4.186 Joules
To convert any temperature scale you can measure the
freezing point and boiling point of water.
if work done is positive, then KE
increases
gas which is diluted means
molecules don't interact = ignore interactions -ex; atmospheric pressure is an ideal gas
slide 4: moment arm
moment arm is the radius --> -force extends a certain angle above or below the moment arm -r⟂ = rsinθ
The direction that we apply a force to a lever arm matters. If you want to open a door, you want to push (or pull)
perpendicular to the door
Thermal Expansion: consequence of temperature change
slide 4: -L0 (left to right) and T (up and down) -L0 + ΔL and T + ΔT -ΔL = αL0ΔT -α is the coefficient of linear expansion -change in length = coefficient of linear expansion x initial length x change in temperature -when heat increases, things expand -thermal expansion = α
Rotational motion of solids
translational motion: -position: x -velocity: v = Δx/Δt -acceleration: a = Δv/Δt -mass (inertia): m -Newton's 2nd law: Fnet = ma -work: W = Fd -KE: K = 1/2mv^2 -power: P = Fv -work-energy: W = ΔK -momentum: p = mv rotational motion: -position: θ -velocity: ω = Δθ/Δt -acceleration: α = Δω/Δt -mass (inertia): I -Newton's 2nd law: τnet = Iα -work: W = τθ -KE: K = 1/2Iω^2 -power: P = τω -work-energy: Wrot = ΔKrot -momentum: L = Iω
We have covered the description of motion of rotating objects. Now we need to consider how forces act on objects to bring about rotation.
want to rotate something? It matters where you apply the force
If you apply a force along a door (towards or away from the center of rotation), the door
won't turn
review:
-larger torque = easier to push -distance from the axis = r (farther away from the axis = easier) -parallel with the door is 180º, want to push it perpendicular -ex: larger screw driver or wrench will be easier (force at a larger distance)
The axis of rotation does not have to pass through the center of mass.
-rod, length L has an end axis: I = 1/3mR^2 -sphere, radius R has a tangent axis: I = 7/5mR^2 -rectangular sheet, Length L has a central axis: I = 1/12mL^2 -rectangular sheet, Length L has an edge axis: I = 1/3mL^2 -axis = center of mass -tables given on exam if needed
Jogger 1 in the figure (Figure 1) has a mass of 66.3 kg and runs in a straight line with a speed of 3.35 m/s. 1.) What is the magnitude of the jogger's linear momentum? 2.) What is the magnitude of the jogger's angular momentum with respect to the origin, O? (radius = 5.00 m)
1.) p = mv --> p = 66.3 x 3.35 --> *p = 222.105 kg⋅m/s* 2.) L = r⟂mv --> L = (5)(66.3)(3.35) --> *L = 1110.525 kg⋅m/s*
A child of mass m stands at rest near the rim of a stationary merry-go-round of radius R and moment of inertia I. The child now begins to walk around the circumference of the merry-go-round with a tangential speed v with respect to the merry-go-round's surface. 1.) What is the child's speed with respect to the ground? 2.) Check your result in the limits I→0. 3.) Check your result in the limits I→∞. Express your answer in terms of some or all of the variables m, R, I, and v.
1.) vg = Iv/(I + R^2 m) 2.) vg --> 0 3.) vg --> v
The L-shaped object in the figure (Figure 1) consists of three masses connected by light rods (top is 9kg, corner is 1.2 kg, and bottom is 2.5 kg) (1 m between the top and the corner and 2 m between the corner and the bottom) 1.) What torque must be applied to this object to give it an angular acceleration of 1.20 rad/s^2 if it is rotated about the x axis? 2.) If it is rotated about the y axis? 3.) If it is rotated about the z axis (which is through the origin and perpendicular to the plane of the figure)?
1.) ∑τ = Iα --> ∑τ = (9 x 1^2)(1.2) --> *∑τ = 10.8 N⋅m* 2.) ∑τ = Iα --> ∑τ = (2.5 x 2^2)(1.2) --> *∑τ = 12 N⋅m* 3.) ∑I = (9 x 1^2) + (2.5 x 2^2) --> ∑I = 19 ∑τ = Iα --> ∑τ = (19)(1.2) --> *∑τ = 22.8 N⋅m*
KE =
1/2Iω^2
nutritionists have a "dietary calorie" equal to
1000 calories food we eat, calories on packages are given in kilo-calories
Tb = ________ K
373.15
Newton's 2nd Law for Rotation
Consider a torque from a force acting at a right angle a distance r from a point mass. τnet = rFtnet = mar = m(rα)r = (mr^2)α net torque = r x net tangential force = mass x tangential acceleration x radius = ....
A torque of 0.97 N⋅m is applied to a bicycle wheel of radius 40 cm and mass 0.65 kg Treating the wheel as a hoop, find its angular acceleration
I = mR^2 I = (0.65)(0.4^2) I = 0.104 ∑τ = Iα 0.97 = (0.104)α *α = 9.3 rad/s^2*
If an object is a point mass, we can define the multiplier of the rotational acceleration as the moment of inertia
I = mr^2
100 C = _____ F
212
Tf = _______ K
273.15
Rotational inertia
If we have many masses, we can calculate the total rotational inertia by adding the rotational inertia of each of the masses. I = ∑m1r1^2 + m2r2^2 + m3r3^2 ....
Chapter 18
Law of Thermodynamics
Proof: consider a sphere
V=4/3πR3 and dV = 4πR2dR If we divide the two, we get: dV/V = 3 dR/R or β = 3α It does not matter what we call our length variable, so the coefficient of volume expansion is 3 times the coefficient of linear expansion. We can also derive this for cubes.
Some useful constants
We need the ability to talk about systems of many particles. Avagadro's number mols
Equilibrium
When we discussed forces in Chapter 4 we said that one condition for equilibrium is that there is no net force acting on an object. We also need to consider rotation. For an object to be in total equilibrium there must be no net force AND no net torque. ∑Fx = 0 ∑Fy = 0 ∑τ = 0 -equilibrium = net force is zero, velocity is constant, and acceleration is zero (in a linear situation) -torque is zero in equilibrium (rotational)
We already know how to find kinetic energy, but we need an expression for the work done on a rotational system.
Wnet-rot = Fnet s = Fnet rθ = τθ s = rθ
The work-energy theorem for rotational motion states that the work done by a rotation equals the change in rotational kinetic energy.
Wnet-rot = ΔK = K2 - K1 = 1/2Iω2^2 - 1/2Iω1^2
Negative torque creates a
clockwise rotational acceleration
If the net work done by external non-conservative forces is 0, then the mechanical energy is
conserved! Ef - E0 = 0 = ΔKE + ΔKErot + ΔPE
Positive torque creates a
counterclockwise rotational acceleration
negative heat when
energy is transferred FROM the system -gives away heat = -
Latent heat (L) of a substance:
heat required to change the phase of the substance (e.g. solid to liquid, liquid to vapor) Q = m L heat = mass x latent heat
Heat capacity (C) of a system:
heat required to raise the temperature of the system by 1 ºC Q = C ΔT heat = heat capacity x change in temperature
Specific heat (c) of a substance:
heat required to raise the temperature of unit mass of the substance by 1 ºC Q = m c ΔT heat = mass x specific heat x change in temperature
If I have an object on a pivot, it matters if I apply a force far away from a pivot or really close to the pivot. To open and close a door, you apply a force
perpendicular to the door. -Rotation depends on the magnitude, orientation, and location of a force applied to an object. We call this phenomenon the *torque* -torque is a vector (can be positive or negative)
To prepare homemade ice cream, a crank must be turned with a torque of 3.80 N⋅m . How much work is required for each complete turn of the crank?
power = τ x ω ω = 2pi/T power x T = 3.8 x 2pi W = 3.8 x 2 pi *W = 23.9 J*
Suppose we allow the gas to change volume in a way that the system is always in equilibrium ("quasi-static process") and the piston exerts a constant pressure P. How much work W is done BY the gas?
slide 20 picture: gas is in the blue and piston is green, this is our system
Result2: Maxwell Speed Distribution
temperature at 80 K and temperature at 300 K
The center of gravity is
the point of an object where the weight will appear to act on a rotating object for calculating a torque.
moment of inertia has a different equation depending on
the shape of the object (rod, sphere, etc.)
In rotations, the power is still defined as
the time rate of change of the work. Since the work is the torque acting over a rotational displacement, the power is the torque acting at a rotational velocity. P = ΔW/Δt = τ(θΔ/Δt) = τω
The magnitude of the force that creates a torque is
the magnitude of a force perpendicular to the rotation.
Result1:Pressure & RMS Speed
-Consider ONE molecule that collides with one wall of a box as shown (yz plane); -Calculate the change in momentum -- only the x-component changes = -2mvx -Calculate the RATE of momentum change = FORCE; Pressure = force per unit area. -Sum such contributions for ALL molecules -Finally, use the random motion of molecules to argue that the average of the square of one component of velocity = (1/3) average of square of SPEED. vRMS = root mean square speed P = N/3 x mv^-2/L^3 = N/3 x mv^2 rms / V PV = 2/3 N (1/2mv^2 rms)
The Kinetic Theory of Gases
-Identical molecules in random motion -Molecules obey Newton's laws of motion -Molecules make ELASTIC collisions with each other and with walls of container -No force on molecules except during collision -Number of molecules N is large but NVm << V, where Vm = volume occupied by each molecule and V = volume of container What are the principal results of this model on slide 23? -
The 0th Law of Thermodynamics:
-If we place two object in contact and leave them for a long time, they reach "thermal equilibrium". -If we then measure the temperature of each of these objects, we must get the same reading. -ex: stick an ice cube in room temperature water = eventually their temperature will be the same (ice will melt, water will cool) = eventually they will reach equilibrium (room temperature)
Some common sense notions:
-There exists such a property as "temperature" that tells us whether an object is "hot" or "cold". -If you mix "hot" and "cold" you get "warm". -Many physical properties depend upon temperature: volume, pressure, "phase" (solid, liquid, gas), length, electrical resistance, etc. -Temperature has something to do with energy. (temperature is energy) Let's quantify these notions! -volume, pressure, phase (solid, liquid, gas), length, electrical resistance, etc. depend on temperature and react to temperature
Finding rotational inertia: Most of the time, in the real world, the moment of inertia has been calculated for most shapes. If we want to find the moment of inertia, we will look it up in a table.
-hoop, radius R has a central axis: I = mR^2 -disk, radius R has a central axis: I = 1/2mR^2 -sphere, radius R has a diameter axis: I = 2/5mR^2 -rod, Length L has a central axis: I = 1/12mL^2 -sphere shell, radius R has a diameter axis: I = 2/3mR^2 -axis = center of mass -tables given on exam if needed
The rotating systems shown in the figure(Figure 1) differ only in that the two spherical movable masses are positioned either far from the axis of rotation (left), or near the axis of rotation (right). 1.) Is the tension in the string on the left-hand rotating system greater than, less than, or equal to the tension in the string on the right-hand rotating system? 2.) What is the best explanation
1.) greater than 2.) The mass in the right-hand system has the greater downward acceleration.
Two spheres of equal mass and radius are rolling across the floor with the same speed. Sphere 1 is a uniform solid; sphere 2 is hollow. 1.) Is the work required to stop sphere 1 greater than, less than, or equal to the work required to stop sphere 2? 2.) Choose the best explanation
1.) less than 2.) Sphere 2 has the greater moment of inertia and hence the greater rotational kinetic energy. I = mr^2 KE = Iω^2
A centimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. (Figure 1) One coin, of mass M1 = 10g, is placed at the zero mark; the other, of unknown mass M2, is placed at the 4.7 cm mark. The center of the ruler is at the 3.0 cm mark. The ruler is in equilibrium; it is perfectly balanced. 1.) Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot? 2.) Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. Why not? 3.) Find the mass M2.
1.) yes; no (Note that the weight of the ruler itself also does not exert a torque with respect to the pivot point since the ruler is uniform and is pivoted at its midpoint.) 2.) Because the distance from the exerted force to the pivot is zero 3.) M1R1 = M2R2 --> (10)(3) = M2(1.7) --> *M2 = 17.6 g*
A torque of 0.12 N⋅m is applied to an egg beater. 1.) If the egg beater starts at rest, what is its angular momentum after 0.55 s ? 2.) If the moment of inertia of the egg beater is 2.2×10^−3 kg⋅m^2 , what is its angular speed after 0.55 s ?
1.) ∑τ = ΔL/Δt --> 0.12 = ΔL/0.55 --> *ΔL = 0.066 kg⋅m^2/s* 2.) ∑τ = Iω/Δt --> 0.12 = (2.2×10^−3)ω / 0.55 --> *ω = 30 rad/s* or α = 0.12/2.2×10^−3 --> α = 54.54 ω = α x t --> ω = (54.54)(0.55) --> *ω = 30 rad/s*
To tighten a spark plug, it is recommended that a torque of 20 N⋅m be applied. If a mechanic tightens the spark plug with a wrench that is 27 cm long, what is the minimum force necessary to create the desired torque?
27 cm = 0.27 m τ = rF 20 = 0.27(F) F = 74.07 N
What is heat?
A change in the energy of a system as a result of a temperature difference between the system and its environment. The energy is from motion of the system that does not move the center of mass. We use the symbol *Q* to denote heat. not a vector, but has a sign (positive or negative)
Thermal Expansion: Isotropic Expansion Example Aluminum washer has an hole with diameter = 0.5 cm at 20 0C. If the washer is heated to 100 0C, what is the new diameter of the hole? (Aluminum α = 23 x 10-6/ 0C)
Expansion (usually, but not always) occurs isotropically -- every length on the object changes in proportion Dnew - Dold =α Dold ΔT --> Dnew = Dold + α Dold ΔT --> Dnew = 0.5 cm + (23 x 10^-6/ ºC) (0.5 cm) (80 ºC) --> *Dnew = 0.50092 cm* -d is diameter -isotropic means it happens in the same way in all directions
example in class: have a plank that is not uniform (center of mass isn't in the middle), where the center of mass m is towards the right side (means the right side is heavier). You apply 2 forces upward on the plank. You want this plank in equilibrium (straight across, not moving). What are the two forces?
F1 = ?, F2 = ?, mass = m, length = L ∑F = 0 ∑τ = 0 F1 + F2 - mg = 0 --> F1 + F2 = mg (means that the 2 forces when pulled up = weight mg) ∑τ = 0 --> (since F2 is on the last 1/4 side of the plank from the center of mass) F1(3/4L) + F2(1/4L) = 0 F2(L/4) = (3L/4) F1 = 0 --> *F2 = 3F1* F2 = 3F1 and F1 + F2 = mg --> F1 + 3F1 = mg --> 4F1 = mg --> *F1 = mg/4*
The ideal gas law
For a system containing a dilute gas of particles, the thermodynamic variables P, V, and T are found to obey the relation: PV = NkBT PV = nRT -R = universal gas constant = 8.31 J/mol.K -kB = Boltzmann constant = 1.38 x 10^-23 J/K pressure x volume = number of particles x Boltzmann constant x temperature if in moles .... PV = number of moles x universal gas constant x temperature
Temperature Scales: Kelvin
Fundamental scale: Kelvin (K) -Lowest possible temperature: 0 K (absolute zero) -Lowest lab temp. (BEC): ~1 nK -Liquid helium: 4.2 K -Liquid nitrogen: 77 K -Room temp: ~300 K -Solar surface: ~3000 K -Helium fusion: ~108 K -Nucleus-nucleus collisions: ~1012 K -The Big Bang: "When there was light": 1039 K
Total KE of gas =
Internal energy U = 3/2NkT *Note: so far, we're only considering TRANSLATIONAL kinetic energy. But the statement above is more general* -internal energy = 3/2 x number of molecules x boltzmann constant x temperature -temperature increases = internal energy increases -n = N (number of molecules) / NA (number of molecules/mole) --> n = [ ] moles
What about Volume Expansion?
It turns out for an isotropic medium we can find the coefficient of volume expansion from the coefficient of linear expansion, i.e. β = 3α.
Average KE of a molecule
KE = 1/2mv^2 = 3/2 (PV/N) = 3/2 kT
Consider a point particle in uniform circular motion:
L = (mr^2)ω = rm(rω) = rmv = rp L = rp -The angular momentum is equal to the radius of the circle of motion times the linear momentum.
If an object with rotational inertia I is rotating with an angular velocity ω, the object has an angular momentum given as:
L = Iω *ω must be expressed in units of radians per second! The units of angular momentum are kg m2/s* -linear momentum was p = mv, so angular momentum is L = Iω -I = mr^2 -ω = r/t
If the particle is traveling at an angle other than 90º from the radius, the angular momentum is given as:
L = rp sinθ = rmv sinθ -if 90º, then just rp = L -look at picture on slide 17 to find θ on circle
Example: Thermal equilibrium An ice cube of mass 10 g at 0 ºC is dropped into a glass containing 250 g of water at 20 ºC. Describe the final equilibrium state of the system. Latent heat of fusion of ice = 80 cal/g; specific heat of water = 1 cal/gºC. (Neglect the heat capacity of the glass.)
Possible scenarios: -All the ice melts; 0 ºC <= Tfinal < 20 ºC -Some ice melts; Tfinal = 0 ºC -All the water freezes; Tfinal = 0 ºC -What is Tfinal? higher temperature has negative heat = heat exchange = lower temperature has positive heat (so set equal to each other) mice Lf +mice cH2O (Tf - 273.15 K) = mH2O cH2O (293.15K - Tf) --> 10g(80cal/g) +10g(1cal/gK)(Tf-273.15K)=250g(1cal/gK) (293.15K-Tf) --> Tf*(250+ 10) = 250*293.15 +10*273.15 - 800 --> *Tf = 289.30 K = 16.15 ºC* -Lf = latent heat of fusion = when ice melts -cH2O = heat required to raise the temperature of the ice from 0ºC to the final temperature (specific heat of H2O)
celsius scale
TC = TK - 273.15
Fahrenheit scale
TF = 9/5 TC + 32
Chapter 16 and 17
Temperature & Heat
We can write Newton's 2nd Law for rotations:
∑τ = Iα -linear: ∑F = ma -for rotation: ∑τ = Iα -no net torque (zero) means zero acceleration
We can write Newton's 2nd Law for rotations:
∑τ = Iα = I (Δω/Δt)
Some preliminary notions
1.) System of interest: an ideal gas in a container 2.) Walls of container may or may not conduct heat 3.) One of the walls may be moveable: it is a piston that exerts a force
similarly: coefficients of area and volume expansion
ΔA = γA0ΔT ΔV = βV0ΔT A is area and V is volume
The net torque changes the rotational momentum of the system. What if we have no net torque?
ΔL/Δt = 0 means Lf = Li (because Lf - Li = 0)
If the force is applied at some angle, the moment arm effectively shortens. The torque on the object is given as:
τ = (rsinθ)F = r⟂F -θ is the angle between the force and the radius of the circle of rotation. -Remember: torque has a magnitude and a direction!
Example: The length of a bicycle pedal arm is 0.152m and a downward force of 111 N is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arms pivot when the arm makes an angle of a) 30º b) 90º and c) 180º with the vertical?
τ = r F sinθ a.) τ = (0.152m)(111 N)(sin 30º) = 8.4 N m b.) τ = (0.152m)(111 N)(sin 90º) = 16.9 N m c.) τ = (0.152m)(111 N)(sin 180º) = 0
Say we apply a force F, some distance r away from the center of mass, perpendicular to the moment arm. The magnitude of the torque is given as:
τ = rF -r = how far out from the center (radius)
Conservation of rotational momentum Our change in rotational momentum is equal to the net external torque. Say we have no external torque:
τnet = 0 = Iα = I(ωf-ωi/t) Li = Lf Iiωi= Ifωf rotational momentum is conserved!
