Physics Leg Flashcards
Under mechanicalproperties of matter what is stress ,strain and Young’s modulus and modulus
State their formulas and sub units
Elastic modulus is used in describing the elastic properties of solids undergoing small deformations.
Suppose a stationary bar is subjected to an axial load. This can be achieved, for example, by fixing one end of the bar and applying a longitudinal force to the other or by applying equal and opposite longitudinal forces to the bar true or false
Stress = ratio of force to cross-sectional area
Cross sectional area- pi r squared. R in m so if it’s in mm,change it to m and if you get diameter instead of radius ,divide the diameter into two because the diameter is twice the length of the radius.
–Stress has dimensions of pressure
–Unit Pa or N m-2
Stress
The stress in the bar is defined as the ratio of force to cross-sectional area. Since the force is perpendicular to the cross-section of the bar, it is designated on the slide with a ┴ subscript. The stress is tensile if the load stretches the bar; it is compressive if the load squeezes the bar. Stress has dimensions of pressure; it is given in units of pascals (Pa) or N m-2.
So As expected by the units, stress is given by dividing the force by the area of its generation, and since this area (“A”) is either sectional or axial, the basic stress formula is “σ = F ┴/A”.
•Strain = ratio of elongation to initial length
It has no unit
–Strain is dimensionless (pure number)
•For small deformations, stress is proportional to strain
Strain
The strain in the bar is the ratio of elongation to length of the bar. For small deformations, strain is to an acceptable degree of accuracy, the ratio of change in size, to original size. Strain, being a ratio of two quantities of the same dimension, is a pure number. For example, tensile strain is ratio of change in length, (Dl = l - l0(lnaught) ) to initial length l0(lnaught); therefore e = Dl/l0 has no units.
Change in length or new length produced : the full length of the wire after the stretch - the initial length of the wire before the stretch
So if the full length is 6 and at first the length was 4, the change in length is 2. So 6-4.
But if they don’t say anything about the full length before the stretch and all they say is the initial length and now it’s been stretched or extended to a new figure then the change in length is the new figure after the stretch
Example , a force is applied to a 4m wire and this produces an extension of 0.24mm
So intial length is 4m and the change in length is 0.24 x 10 to the power -3.
So strain is 0.24 (x 10 to the power -3 )m divided by 4m
Basic strain formula: ε = ΔL/Lnaught (the initial length)
•Modulus = ratio of stress to strain
•For tensile and compressive loads, the modulus is Young’s modulus E
–E is characteristic of a given material
. Modulus
For small deformations, stress is proportional to strain. The ratio of stress to strain is called modulus.
Elastic moduli are given in units of Pa or N m-2. A material is characterized by its elastic moduli: Young’s, bulk, and shear moduli. We shall see that any two of these can be used to calculate the third.
Young’s Modulus
For tensile and compressive stresses, the modulus is Young’s modulus.
How easy it is to stretch a material
Young’s modulus is characteristic of a given material. See Table 11.1 in Young (not the same person after whom we have Young’s modulus) and Freedman for approximate values of Young’s modulus for several materials.
Young’s modulus equation is E = tensile stress/tensile strain = (FL) / (Axchange in L)
Or E= σ/ε
Questions:
Distinguish between tensile ,compressive and shearing loads
Relate the three moduli of elasticity(stress,strain,Young’s modulus)
Solve this question:
A composite wire of 3mm uniform diameter consisting of a copper wire of length 2.2m and steel wire of length 1.6m stretches under a load by 0.700mm. CALCUKATE the load
Take Young’s modulus for copper to be 1.1 x10 to the power 11 N/m raised to the power 2 and for steel its 2.0 x 10 to the power 11 N/m raised to the power 2
If they say load or tension,they still mean force.
Example 2:
A copper wire LM is fused at one end M to an iron wire MN
The copper wire has a length of 0.900m and cross section of 0.90x10 to the power -6 msquared
Iron wire has length 1.400m and cross secretion 1.30x 10 to the power -6 msquared . The compound wire is stretch; it’s total length increases by 0.0100m
Find the tension applied the compound wire,extension of each wire and the ratio of the extensions of the two wires
YM of copper is 1.30x10 to the power 11 Pa and YM of iron = 2.0 x 10 to the power 11 Pa
Solution:
A composite wire is a wire is a wire made up of two different materials
Both wires don’t have to be of the same cross sectional area.
So change in length of the composite wire is the change in length in the first wire + the change in length of the second wire
Young’s modulus = stress / strain
= FL/Ax change in L
Change in L= FL / AxYoungs modulus
Intial Length of copper = 2.2m
Area of copper is= pi x r squared
r=1.5x(10 to the power -3)m
YM for copper is 1.1 x10 to the power 11 N/m raised to the power 2
Intial Length of steel:1.6m
Area of steel: pi x r squared
r=1.5x(10 to the power -3)m
Same for copper cuz the diameter was uniform.
YM for steel: 2.0 x 10 to the power 11 N/m raised to the power 2
Chnage in L = change in L for copper + change in L for steel.
Chnage in L=0.7 (x 10 raised to the power -3 )m
0.7 x 10 raised to the power -3 )m= change in L for copper + change in L for steel.
Change in L for copper =FL / AxYoungs modulus
= F x 2.2m/ (pi x r squared
r=1.5x(10 to the power -3)m ) x 1.1 x10 to the power 11
Change in length for copper is 0.000002829F
Chnage in length for steel: F x 1.6m/ (pi x r squared
r=1.5x(10 to the power -3)m ) x 2.0 x10 to the power 11
Chnage in length for steel = 0.000001132F
0.7 x 10 raised to the power -3 )m= 0.000002829F + 0.000001132F
0.7 x 10 raised to the power -3 )m= 0.000003961F
F= 0.7 x 10 raised to the power -3 )m/ 0.000003961
The load is F= 176.733 or 177N
Answers for second problem:
Tension is 780N, extension for copper is 0.006m and for iron is 0.004m
The ratio of the extensions of copper to iron is 1:5
What is bulk stress,strain and modulus
State their formulas
Bulk modulus is used to measure how incompressible a solid is. Besides, the more the value of K for a material, the higher is its nature to be incompressible. For example, the value of K for steel is 1.6×1011 N/m2 and the value of K for glass is 4×1010N/m2. Here, K for steel is more than three times the value of K for glass. This implies that glass is more compressible than steel.
True or false
Solve this question
An aluminum (B = 70x10to the power 9 N/m to the power 2) ball with a radius of 0.5m falls to the bottom of the sea where the pressure is 150 atm. a. What is the original volume of the ball? b. What’s the Chnage in volume? c. What’s the new volume of the ball ?
d. What is the bulk stress on the ball?
e. Calculate the bulk strain on the ball
A body immersed in a fluid experiences pressure over its surface thus decreasing its volume as compared to its volume before it was immersed in the fluid therefore bulks modulus is negative because the volume of the object decreases in the fluid
Chnage jn volume = -1/B multiplied by (F/A ) x Initial volume
For fluids,pressure is the equivalent of stress since both is force /area
Therefore Chnage in volume = -pressure x initial volume divided by B
Bulk modulus (B)= stress /strain = P/(-Chnage in V/intial volume) There is a force everywhere perpendicular to surface of the body A change in pressure causes a change in volume Increase in pressure decreases volume Bulk stress = change in pressure Volume strain = fractional change in volume Negative because increase in pressure decreases volume Bulk modulus B = ratio of bulk stress to bulk strain B is positive: increase in pressure decreases volume (negative change); decrease in pressure (negative change) increases volume (positive change) Compressibility k = reciprocal of bulk modulus body immersed in a fluid is subject to forces perpendicular to its surface. These forces act inward on the surface, tending to squeeze the body and causing a deformation. Since the forces act on all sides, the deformation occurs on all sides too, resulting in a change in volume. The force per unit area is just the fluid pressure. It is well known that pressure varies with depth; however, for a small body, the pressure is nearly uniform over the surface of the body. A change in pressure is accompanied by a change in volume. Bulk stress designates the change in pressure; bulk strain is the ratio of change in volume to initial volume. Note that an increase in pressure is accompanied by a decrease in volume whereas a decrease in pressure is accompanied by an increase in volume; therefore, bulk stress and bulk strain always have opposite signs. Bulk modulus is the ratio of bulk stress to bulk strain and is the appropriate elastic modulus for materials undergoing small bulk deformations. Note the negative sign in the definition. Bulk modulus is a positive number; since bulk stress and bulk strain have opposite signs, a negative sign is needed in the definition. The reciprocal of bulk modulus is the compressibility of the material. Materials which are easily compressed have high compressibility and low bulk modulus; conversely, relatively incompressible materials have high bulk modulus.
bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.
Bulk modulus: bulk stress(change in pressure)/bulk strain(change in vol)
B = -(ΔP /(ΔV/Vnaught))
K=1/-B= therefore K =-1/Vnaught x ΔV/ΔP
Where:
B: Bulk modulus
ΔP: change of the pressure or force applied per unit area on the material
ΔV: change of the volume of the material due to the compression
V: Initial volume of the material in the units of in the English system and N/m2 in the metric system.
Answer to problem
Explain shear stress,strain and modulus
For the mechanical properties watch videos from Jeff Siyambango on YouTube
Solve this question on shear modulus
Two parallel and opposite forces each of 4000 N are applied tangentially to the upper and lower faces or a cubical metal block 25cm on each side
Find the angle of shear,and the displacement of the upper surface relative to the lower surface
Shear modulus for the metak is 0.80 x10 to the power 11 Pa. (Fi or phi is 8x10 to the power negative something,2.0 x 10 to the power something m)
The formula for shear stress is tau = F / A, where ‘F’ is the applied force on the member, and ‘A’ is the cross-sectional area of the member.
The force moves or displaces the object on a surface from its initial distance or position.
The area parallel to the force is what is used to calculate the stress or the area seen after the object has been shifted from its original shape or distance is what is used
The force is tangential to that area
A third elastic modulus is connected with shearing loads. The block of material shown on the slide is subjected to two equal and opposite forces F|| parallel to a pair of opposite faces. These forces deform the block from the rectangular parallelepiped shown to a rhombic. Shear stress is the ratio of tangential force F|| to area A over which it acts. Deformation in this case is the displacement x of the upper relative to the lower face. Shear strain is defined as x/h, h being the initial height of the block. Note that shear strain equals the tangent of the shear angle q. If the shear angle is small as is true in small deformations, the shear strain is approximately equal to the shear angle as given in radians (not in degrees).
Shear modulus, also called modulus of rigidity, is the ratio of shear stress to shear strain.
Shear stresses occur when forces act tangential to the surface of a body
This causes displacement of one face relative to a parallel, opposite face
Shear stresses also occur in torsion
Shear stress = ratio of tangential force to area of surface
Shear strain = ratio of displacement to initial transverse height
Equals tangent of shear angle
If strain or shear is small and the change in x is also small, the angle will be very small so the tangent is approximately equal to shear angle in radians so tan theta ~ theta
Therefore Tan theta of the angle is approximately equal to Chnage in x / initial height of object
Arc length = radius multiplying the angle
So Chnage in x if the shear is very small it becomes the arc length
Therefore Chnage in x= h times the angle
So the angle= change in x/ h
Shear stress, often denoted by τ (Greek: tau), is the component of stress coplanar with a material cross section. It arises from the shear force, the component of force vector parallel to the material cross section. Normal stress, on the other hand, arises from the force vector component perpendicular to the material cross section on which it acts.
SI unit
pascal
τ = F/A
Shear modulus = ratio of shear stress to shear strain
Also called modulus of rigidity
Answer:
Shear modulus = FL/Chnage in x times A
Chnage in x= FL/ shear modulus x Area
= 4000 multiplied by 25x10 to the power -2 /0.80 x10 to the power 11 multiplied by (25x10 to the power -2 )squared ( the area of the cube that is parallel to the force will be L squared since that area on that specific part of the cube is a square shape but if you were finding the area of a cube it’ll be the area of a specific face times 6 cuz a cube has six faces so 6 times L squared)
= 2.0 x 10 to the power -7 m
This value is small so the approximations will hold
Tan phi = Chnage in x / h
= 2.0 x 10 to the power -7 m / 25x10 to the power -2
= 8x10 to the power -7 radians
This will be approximately equal to the angle phi
What is Poissons ratio
Tensile strains are positive but compressive strains are negative .
The negative sign is Poissons ratio is there in case the lateral strains are opposite in sign to the longitudinal strains we’ll get a positive Poissons ratio
Poissons rule only applies for isotopic materials which are materials that have the properties in all directions
When you apply a load to a material in one direction the material will deform in the lateral direction.
Poissons ratio tells us how much the material will deform in the lateral directions
When a tensile force is applied longitudinally,there is a change in length of a cuboid in three dimensions the height(z) width (y)and length (x)
So strain for length is Chnage in length divided by original length, for height is Chnage in height divided by original height, for width is change in width divided by original width
Also when we apply the longitudinal load it turns out that the resulting strains (strain for height and strain for width) in the lateral directions are equal. They (strain for height can be proportional and strain for width can also be proportional) proportional to the strain in the longitudinal direction(strain for length)
The ratio between the longitudinal strain and the lateral strain is a material constant
The material constant is Poissons ratio v
So v= -strain in the lateral direction divided by the strains in the longitudinal direction
A strain in one direction is accompanied by proportionate strain in a perpendicular direction
Lateral strain is always the opposite sign : the material contracts laterally when there is axial elongation
Poisson ratio: v=ratio of transverse to axial strain
Positive number - 1/2
Elastic moduli are related through v
Elongation of a material in one dimension is always accompanied by contraction in the transverse dimension. If a rectangular block is stretched as shown in the diagram, the two pairs of sides perpendicular to the direction of elongation contract, the contraction being proportional to the longitudinal elongation when deformations are small. More precisely, the transverse (or lateral) strain, defined as change in lateral size divided by initial lateral size, is proportional to the axial strain. The magnitude of the proportionality constant is Poisson’s ratio, defined as the negative ratio of transverse strain to longitudinal strain. Poisson’s ratio is therefore a positive number, its value is less than ½, and it is the same for either transverse dimension.
The bulk and shear moduli are related to Young’s modulus through Poisson’s ratio by the given expressions. Indeed, only two constants are needed to describe the mechanical properties of a material. For example, given Young’s modulus and Poisson’s ratio, the other moduli can be calculated as shown on the slide. Two moduli could be given; then Poisson’s ratio can found and the third modulus calculated.
The relations shown suggest the necessity for Poisson’s ratio to be less than ½ and greater than -1. Poisson ratios are positive and less than ½ in practice. For steel it is about 1/3
Solve these questions
• A 0.5 m length of wire is clamped at its upper end and a 20 kg mass is attached to its lower end. The wire stretches by 4.0×10-4 m. If the diameter of the wire is 2.0 ×10-3 m,
determine Young’s modulus of the material.
• When an oil specimen of initial volume 600 cm3 is subjected to a pressure increase of 3.6×106 Pa, its volume decreases by 0.45 cm3. What is the bulk modulus of the oil? What is the compressibility of the oil?
• The shear force on the rubber shoe sole of an athlete is 800 N over an area of 0.021 m2. The shear modulus of the sole is 1.8 × 107 Pa. Determine the shear strain of the sole.
• The Young modulus and Poisson ratio of an alloy are 30 GPa and 1/3 respectively. Determine
– its bulk modulus
– its shear modulus
– the transverse stress in a cube of side 10 cm when a 10 N force is applied across opposite faces
a. Young’s modulus = 78 GPa
b.Bulk modulus = 4.8 × 109 Pa
Compressibility = 2.1 × 10-10 Pa-1
c. Strain = 2.1 × 10-3
d. • B = 30 GPa
• S = 11.25 GPa
• Transverse stress = 333 Pa
Under hydrostatics what is a fluid,what is gauge pressure? What is density? What is archimedes principle
A fluid is a substance that cannot resist shear stresses
Pressure in a static fluid is transmitted equally in all directions
Gauge pressure is absolute pressure minus atmospheric pressure
Pgauge=Pabs-Patm
Density is the ratio of mass to volume
An incompressible fluid has constant density
Archimedes Principle: a floating body displaces its own weight of fluid or the buoyant force or thrust on an object is equal to the weight of the fluid displaced by the object or the weight loss is equal to the weight of th fluid the object displaces
Apparent weight = weight of object in air - thrust force
Pressure p due to head h of a static fluid is proportional to density and head
Pressure (p) = density x g x h
Density = mass /volume
Mass of displaced liquid = density x volume
Weight of displaced liquid= mass x acceleration due to gravity
Weight = mass x g = (density x volume) x g
Force = density x g x volume
When an object is under water or under another fluid it experiences a buoyant force upwards that is equal to the weight (pulling it downwards) displaced by the object
Weight(N) = mass (kg) x acceleration of gravity (ms to the power-2)
Under continuity equation, what does the principle do conservation of mass state
What is Bernoullis equation and what does it apply to
What principle is this equation derived from
Use the efficient engineer on YouTube to understand this better
Stagnation point is the point where fluid velocity is reduced to zero and stagnation pressure is the pressure measured by the meter true or false
Is calculation of flow rate part?
Continuity equations fluids flow is a mathematicla expression of the principle of conservation of mass and it states that for a steady flow the mass flow rate into the volume must equal the mass flow rate out
Mass flow rate is the movement of mass per unit time (kg/s) = density(kg/m to the power 3) x velocity (m/s) x area (m squared)
Volume flow rate = vA
v-flow velocity m/s
Area- msquared
Consider two sections of a flow tube: one section has cross-sectional area A1 and the other has area A2
If the speed of fluid is v1 in section 1 and v2 in section 2, conservation of mass requires
density 1 x v1 x A1= density 2x v2 x A2
For an incompressible fluid, density is constant so A1v1=A2v2
Area of a rectangle : width x height
Area of a sphere: 4 pi r squared
Area of a cube: 6xsquared
The equation Applies to steady flow of an inviscid, incompressible fluid
It is used to compare the flow at at two different locations or points . So if used to compare the flow of a fluid at two different locations in a pipe the equation is P (pressure) 1+ 1/2density x velocity squared) 1 + densitygh1= P (pressure) 2 + 1/2density x velocity squared) 2+ densitygh2
Since there’s no significant change in elevation between point 1 and point 2 ,the potential energy terms cancel each other out thus making the equation now look like this :
P (pressure) 1+ 1/2density multiplied by velocity squared) 1 = P (pressure)2 + 1/2density multiplied by velocity squared) 2 or P1-P2=density /2 multiplied by ( V squared 2 - V squared 1 )
Since the fluid is incompressible,the mass flow rate at point 1 and 2 is equal obeying the continuity equation
This principle is only applicable for isentropic flows (when the effects of irreversible processes such as turbulence and non adiabatic processes such as heat radiation are small and can be neglected
An increase in the speed of a fluid occurs simultaneously with a decrease in static pressure of a decrease in the fluids potential energy or for horizontal flow,pressure decreases when flow speed or velocity increases.
The energy required to increase velocity comes at the expense of the pressure energy
It is derived from the principle of conservation of energy. This states that in a steady flow the sum of all forms of energy (pressure energy,kinetic energy,potential energy) in a fluid along a streamline is the same or constant at all points on that streamline.
It can only be applied on a streamline body.
P (pressure) + 1/2density x velocity squared) + p(density)g(acceleration due to gravity)h(height or elevation of a fluid just above its reference level)= constant
This equation states that the sum of the three terms remain constant along a streamline body
The first term is static pressure which is the pressure p of the fluid
The second term is dynamic pressure which represents the fluid kinetic energy per unit volume
The last term is the hydrostatic pressure which is the pressure exerted by the fluid due to gravity
Solve this question pressure is 200kPa at point 1.Find the pressure at point 2 when an inviscid,incompressible fluid of density 800kg/m cube flows at a rate of 0.03mto the power 3/s
What is speed of efflux and Torricellis theorem?
An open tank with cross section A subscript t contains fluid of density to a depth (h) above a small hole of area A subscript h . Because the top is open,it is open to the atmosphere.
Solve this question
A sealed storage tank contains water to a height of 25m
The air above the water inside the tank has a gauge pressure of 4.5atm. How fast will water flow out of a hole that is located at the bottom the tank?
Use the organic chemistry tutor to solve this question
Using Bernoullis principle:
P (pressure) 1+ 1/2density multiplied by velocity squared) 1 = P (pressure)2 + 1/2density multiplied by velocity squared) 2 or P1-P2=density /2 multiplied by ( V squared 2 - V squared 1 )
Removing the potential energy equation, it’ll now be
P (pressure) 1+ 1/2density multiplied by velocity squared) 1 = P (pressure)2 + 1/2density multiplied by velocity squared) 2 or P1-P2=density /2 multiplied by ( V squared 2 - V squared 1 )
= 200 x 10 to the power 3Pa + 1/2 x 800kg/m cube
Look at the slide with this example on it
vefflux = square root of 2gh
h is the distance between the water level in the tank and the level where the water leaves
This formula only works if the tank is open to the atmosphere
You can also use Bernoullis principle to get the Toricellis theorem
Gauge pressure is the pressure relative to the atmospheric pressure since the atmospheric pressure at seal level is 1atm. The absolute pressure will be 4.5 higher than the atmospheric pressure so the absolute pressure inside the tank is 5.5atm
When the water leaves the tank it’s exposed to an atmospheric pressure of 1atm
Using Bernoullis principle:
P1 + densitygh1 + 1/2densityvsquared1 =P2 + densitygh2 +1/2densityvsquared2
Under thermodynamics State the equation of state of an ideal gas
Some Ideal Gas Properties
•Effusion rate of the same gas at different temperatures
rA(T1)/rA(T2)=
Effusion rate of the different gases at the same temperature
rA/rB=square root of mB/mA
pV=nRT=NkBsubscript T
What is isolated,closed,open systems
States of systems are characterized by what
Isolated systems do not exchange matter or energy with their surroundings and are bounded by rigid walls
Closed systems cannot exchange matter
Open systems can exchange energy or matter with their surroundings
States of a system are characterized by definite values of certain quantities; for example, volume, temperature, pressure
Processes:
Cyclic: system returns to initial state
Isochoric: constant volume
Isobaric: constant pressure
Adiabatic: no heat exchange
Isothermal: constant temperature
The internal energy U of systems is a function of the state of the system
It may depend on volume ,number of particles and temperature
U Changes in a Chnage of thermodynamic state
Delta U=0 in a cyclic process
The first law (fixed number of particles):change in internal energy is heat absorbed minus work done by the system
dU=dQ-dW
-heat absorbed is positive (increases U)
Work done when volume chnages is dW=PdV(system works when it expands ) so dU=dQ-PdV
Work and heat are not functions of state :they depend on path
For ideal gas:PV=nRT so W=differentiation sign of PdV = differentiation sign of nRT/V multiplied by dV
Name some ideal gas properties
Determine the work in the following cyclic process when traversed in both directions
For a gas at temperature T rms speed is proportional to square root of T/M
Recall average molecular velocity is zero(no net translation) but average speed is not zero
Average speed is proportional to rms speed
Justification of Grahams law:
Effusion rate r is proportional to speed hence inversely proportional to square root of mass
pV=nRT=NkBT
vrms =square root of 3kBT/m
vavg=square root of 8kBT/pi m
Effusion rate of the same gas at different temperatures:rA(T1)/rA(T2)= square root of T1/T2
Effusion rate of the different gases at the same temperature=
rA/rB= square root of mB/mA
The triangle is ABC
WABCA = (PB – PA)(VC – VA)/2
WACBA = (PB – PA)(VA – VC)/2 = – WABCA
What is the second law of thermodynamics
How can it be stated
What does the second law determine
1.Determine the entropy change when 1 kg of ice melts at 0 ℃. Specific latent heat of ice is 3.34×105 J/kg.
2.Determine the thermal efficiency of a Carnot engine operating between 50 ℃ and 200 ℃.
“It is impossible for any system to “It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began.” a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began.”
The second law can also be stated as dS greater than or equal to zero
In any process,entropy either increases or stays constant
Heat is related to entropy through TdS is greater than or equal to dQ
Entropy is a function of state but heat is not
For a reversible process ,TdS=dQ
The second law determines which processes can occur spontaneously
It also determines the efficiencies of engines
The most efficient engine (Carnot) operating between reservoirs at T subscript H and T subscript C has efficiency
n= T subscript H — T subscript C divided by subscript H
This is = 1 - T subscript C divided by T subscript H
Answers :
- Delta S = Q/T = 1223 J/K
- n = 1 – TC/TH = 32%
Explain circular motion
Velocity is always tangential to path true or false
For rectilinear motion,a is parallel to v and can be zero while for circular motion,a has a radial component and is non zero
Circular motion, that is, motion on a circular path, is a special case of motion on a curved path. Circular motion differs from rectilinear motion in the following respects. In rectilinear motion, acceleration is parallel or anti-parallel to velocity and can be zero as in uniform motion (that is, motion with constant velocity–constant speed and constant direction). In circular motion, acceleration is never zero, even when speed is constant, and there is always a component of acceleration perpendicular to the direction of motion; therefore, acceleration is never parallel to velocity in circular motion.
In the figure at left showing rectilinear motion, the velocity vector remains parallel to itself at all instants—direction is constant. In the case of circular motion as in the figure at right showing motion on a circle of radius R, the direction of motion is different at different instants. Acceleration is the rate of change of velocity: any change in velocity, whether in magnitude or direction or both, is brought about by acceleration. Therefore, since the direction of motion must change in such a manner that a circle is described, circular motion always involves acceleration.
In any type of motion, rectilinear, circular, or curvilinear, instantaneous velocity is tangential to path. To change the direction of curvilinear motion of a body, it must be accelerated in a direction perpendicular to its instantaneous velocity. Acceleration in circular motion therefore always has a component perpendicular to velocity.
State the types of circular motion
What is centripetal acceleration
For uniform circular motion there is constant speed,zero tangential acceleration and for non uniform motion there is non zero speed and tangential acceleration
How is uniform circular motion described
What is period
What is frequency
What is angular velocity
How can speed and centripetal acceleration be expressed
Two types of circular motion can be distinguished: uniform and non-uniform circular motion. A body in uniform circular motion moves with constant speed but in a direction that varies around the circle. In the non-uniform case, speed is not constant.
Note that, acceleration can be resolved into components parallel and perpendicular to instantaneous velocity. Changes in speed are brought about by the parallel (tangential) component atan whereas changes in direction are brought about by the perpendicular (radial) component arad (Young and Freedman, Section 3.2). Speed increases when tangential acceleration is in the same direction as velocity and decreases when the directions are opposite.
Radial acceleration is known as centripetal acceleration and is always directed towards the centre of the circle. (Centripetal comes from Greek and means “seeking the centre.”) In the resolution of the acceleration given above, the radial unit vector points away from the centre of the circle; therefore, the centripetal term must carry a negative sign so it points to the centre. The magnitude of acceleration is naturally the square root of the sum of squares of the radial and tangential components.
From the forgoing, it is clear that in uniform circular motion, tangential acceleration is zero whereas it is non-zero in non-uniform circular motion.
There are two types of circular motion: uniform and non-uniform
Resolve acceleration into tangential (atan) and radial (arad) components
Tangential accel. changes speed
Radial accel. changes direction (- sign implies toward O)
Uniform circular motion: atan = 0 hence constant speed E.g.
Vehicle rounding a constant-radius curve at constant speed
Satellite in circular orbit
Non-uniform circular motion: atan ≠ 0 E.g.
Skater on a vertical circular track
In general, magnitude of acceleration is
Square root of a squared subscript tan + a squared subscript rad
arad = v squared/R
In describing uniform circular motion, it is useful to introduce period, frequency, and angular velocity. The period T is the time for a complete trip around the circle. For example, if ten revolutions are completed in 1 s, the period is T = 1 s/10 = 0.1 s.
Frequency is the number of complete cycles in unit time: f = 1/T. The angular frequency w = 2pf is the magnitude of the angular velocity. It is given in units of radians per second. Angular velocity w is a vector directed perpendicular to the plane of the circle and oriented in such manner that if the fingers of the right hand are curled around the circle in the direction of motion, the thumb points in the direction of w.
Speed and centripetal acceleration can be expressed in terms of angular speed by noting that distance travelled in one trip around the circle is 2pR, the circumference of the circle. Speed and centripetal acceleration are then given by the expressions shown on the slide.
Note that the expression for centripetal acceleration also applies to any curvilinear motion, including non-uniform circular motion. At any point on the path of a body where the instantaneous speed is v and the local radius of curvature of the path is R, there is a component of acceleration perpendicular to the direction of motion, directed towards the centre of curvature of magnitude
arad = v2/R.
What is period,frequency
Angular frequency under circular motion
Time for one complete revolution is the period T
The number of revolutions in unit time is frequency f = 1/T
The angular frequency w = 2pf =2p/T
w is also called angular speed (unit: radians per second)
Angular velocity is a vector pointing perpendicular to the plane of the circle, in a right-handed sense
In one revolution, distance travelled is the circumference 2pR of the circle
Speed v = 2pR/T = 2pRf = Rw
Centripetal acceleration arad = v2/R = Rw2 = 4p2R/T2 = 4pi2f2R
arad= v squared /R also gives radial acceleration when speed is v at a point of radius R on any curvilinear path
