PHY 115 Units and Dimensions 2019 PDF
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University of Ilorin
2019
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This document is a past paper for a physics course, likely PHY 115, at the University of Ilorin. It covers topics in units and dimensions, including fundamental and derived quantities, offering examples and problems for calculation. The document features formulas and calculations related to the physics subject of units and dimensions.
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Physics Department, University of Ilorin PHY 115 UNITS AND DIMENSION Observation and measurement of physical quantities / parameters are the core of scientific activities. Measurement is the act or process of assigning size, or value to a physical quantity. Quantities are measured when we are able...
Physics Department, University of Ilorin PHY 115 UNITS AND DIMENSION Observation and measurement of physical quantities / parameters are the core of scientific activities. Measurement is the act or process of assigning size, or value to a physical quantity. Quantities are measured when we are able to quantify it by assigning value and unit to such quantities or parameters. This is done by the use of measuring equipment- which are mostly based on one physical law or the other. Quantities are categorised into two namely: fundamental quantities and derived quantities (likewise the Units- fundamental units and derived units). Fundamental quantities are quantities upon which other quantities are based while derived quantities are quantities that are obtained from fundamental quantities. Fundamental quantities, units and dimensions Quantity Unit Dimension Length Meter(m) L Time Second(s) T Mass Kilogramme(kg) M Temperature Degree Kelvin(oK) Ζ or K Electric current Ampere(A) I or A Electric charge Coulomb(C) Q Mole(Amount of substance) Mole(mol) N Luminous intensity Candela(cd) J or C The underlined dimensions are the one adopted for this course. Derived quantities, Units and Dimension Quantity Equation Unit Dimension Velocity πππ πππππππππ‘ π π πΏπβ1 π‘πππ Acceleration π£ππππππ‘π¦ π π 2 πΏπβ2 π‘πππ Momentum πππ π Γ π£ππππππ‘π¦ πππ π ππΏπβ1 Force πππ π Γ πππππππππ‘πππ πππ π 2 ππΏπβ2 Example π1 π2 Electrostatic Force = πΉ = π π2 1 F=[MLT -2]=[ππ2 πΏβ2 ] This implies: [ππ2 ] = [ππΏ3 πβ2 ] [k] = ππΏ3 πβ2 πβ2 Work done = πΉ Γ π = ππ = ππΏ2 πβ2 The relationship between Work done and voltage due to separation of charges is ππππ ππππ = ππ , but the dimension of work is [ππΏ2 πβ2 ] Therefore, [ππ] = [ππΏ2 πβ2 ] Implying [V] = [ππΏ2 πβ2 πβ1 ]. The quantity of heat energy transfer π = πππ, obtain the dimension of the specific heat capacity (c). (Please note that Q is used as the dimension of charge above, quantity of heat is also represented with symbol Q - but not the same as the dimension of charge) The quantity of heat is in Joules β dimension of Q, [π] = [ππΏ2 πβ2 ] Using π = πππ, implies Q ο½ [πππΎ] = [ππΏ2 πβ2 ], or ππΏ2 πβ2 [c] = = πΏ2 πβ2 πΎβ1 ππΎ Optics πππ π Refractive index π = πππ π is dimensionless. Wavelength ο¬ [L] has the dimension of length and the unit is meter. Questions πΊππ 1. The gravitational force F is given by πΉ = , obtain the π 2 dimension of G. 2. Acceleration due to gravity (π), below the earth surface, is given ππΊπ by π = , obtain the dimension of b. π3 2 3. The force a current carrying wire will experience when placed in a magnetic field of magnitude B is FB ο½ BIL , where I is the current passing through the wire and L is the length of the wire that is inside the field. Obtain the dimension of B. SCALAR AND VECTOR QUANTITIES Physical quantities can be grouped into two namely: scalar and vector quantities. SCALAR β A quantity defined only by magnitude e.g. Distance, Speed, Mass. Distance β Shortest space ππ₯ ππ π between two points. The unit is in meter or foot. Speed(s) β Rate of change of distance with time and is equal to (distance divided by time) Mass β Quantity of matter in a body and is measured in kg VECTOR β A quantity defined by both magnitude and direction e.g. displacement, velocity, acceleration, force. Displacement β Distance in a defined direction. π β ΞΈ ---------------------- Velocity β Time rate of chance of displacement or change of distance with time in a given direction. π£ = π/π‘ = ππ /ππ‘ = ππ₯/ππ‘ πβ ΞΈ ---------------------- Acceleration - rate of change of velocity with time. π = ππ£/ππ‘ πβ 3 Force β Rate of change of momentum with time. This can be shown (later) to be the product of mass and acceleration. Addition and Subtraction of Scalars - Addition and subtraction of scalar quantity is done like numbers. Sum of d1 and d2 = d1 + d2 d1 d2 Difference between d4 and d3 is equal to d4-d3 d3 d4 β d3 d4 Note β Scalars have magnitude only e.g. mass, length, time, density, energy while Vectors have magnitude and direction. Displacement is characterized by length and direction. π΄π΅β is net the effect and is independent of the path taken to go from A to B. Vectors: Considering the net displacement say from A to C through AB followed by BC. The sum is not an algebraic sum Other example includes: b C Force, Velocity, Acceleration B r E Electric field strength a B Magnetic induction A The symbols for vectors are bold face letters or letters with arrow, i.e., πΈ or π΄β. The magnitude is represented by the modulus, i.e., the magnitude of vector π΄β is |π΄β| Vectors can be expressed in vector notation or in magnitude and the angle the vector makes with positive x-axis. In vector notation, a 2D vector, e.g., π΄β can be written as π΄β = π΄π₯ π + π΄π¦ π 4 where Ax is the x-component of the vector, Ay is the y-component of the vector while π and π are the unit vectors. In terms of magnitude and angle / direction, vector π΄β can be expressed as |π΄β |, ΞΈ. Magnitude of vector π΄β Ay Aο½ Ax2 ο« Ay2 , ο± ο½ tan ο1 Ax Ay A ο± Ax x- and y- components of a vector can be obtained from the magnitude and angle representation of the vector. Ax ο½ A cos ο± , Ay ο½ A sin ο± Therefore, vector A can also be expressed as π΄β = π΄ cos π π + π΄ sin π π Addition of Vectors The addition of two vectors can be written as r ο½a ο«b PROPERTIES OF VECTORS 5 Cumulative Law: πβ + πβ = πβ + πβ = πβ b r a a b π¨ππππππππππ π³ππ β π.π.πππππππππππ‘ ππ πππππ ππ ππππ’ππππ, π‘βπ π π’π ππ π£πππ‘πππ ππ π‘βππ πππ πβ + πβ + πβ = πβ + πβ + πβ πβ πβ + πβ πβ + πβ πβ πβ + πβ + πβ Subtraction - For a vector b one can define another vector ο b that is of the same magnitude with b but of opposite direction as shown below. β πβ + πβ Then, a ο b ο½ a ο« (ο b ) πβππ πβ β πβ = πβ + β πβ ππππ‘ππ ππ βππππ§πππ‘ππ πππ ππ‘π ππππππ‘π’ππ 5 π’πππ‘π π£πππ‘ππ π ππ 450 ππππ π‘βπ βππππ§πππ‘ππ Example: vector a is along the horizontal and its magnitude is 5 units, vector b makes angle 45Β° with the horizontal and its of magnitude 4 units while 6 vector c makes angle 30Β° with the vertical axis and its magnitude is 3 units. Compute aο« b ο c. Unit Vector: This is a vector of magnitude 1 in a particular direction. For example, ο in 1D, vector πβ = 2.5π = 2.5π, where 2.5 is the magnitude and a or π is the unit vector. πβ π a π πβπ U=1 π Sum of Two Vectors (πβ πππ πβ) πβ = πβ + πβ π»ππ ππππππ πβ + πβ πππ πππππ ππ πππππ ππππππππππ πππ ππππππππππ πππ πππππ πππ πππ πππππππππ ππππππ ππ πβ, i.e., ππ = ππ + ππ , ππ = ππ + ππ ππ¦ π= πππ + πππ , π‘πππ = (in magnitude and direction / angle) ππ₯ πβ = ππ₯ π + ππ¦ π (in vector notation) EXAMPLE β Three coplanar vectors are expressed with respect to a certain rectangular coordinates system of a given reference frame as πβ = ππβ β π πβ =β ππ + ππβ πβ =β ππβ π»ππ ππππππππππ πππ πππππ ππ πππππππππ πππππ. ππππ πππ ππππππ π πππππ ππ πππ πππ ππ πππππ πππππππ. ππ = ππ + ππ + ππ = π β π + π = π 7 ππ = ππ + ππ + ππ =β π + π β π =β π πβ = πβππ + πβππ πβ = πβ β ππβ and can be represented as shown below πΊππππ ππππππππ π π = πππ + πππ = π+π ΞΈ = π = π.ππ -2 The angle the vector makes with positive x-axis is 360ο° ο ο±. 1 π΄ππππ ππππ π€ππ‘β πππ ππ‘ππ£π π₯ β ππ₯ππ ππππ π’πππ πππ’ππ‘ππππππππ€ππ π 2 πππβ1 ( β ) = 296.57Β° 1 MULTIPLICATION OF VECTORS One can not add vector and scalar in each task. One can add and multiply vectors with vectors. One can also multiple a vector by a scalar. Three Kinds β a) Multiplication of a vector by scalar b) Multiplication of 2 vectors yielding scalar (dot product) c) Multiplication of 2 vectors yielding vectors (cross product), etc. VECTOR-SCALAR MULTIPLICATION Scalar k, can be used to multiply vector πβ to have Product = π πβ A new vector with magnitude (k | a |) will have the same direction as πβ 1 β π πππ€ π£πππ‘ππ π€ππ‘β ππππππ‘π’ππ π π‘ππππ ππ πβ. ππππ ππππππ‘πππ ππ πβ. πΆβπππ β ππ π ππ + π£π To divide by scalar, multiply by reciprocal of k 8 1.πβ π VECTOR-VECTOR MULTIPLICATION I. Scalar (or dot.) II. Vector (or cross Γ ) SCALAR PRODUCTS πβ.πβ = ππ cos π π€hπππ π ππ π‘hπ ππππππ‘π’ππ ππ πβ, π ππ π‘hπ ππππππ‘π’ππ ππ πβ πππ cos π ππ π‘hπ πππ πππ ππ π‘hπ πππππ πππ‘π€πππ π‘hπ π‘π€π π£πππ‘πππ . πhπ’π , π πππππ πππππ’ππ‘ πππ ππ ππππππππ ππ π‘hπ πππππ’ππ‘ ππ π‘hπ ππππππ‘π’ππ ππ πππ π£πππ‘ππ πππ π‘hπ πππππππππ‘ ππ π‘hπ ππ‘hππ ππ π‘hπ ππππππ‘πππ ππ π‘hπ ππππ π‘. For example, the dot product of vector a and vector b , i.e., π = πβ.πβ can be obtained if a and b are explicitly defined. For a ο½ a x i ο« a y j and b ο½ bx i ο« by j , a. b ο½ ( a x i ο« a y j ).(bx i ο« by j ) ο½ a x bx ο« a y by Since, i.i ο½ j. j ο½ 1, and i. j ο½ j.i ο½ 0 This implies that a xbx ο« a y by ο½ a.b cosο± The equation above can be used to obtain the angle between two vectors. Example 1: Calculate the angle between two vectors ο 3i ο« 4 j and 2i ο« 3 j Solution: Using ππ₯ ππ₯ + ππ¦ ππ¦ = π.π cos π 9 ο 6 ο« 12 ο½ 32 ο« 4 2. 2 2 ο« 32 cos ο± 6 6 ο½ 5 ο΄ 13 cos ο± ; ο± ο½ cos ο1 ο½ cos ο1 0.3328 18.0277 ο± ο½ 70.56ο° VECTOR PRODUCT πβ = πβ Γ πβ Magnitude of vector πβ can be written as c ο½ ab sin ο± where ΞΈ is the angle between a and b Example Calculate the angle between two vectors ο 3i ο« 4 j and 2i ο« 3 j using magnitude c ο½ ab sin ο±. Solution i j k (ο3i ο« 4 j ) ο΄ ( 2i ο« 3 j ) ο½ ο 3 4 0 ο½ i (0) ο j (0) ο« k (ο9 ο 8) ο½ ο17k 2 3 0 Therefore using c ο½ ab sin ο± , note that magnitude of ο17 k is 17. 17 17 ο½ 18.0277 sin ο± , ο± ο½ sin ο1 ο½ sin ο1 0.9430 ο½ 70.56 18.0277 ο± is the angle between a and b. By definition, the direction of πΆ, the cross β product of a and b is perpendicular to the plane formed by a and b or the plane that contains a and b 10 c c πβ b πβ b π πβ π' = πβ Γ πβ πβ Γ πβ is pronounced as "πβ cross πβ" πβ Γ πβ β πβ Γ πβ πβ Γ πβ =β πβ Γ πβ For cross product, i ο΄ j ο½ k , j ο΄ k ο½ i , k ο΄ i ο½ j , i ο΄ k ο½ ο j , j ο΄ i ο½ ο k , k ο΄ j ο½ οi πππππ’π π ππππππ‘π’ππ ππ πππ πππ = πππ πππ ππ’π‘ π‘βπ ππππππ‘ππππ πππ πππππ ππ‘π. Examples of such cross products include β Torque, angular momentum, force of a moving charge in a magnet, flow of electromagnetic energy. Vector Products Tensor β generated by multiplying each of three components of one vector by the three components of another vector. Tensor 2nd Rank β Has nine numbers associated with it. Vector β Three numbers Scalar β One Example of Tensors β Mechanical and Electrical Stress, Moments, Products of Inertia, Strain. PROPERTIES OF VECTORS What happens to the laws of physics when simple operations such as translation and rotation of coordinates are performed? Coordinate system π₯,π¦,π§ ππππ‘πππ πβ, πβ, πβ 11 π ππππ‘ππππ βππ πππ‘π€πππ π‘βππ πβ = πβ + πβ π΅π¦ πππππππ πππππππ‘πππ ππ π π’π ππ₯ = ππ₯ + ππ₯ ππ¦ = ππ¦ + ππ¦ ππ§ = ππ§ + ππ§ Consider a New coordinate system π₯,π¦,π§ with properties I. Origin does not coincide with the origin of the first coordinate system π₯,π¦,π§ β Translation II. Its three axes are not parallel to the corresponding axes in the first system β Rotation Representation of vectors π₯,π¦,π in the new system would in general prove to be different. Letβs put them in primes. The relationship between them however would be: ππ₯ ' = ππ₯ ' + ππ§ ' π π¦ ' = π π¦ ' + ππ¦ ' ππ§ ' = ππ§ ' + ππ§ ' And, the relationship πβ = πβ + πβ π π‘πππ βππππ Consequently, it may be said that relations among vectors are invariant (unchanged) with respect to translation or rotation of coordinates. I.e. the laws of physics are unchanged when we rotate or translate the reference system. M y x i j Z k yβ π Z β π π xβ Mβ Mβ 12 (a) is left handed (b) is right handed The other is a mirror image of one ππ π π Γ π =β π π πΓπ=π VIOLATION (1956) Decay of some elementary particles showed that the result was independent of the handedness whether left or right. i.e., the experiment and its mirror image would yield different results. This leads to question on the symmetry of physical laws. 13