Mechanics

                          UNITS AND DIMENSIONS

Units “ To measure or represent any physical quantity we need units”

Exception: Although there are few physical quantities which do not

need any unit like refractive index, Relative density or specific gravity etc.

Physical quantity            

 nu = Constant

 

Example: In a particular system, the unit of length,

mass and time  are chosen to be 10cm, 10gm and 0.1 sec

respectively. The unit of force in this system will be

equivalen to:      

1. 0.1 N
2. 1 N
3. 10 N
4. 100 N

Sol. Unit of force Þ F = ma = Kg metre /sec²

Here   

F = 0.1 N     So the Correct option is (a)

Example: The density of a material in CGS System of unit

is 4gm/cm³. In a system of units in which unit

of length is 10cm and unit of mass is 100 gm the

value of density of material will be     

(a) 0.04

(b) 0.4

(c) 40

(d) 400

Solution. Given n₁ u₁ = 4 gm/cm³

n₂ u₂ = n₁ u₁

 

n₂ = 40                                            Answer (a)

FUNDAMENTAL PHYSICAL QUANTITIES, UNITS, SYMBOLS

SUPPLEMENTARY UNITS

Derived Units: The units which may derive by

fundamental or Supplementry units

Ex. Force= ma = Kg mt/sec²

Conventions adopted while writing a unit

1. Even if a unit is named after a person it should

not be Capital initial letter.

Newton (no)      Joule (no)            Ampere (no)  

     or                      or                            etc.

newton (yes)     joule (yes)          ampere (yes)    

2. In symbol for a unit named after a person

Newton ® N

Ampere ® A

Joule ® J

Watt ® w

3. The symbol or units are not expresse

d in plural form : 50 m (yes) , 10 joule (yes)

                          50 ms (no)  , 10 joules (no)

4. Not more than one Solidus is used

1 poise = 1g/s  cm or 1 gs^-1

 ERRORS

The result of every measurement by any measuring instrument contains some uncertainty this uncertainty is called errors.  The errors in a measurement is equal to the difference between the true value and the measured value of any quantity.

Errors = True value – Measured value

Absolute Errors:- the several values obtained in an experiment measured are a₁, a₂, a₃….an the arthmetic mean of these values will be

Now error in first measurement = Da₁ = a_mean –a₁

Similarly in second measurement = Da₂  = a_mean –a₂   and so on

 Relative error:Relative error=∆ameanamean

 

Combination of errors (maximum possible error) Addition or subtraction

Product or multiplication 

Let

 

Division

 

     

 

(4) Powers

Let C = aⁿb^m

CONCEPT WITH EXAMPLES

Role of constants

The radius of sphere is measured to be (2.1 ± 0.5) cm and is surface area with error limits.

Solution:

Surface area = A = 4pr²

 

 

Now% error

 

= 47.62 %

Other way

= 4 p r²

 

= 55.4 cm²

 

 

 

= 26.4 cm²

Now (A ± DA)

 = (55.4±26.4)  

FINDING QUANTITY 

Percentage error in determining of acceleration of gravity with the help of simple Pendulum time period .  Given error in length of pendulum 4% while in The time period it is 2% then percentage error in g = ?

 

Solution

LEAST COUNT AND DIFFERENT UNITS

If a particle of mass m = 25.0 Kg is moving in a circular path of radius r = 50.00 cm with constant speed of v = 10.0 Km/hr them find error in force required by particle.

Now Given

m = 25.0 Kg

    = L.c. = 0.1

V = 10.0 Km/h

    = L.c = 0.1

DV = 0.1, V = 10

R = 50.00 cm

Dr = 0.001, r = 50cm

NOTE: never change units in error.

Least count

25.00 – L.c = 0.01

16.50 – 50 –L.c = 0.01

76.03 – 03 –L.c = 0.01

15.0002 –L.c = 0.0001

17.030 – L.c = 0.01

 

Solution:

 

 

 

 

 

= 1.21 x 2= 2.42% Ans.

UNIQUE CONCEPT

Calculate focal length of a spherical Mirror from the following observations Object distance u = (50.1 ± 0.5) cm. and image distance v = (20.1 ± 0.2) cm.

 

 

 

   = 14.3 cm

Now

 

     

 

    

 

      ± 0.4 cm

Note:  Similar case in resistance connected in parallel combinations

    

SIGNIFICANT FIGURES

The number of significant figures of a numerical quantity is the number of reliably known digits it contains:

Rules:

1. Zeros at the beginning of a number are not significant.

Ex. 0.0523 – there S.f. (5, 2, 3)

2. Zeros within a number are significant.

Ex: 2056 –Four S.f.    (2, 0, 5, 6)

3. Zeros at the end of a number after decimal points are significant.

Ex. 3702.0 –five S.f. (3, 7, 0, 2, 0)

Law of homogeneity

Here we can jump through +, - , = (signs)

 

and a,b, c- are Constant then find dimensional formula of a, b, c-

 

Apply Law of homoginity

Dimensional formula of v = dim-formula of at = v

 

Similarly we can say t = c Þ C = [T]

 

b = [L]

Example: The equation of state of some gases can be expressed as here p is pressure, V is the volume, T is temp. then the dimensional formula of Constants a and b will be ?

 

Apply Law of homoginity Dimensional formula of p = Dim-formula of 

a = PV² = ML ^-1T^-2L⁶ =[ML⁵T^-2]

Similarly dimensional formula of v = Dim. Formula of b b = v

= [L³]

ANS a = [ML⁵ T^-2]

b = [L³]

Application No-3

Dimension less quantities or functions

(i) Quantities having units but dimensionless

So all trigonometric functions will be dimension less i.e. sin x, cos  x, tan x – etc.

(ii) Quantities neither having units nor dimension 

Refraction index

Reynolds number, Relative density etc.

(iii) Dimension less mathematical functions

(a) Logarithmic functions: log_e x or log₁₀ x

(b) Exponential functions: e^x, a^x

 

(a) P

(b) R

(c) T

(d) V

 

 

 

 

                    Correct answer is (a)

 

Example Nuclear force between nucleons is given by   Here F is force, r is distance, and c, k are Constants then find dimensional formula of C and k = ?

 

Here e^-kr is dimension less

So C = Fr² = M L T^-2 L² = [M L³ T^-2]

No because e^-kr is dimension less

So k = 1/r

 = [L^-1]

Ans: dimensional formula of

C = [M L³T^-2]

K = [L^-1]

SYSTEMS OF UNITS

(1) MKS AND MKSA SYSTEM

Here in this system M® meter, K® Kilogram, S ®Second and A® ampere

Physical quantity Unit in MKS or MKSA System

 

 

…….. and so on

(2) SI SYSTEM (INTERNATIONAL SYSTEM)

Here in this system unit of physical quantities are use to be named after different great Scientists.

Physical quantity 

Unit in SI System

1. Force ® newton or N
2. Energy ® joule or J
3. Power ® watt or W

(3) CGS SYSTEM

Here  C® Centimeter, G ® gram, S ® Second

Physical quantity  Unit in CGS System

 

 

 

….and so on

(4) FPS SYSTEM

Here  F ® foot, P ® pound , S ® second

Force ® pd ft/ sec², Similarly other physical quantities

Dimensional and dimensional formula

Dimensional formula

Almost all physical quantities can be expressed in terms of the seven fundamental units in symbolic form.

Dimensions 

Dimensions of a physical quantity are the powers to which we must be raised to fundamental quantity represent the given physical quantity.

……. And so on

APPLICATIONS OF DIMENSIONAL FORMULAS

Application No. (1)

To establish results

Suppose any mass m is moving along a circular path of radius R with uniform speed of V. If the force required to do so is F then establish the result or relation between F, m, v and R.

Given ,

F = k  m^x  v^Y R^Z

F a m^x  v^Y R^Z

[M L T ^-2] = K [M] ^x [LT ^-1] ^y [L] ^z

[M LT^-2] = K [M^x L^y+z T^-y ]

Comparing both sides

X =1, y + z = 1, - y = - 2 Þ y = 2

2 + z = 1

Z = 1 - 2 = - 1

By experiment Constant K = 1

So we have result F = m¹ v² R^-1 Þ

Example 1 Given a simple pendulum of very small bob of mass connected with a light String of length l. Also given acceleration due to gravity is g and time period T of simple pendulum depends on mass of the bomb, length of the String l and acceleration due to gravity g then establish   result for T.

Given

T a m ^x l ^y g ^z

T = k m^x l ^y g ^z

[M° L° T¹] = K [M]^x [L]^y [LT^-2]^z

Þ [M° L° T¹] = k [M ^x L ^{y + z} T ^-2z]

Compare both Side

x = 0, i.e. Time period will not depend on mass of the bob Y + z = 0

Compare both Side

x = 0,i.e. Time period will not depend on mass of the bob Y+ z = 0

 

 

By experiment k = 2p

 

Example: In a system of units , if force F, acceleration A and time T are taken as fundamental units then the dimensional formula of energy is

(a) [F A² T]

(b) [F A T²]

(c) [F² AT]

(d) [FAT]

Solution:  Let energy denote by E

Then E a F^x A^y T^z

E = F^x A^y T^z

[M L² T^-2] = [M L T^-2]^x [ LT^-2]^y [T]^Z

[M L² T^-2] = [M^x L^x+y T^{-2x-2y +z}]

Compare both Sides

x =1 ,         - 2x – 2y + z = -2

x + y = 2    - 2x1 -2 x1 + z = -2

Þ Y = 1    - 4 + z = -2

 Z = 4 – 2  

  Þ Z = 2

Therefore dimensional formula of energy E in Terms of FAT is  [ F¹ A¹ T²] Þ [ FAT²]  Ans B                                 

Application No.2: To check the validity of any result.

 VERNIER CALIPERS

To measure length accurately upto 0.1 mm or 0.01 mm Vernier Calipers and Screw gauge are used

 VERNIER CALIPERS

Parts of Vernier Calipers

(i) Main scale

(ii) Vernier Scale                   

(iii) Metallic Strip

PRINCIPLE OF VERNIER CALIPERS

N VSD = (n -1) MSD

 

1 MSD -1VSD = 1 MSD  

 

LC or VC ® Smallest distance that can be accurately measured with

Vernier Scale

 

READING A VERNIER CALIPERS

If we have to measure a length AB, the end A is coincided with the zero of main scale, suppose the end B lies between 1.0 cm and 1.1 cm on the main scale. Then,

 

Let 5^th division of Vernier scale coincides with 1.5 cm of main scale.

Then,           

Thus, we can make the following formula,

Total reading = N + n x VC

Here, N = main scale reading before on the zero of the Vernier scale.

n = number of Vernier division which just coincides with any of the

main scale division.

Zero Error and Zero Correction

If the zero of the Vernier scale does not coincide with the zero

of main scale when jaw B touches A and the straight edge of D

touches the straight edge of C, then the instrument has an error called zero error. Zero error is always algebraically subtracted from measured length.

Zero correction has a magnitude equal error but its given is opposite to that of the zero error. Zero correction is always algebraically added to measured length.

Zero error ® algebraically subtracted

Zero correction ® algebraically added

POSITIVE AND NEGATIVE ZERO ERROR

If zero of Vernier scale lies right of the main scale the zero is positive and if it lies to the left of the main scale the zero error is negative (when jaws A and B are in contact)

Positive zero error = (N + x x VC)

Here, N = main scale reading on the left of zero of Vernier scale.

X = Vernier scale division which coincides with any main scale division.

When the Vernier zero lies before the main scale zero the error is said to be negative zero error. If 8^th  Vernier scale division coincides with the main scale division, then

Negative zero error = -[0.00 cm + 8 x VC]

                       = -[0.00 cm +8 x 0.01 cm]

                          = -0.08 cm

SUMMARY

1. VC=LC=1MSDn=Smallest division on main scale number of division on vernier scale=1 MSD-1VSD  

2. In ordinary Vernier calipers, 1 MSD = 1mm and  n = 10

 

3. Total reading = (N + n x VC)

4. Zero correction = -zero error

5. Zero error is algebraically subtracted white the zero correction is algebraically added.

6. If zero of Vernier scale lies to the right of zero of main scale the error is positive. The actual length in this case is less than observed length.

7. if zero of Vernier scale lies to the left of zero of main scale the error is negative and the actual length is more than the observed length.

8. Positive zero error = (N + x x VC)

Example

The smallest division on main scale of a Vernier calipers is 1 mm and 10 Vernier division coincide with 9 main scale divisions. While measuring the length of a line, the zero mark of the Vernier scale lies between 10.2 cm and 10.3 cm and the third division of Vernier scale coincides with a main scale division.  

(a) Determine the least count of the calipers

(b) Find the length of the line

Solution:  

 

 

SCREW GAUGE

Principle of a Micrometer Screw

The least count of Vernier calipers ordinarily available in the laboratory is 0.01 cm. when lengths are to be measured with greater accuracy, say upto 0.001cm, screw gauge and speedometer are used which are based on the principle of micrometer screw

If an accurately cut single threaded screw is rotated in a closely fitted nut, then in addition to the circular motion of the screw there is a linear motion of the screw head in the forward or backward direction, along the axis of the screw. The linear distance moved by the screw, when  it is given one complete rotation is called the pitch (p) of the screw. This is equal to the distance between two consecutive threads as measured along the axis of the screw. In most of the cases, it is given one complete rotation is called the pitch (p) of the screw. In most of the cases, it is either 1 mm or 0.5 mm. A circular cap is fixed on one end of the screw and the circumference of the cap is normally divided into 100 or 50 equal parts. If it is divided into 100 equal parts,

then the screw moves forward or backward by  of the pitch, if the circular scale is rotated through one circular scale division. It is circular scale division. It is the minimum distance which can be accurately measured and so called the least Count (LC) of the screw.

Thus,                   

If pitch is 1 mm and there are 100 divisions on circular scale then,

 

= 0.001cm = 10mm

Since, LC is of the order of 10mm, the screw is called micrometer screw.

SCREW GAUGE

SCREW GAUGE

Screw gauge works on the principle of micrometer screw. It consists of a U –Shaped metal frame  M. At one end of its is fixed a small metal piece A. it is called stud and it has a plane face. The other end N of M carries a cylindrical millimeter depending upon the pitch of the screw. This scale is called linear scale or pitch scale.

A nut is threaded through the hub and the frame N. through the nut moves a screw S. the fount face B of the screw, facing the plane face A is also plane, A hollow cylindrical cap K is capable of rotating over the hub when screw is rotated. As the cap is rotated the screw either moves in head scale. In an accurately adjusted instrument when the face A and B are just touching each other. Zero of circular scale should coincide with zero of linear scale.

To measure Diameter of a given wire using a screw gauge

If with the wire between plane faces A and B, the edge of the cap lies ahead of N th division of linear scale, and nth division of circular scale lies over reference line.

Then,                 Total reading = N + n x LC

ZERO ERROR AND ZERO CORRECTION

ZERO ERROR AND ZERO CORRECTION

If zero mark of circular scale does not coincide with the zero of the pitch scale when the faces A and B are just touching each other, the instrument is acid to possess zero error. If the zero of the circular scale advance beyond the reference line the zero is positive and zero correction is positive. If it is left behind the reference line the zero is positive and zero negative. For example, if zero of circular scale advance beyond the reference line by 5 divisions, zero correction = + 5 x (LC) and if the zero of circular scale is left behind the reference line by 6 divisions, zero corrections = -5 x (LC)

Example The pitch of a screw gauge is 1 mm and there are 100 divisions on circular scale. When faces A and B are just touching each without putting anything between the studs 32^nd divisions of the circular scale (below its zero) coincides with the reference line. When a glass plate is placed between the studs, the linear scale 4 division and the circular scale reads 16 divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden from circular scale when A and B touches each other.

Solution:     

                                          = 0.01 mm

As zero is not hidden from circular scale when A and B touches each other. Hence, the screw gauge has positive error.

 

Linear scale reading = 4 x (1mm) =4 mm

Circular scale reading = 16 x (0.01 mm) = 0.16mm

Measured reading = (4 + 0.16)mm = 4.16mm

Absolute reading = Measured reading = -e         = (4.16 – 0.32) mm = 3.84 mm

Therefore, thickness of the glass plate is 3.84 mm.

VECTOR AND SCALARS

Physical Quantity

May have numerical value, units and any specified direction.

Note:-  Some physical quantities only having numerical value not specified direction.

Ex:-  Refractive index, strain etc.

We can say any physical quantity Must/may have numerical value(n), unit (u) and specified direction.

PHYSICAL QUANTITY (N + U + DIRECTION) MAY/MUST HAVE

Note:

Scalars or zero order tensors: Any physical quantity have only one component.

Vectors or first order tensors: Any physical Quantity have component greater then one but less than or equal to four.

REPRESENTATION OF VECTOR:

Types of vectors

(1) Polar Vectors:

Vectors related to linear motion of any object Ex. Displacement, velocity, Force etc.

(2) Axial vectors:

Vector represent rotational effect and are always along the axis of rotation Ex: Angular velocity, torque, angular momentum etc.

(3)  Null vector or zero vectors:

Vector whose magnitude is zero and direction in determinant.

(4) Unit vectors:

Vector having magnitude equals to one (unity) but must be some specified direction representation of unit vector – , A  

 

STANDARD UNIT VECTOR

These are

Direction should be on your copy

UNDERSTANDING OF UNIT VECTORS

1 –D Motion: Motion along x or y or z

A car is moving with 60 k/m toward or due east then speed (scalar) = 60 km/h

Velocity (Vector) = 60 km/h

2 –D Motion:

A car is moving with 60 k/m due north – east then

 

 

2 –D vector

 

With x axis or with horizontal

 

With Y-axis or with vertical

3 –D Vector

 DIRECTION

 

With x-axis

 

With y-axis

 

With z-axis

ADDITION AND SUBSTRACTION OF VECTORS

Law of parallelogram

 

 

 

 

 

 

 

 

 

 

 

 

            RESOLVING OF ANY VECTOR

Example. Two equal vector have a resultant equal to either of the two. The angle between them is                                                                                                                    

(a) 90^o                                                      (b) 60^o

(c) 120^o                                                    (d) 0^o

Solution: By using expression R² = P² + Q² + 2PQ cosa

 

x² = 2x² (1 + cosa)

cos a = -1/2 = a = 120^o                                                                             Answer is (c)

Example: Two vector having equal magnitude of x units acting at an angle of 45⁰ have resultant   units the value of x is                                                             

(a) 0                                        (b) 1

(c)                                       (d)

Solution: Using the expression R² = P² + Q² + 2PQ cosµ

 

 

 

 

 

Þ x² =1                                                                    Ans (b) Þ x = 1 Ans (b)

 

 

(a) 0°                                                   (b) 180°

(c) 90°                                                 (d) 120°

 

 

= 4PQ cos q = 0

   Cos q = 0

   q = 90^o                                                                                   Answer is (c)

          LAMI’S THEOREM

        

 

KINEMATICS

Rest and motion:-

Rest: “if position of any object or particle is not changing with respect to any observer then this object is said to be at rest”

Motion: “if position of any object or particle is changing with respect to any observers then this object is said to be in motion”

“Absolute rest and absolute motion is impossible”

Sate of rest and motion depends upon reference frame (State of observer)

Type of motion

(1) One dimensional (1-D) motion:

If position of any object is changing any of x axis, or y-axis or z axis.

All 1-D motion will be always straight line motion”

(2)  Two dimensional (2-D) motion (Motion in a plane)

X-Y plane

Z- X plane

Y-Z plane

“  If position of Any object is changing in x –y or y-z or z-x  plane”

Example:

 

2-D motion may be straight line motion”

“Writing on board or copy is also 2-D event”

(3) Three dimensional (3-D) motion:-

“ if  position of any object is changing along all 3 – axis (x-y z)”

Example:-

 

Human can move in 1-D, 2-D, or 3-D motion”

BASIC TERMS IN KINEMATICS

(1) Displacement:

“length of straight line drawn between initial and final position

of any object”

“Shortest path between initial and final position of any object”

It is vector quantity so it may be +, - , zero.

(2) Distance:-

“length of actual path travelled by any object” It is scalar quantity so it will be always positive.

Distance   ≥  Displacement

For straight line motion distance will equal to displacement.

UNDERSTANDING OF DISPLACEMENT AND DISTANCE

Example:-

 

“A person is moving 3 km towards east then take left turn and move 3 km towards north then person take a right turn and move 1 km further due east and stops.

(i) What will be the total distance travelled by person.

(ii) What will be the net displacement of the person.

Solution: As we study in vectors and scalars.

(i) Distance: length of actual path so distance = 3 km + 3 km + 1 km = 7 km

(ii) Displacement: length of straight line

First method:

Second method

Applying Pythagoras                                         = 5 km

Example: 2 An object is moving in a circular path of radius of  

meter shown in figure Find magnitude of distance

and displacement.

(i) From A to B

(ii) From A to C

(iii) From A to D

(iv) From A to A

Solution

A to B

= 5 meter

A to D

= 15 meter

A to C

= 10 meter

A to A

Distance = 2pR

= 20mt

Displacement = 0

1-D MOTION

2-D MOTION OR MOTION IN A PLANE Þ

Motion of any object in any of two axis involve (xy or yz or zx)

“Projectile Motion may be 2-D or even 3-D also but in general it is use to be 2-D Motion”

-D PROJECTILE MOTION

VERTICAL MIRROR

Vertical Mirror ®  Gravity acts in vertical direction so we can use equations of motion

Time taken by the image of ball going up = time taken going down = t (Let)

Use  v = u – g t

O = u sinq - g t

Total time

 

 

Now to get Maximum height

Using v² = u² – 2g h

(O)² = (usinq)² – 2g H

HORIZONTAL MIRROR

Velocity remains constant so we can not use equation of Motion

Note: Untill any external reason present to change ucosq, like air flow ucosq remains Constant

Now ucosq = Const

 

R = (u cosq) T

 

Now Sin2q = 2 sinq Cosq

  

Now we have

 

 

 

 

 

CONDITION

“These three results only when can be use if initial point of projection and final point of projection are at same level.

Example: A body of mass m is projected upward with initial velocity                                  then find time of flight, Maximum height attained and range attained by the body (g = 10m/s²)

Solution

 

 

 

  

  

                                                             

Maximum range: To get maximum horizontal distance covered by any mass in projectile motion we must have a unique specified angle.

 

(Sin2q)_max = 1

2q = 90°

q = 45°

To get maximum range angle of projection should be 45 °,

Note: Here we are neglecting the effect of air resistance.  If we Consider air resistance then this angle q should be little bit less than 45°

SAME RANGE

Mathematically there must be two different angle of projection for which we will get same range

Ball 1

  

Ball 2

 

Now if we assume

        a + b = 90° 

then b = 90° - a

 

 

     

If sum of angle of projection is = 90° and initial speed is same then both balls will have same range.

Note: Here in this Case only range will be same not time of flight and maximum height

R₁ = R₂

T₁ ¹ T₂

H₁ ¹ H₂

In case of same range relation of time of flights

Here we know a + b = 90°

Ball 1

 

Ball 2

 

 

 

Now

  

 

 

In case of same range relation of maximum heights

Here again a +b = 90°

Ball 1

 

Ball 2

 

=  

 

 

 

 

 

 

 

RELATIVE MOTION

Relative is a most common word use to compare any property of any object with respect to other object “

    X_AB = X_A ─ X_B,

        ↓

X of A with respect to B

Here x may be any physical or non physical quantity

Sign convention 

V_AB = V_A - V_B

= + 10 – (16)

V_AB = - 6m/s , B says that it seems to be that car A is moving left with 6 m/s

But if V_BA = V_B - V_A

= (16) – (+10)

V_BA = 6m/s

A says  that if seems to be that car B is moving right with 6 m/s

And if V_BC = V_B-V_C

= (+16) – (-18)

= +34 m/s

C says that it seems to be that car B is moving right with 34 m/s

Also V_CA = V_C –V_A

= - 18 – (+10)

= - 28 m/s

A says that it seems to be that car C is moving left with 28 m/s

Applications and Understanding of relative motion

Case : 1 Catching, collision problem

As shown in figure a cop is chasing to a thief . After how long the cop will catch the thief?

Solution : First of all we will shift our frame on car so now everything will be relative to  car

Initial velocity of car w.r. to bike u_CB = u_C - u_B

 = 5 - 2

u_CB = 3m/s

Acceleration of car w.r. to bike a_CB = a_C- a_B

= 2 -5

= -3 m/s²

Distance of car w.r. to bike S_CB = S_C – S_B

= 0 – 500

=  - 500 mt

Note : Here we are shifting our frame on car so it would be at rest

 

 

 

 

 

Case II:- Lift or elevator in relative motion.

In this case it is given that boy is dropping a coin inside the lift at the same time lift also start moving down ward with acceleration 5 m/s², if height of coin from the flower is 10 meter. Then find. Time after which coin strike the floor of lift. 

Solution: Here again first of all we will shift our frame inside the lift (or on the lift).

So now lift is assume to be at rest now.

Initial velocity of coin w.r. to lift u_cl = u_c – u_l

Because coins & lift start simultaneously = u – u = 0

Acceleration of coin w.r. to lift           a_cl = a_c – a_l

Sign convention                                  = -g -5

                                                              = -10 -5

                                                        a_cl = -15m/s²

distance covered by coin w.r to lift h_cl   = h_c –h_l                                                     

                                                                = –10 – 0

                                                                = –10 meter

Now

5  

 

If Any object moves such that it covers an angle q at the fixed point (center of the circle) and its gap from the fixed point remains constant (radius of the circle)

 

Here R is constants

 

     

Now finding linear speed of

(i) hour hand (30 cm)

(ii) Minute hand (90 cm)

(iii) Second hand (60 cm)

Solution:- (i) time period of hour hand T = 12 hr

               = 12 x 3600 sec

                                                                v = r w

 

(ii) Minute hand:- time period = T = 60 min = 60 x 60 sec

 

(iii) Second hand:-

Time period= T = 60 sec

v = r w

Types of circular motion: (on the basis of speed)

(1) Uniform circular motion: (U C M)

Only direction of velocity is changing, magnitude remains unchanged.

 

Now acceleration due to change in direction of velocity

Using law of parallelogram

 

 

 

= 2v² - 2v² cosq                                 

= 2v²  (1- cosq)

 

 

Now

 

 

sinq » q if  q <<<<

 

 

a = v w,

 

 

 

Direction of this acceleration is towards center

   

Centripetal acceleration or Radial acceleration

(2) Non uniform circular motion (Non U C M)

Here in this case direction as well as magnitude both are changing continuously therefore here two different named acceleration will be as,

(i) a_c = a_R (centripetal or radial acceleration)

(ii) a_t (Tangential acceleration)

Now,

Tangential acceleration. 

“Rate of change of magnitude of velocity”

v = R w

 

 

 

So a_t ^ a_c

 

Net acceleration for non-uniform circular motion.

Example:- A particle moving in a circular path of radius 2 meter and its velocity varies as  v = 10t^2. Then net acceleration of the particle at t = 2 sec.

Solution: Given v = 10t²

 

= 20 x 2

= 40 m/s²

 

 

Centripetal force:- the force require to move in a circular path for any object with respect to inertial frame is called centripetal force.

 

Note:- Centrifugal force will also have same magnitude but direction opposite to centripetal force.

Direction of this force is towards the center.

HORIZONTAL CIRCULAR MOTION

(1) Only banking

Here N sinq is C.P. provider

 

But N cosq = mg

 

     

            Safe speed

(2) Only friction 

Here now m N is C.P. provider

 

N = mg

 

 

 

Safe speed

Friction and Banking both:- To prevent inward sliding

 cosq + m N sinq = mg

Divide

          

TO PREVENT OUTWARD SLIDING

N cosq - m N sinq = mg

          

 

v₁ < V_safe < v₂

CONICAL PENDULUM

T sinq = mrw²

T cosq = mg

 

 

Time period of conical pendulum

 

 

           

Force:- Any pull or push”,     unit: kg met/sec² or Newton (N)

Types of forces:-
1- Field Forces:- No need of contact

Always attraction type

 

Direction:- Always along (-y) axis or straight vertically down ward.

(2). Contact forces: There must be some contact

(i) Normal force (N):- This force act when two surfaces of different object are in contact.

Direction:- perpendicular to contact surfaces.

(ii). Frictional force(f):- This force also act when two surfaces of different objects in contacts.

Direction:- Parallel to contact surfaces

Also F = m N here m is coefficient of friction.

Now

N → Perpendicular to contact surface

So F ^ N  

 

Understanding of frictional force

The force which resist the motion”

m_S > m_k      m can be less than or greater than one

If F_ext = 0, f = 0

Example:- 1

Now

f_s = m_sN

 = m_s mg

 = 0.6 x 10 x 10

f_s = 60 N

f_k = m_kN

 = 0.2 x 10 x10

f_k = 20 N

Amount of frictional force = 0

Because f_ext = 0, = f = 0

 

Example:- 2

m = 10 kg As in case (i)  f_s = 60 N, f_k = 20 N

f = f_s = 40 N

Body is at rest

 

Example:- 3

As in case (i) f_s = 60N, f_k = 20 N

f = f_s = 60 = F_ext →    limitation friction

Body is at rest but tends to move

Example:- 4

As in case = f_s = 60N, f_k = 20N

Body is  in moving condition

10a = 85 - 20

f = f_k = 20 N

(3) Attachment forces indirect force

 

Direction: Direction of tension (T) is always away from the body. 

Example:-                                  

10 g –T = 10 a   ---- (1)

N = 5 g      ------(2)

T = 5a --------(3)

Solving 1& 310 g - 5a = 10a

15a = 10g

            

 

T = 5a

                 

Net force on the pulley

Example:-

f_s = m_s N

   = 0.5 x 5g

f_s = 25 N

f_k = m_k N

    = 0.2 x 5g

f_k = 10 N

Pulling force for  5 kg

10 g = 100 N

So both block will move

10 g –T = 10a -----(1)

T -m_kN = 5a ------(2)

T – 10 = 5a

N = 5g -----(3)

Solving  (1) & (2)

10g –T = 10a

1-10 = 5a

90 = 15a Þ2

Similarly

T -10 = 5 x 6                        T= 40N

WORK DONE

(i) By Constant force

W = (f cosq) S

Work done by constant force = (force). (Displacement in direction of force

                                                = (displacement). (force in direction of displacement)

W=F.S

W = FS Cos

 

Here W is Scalar but it may be +, ─, and zero

Work done will be equal to zero if

(i) Cosq = 0 = q = 90^o = F ^ S or F ^ V

Ex. Work done by centripetal or centrifugal force is always zero.

Fm=q (V×B)  

 Fm⊥V, Fm⊥B

(2) Displacement = 0, S = 0  in circular motion

Negative work done

If cosq  → negative, 180^o ³ q ³ 90^o

W → ─

Note: Work done by frictional force may be negative.

Example:- It is given that a particle is moving from initial position (2,3,5) to final position (4,1,2) under the action of constant force F=2i+3jN.    what will be the work done by the constant force. 

S��lution-Initial position r1=2i+3j+5K as given (2,3,5)  

Final position r2=4i+j+2k as given (4,1,2)  

Now S=r2-r1=2i-2j-3k  

F=2i+3j  

Now W=F.S.  

=2i+3j. (2i-2j-3k)  

= 4 -6 + 0.

W = -2j                                                                             Ans

(ii) Work done by variable force

Case I: Applying integration force

W=F.dx=F.dy=F.ds  

or W= F.ds or W=F.ds  

Example:- if a variable force F depends on displacement X as F = (3x² -5) due to this force body displaces from x = 2 mater to x = 5 meter find the work done by this variable force. 

Solution: Given F = 3x² – 5,     displacement from x = 2 meter to x = 5 meter 

W=25fdx=253x2-5dx  

     =3 25x2dx-525dx  

3x3325-5X25   

      =33×53-23-55-2  

      = [125 – 8] – 5 x 3

W = 102 J

Case II: Applying graphical method.

W = F.S   if we have group F vs S then area under the S Curve will give us work done.

Work done = Area under the F – S curve

Area  = 1/2 x basic height + length x width

         = 1/2  x  S  x  F + F x S

      = 3/2 FS

W = 3/2 FS

Example:- Spring force F = - Kx {Here K is Force constant} it a spring elongate by 5cm and force constant  K is 100N/meter then find work done on stretching the spring.

Solution:

Now F = -K x

            = -100x

W=Fdx  

   =-10000.05xdx  

-100x2200.05  

-10020.052=-50×25×10-4  

= 1205 x 10^-14

W = - 0.125 J                                                      Ans.

Question: An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined at 30° to the horizontal. A 10 kg block M is released from rest at the top of the incline and brought to rest momentarily after compressing the spring by 2 m. If g = 10 ms^-2, what is the speed of mass just before it leave the spring?

a 20 ms-1                                          b 30 ms-1

c10 ms-1                                           d 40 ms-1

Solution: F = kx

∴    k=Fx=1001  

= 100 N/m

E_i = E_f

∴12×10×v2=12×100×22-1010(2sin30°)  

Solving we get,

v=20 m/s                                                                                                       Answer (a)

 

KINETIC ENERGY

Scalar quantity, always will be positive   K = ½ mv²

Work energy theorem: It is given that a mass is having initial velocity u accelerate and final velocity V and covered a distance S. Here v > u

Initial kinetic energy K1=12mu2  

Final kinetic enrgy=K2=12mv2  

Using equation of motion

v² = u² + 2a s

12m v2-u2=2as12m  

12mv2-12mu2=maS  

K₂-K₁ = FS

ΔK = W

Question: A body of mass m = 10^-2 kg is moving in a medium and experiences a frictional force F = -kv². Its initial speed is v₀ = 10 ms^-1. If after 10s, its energy is 18mv02,  the value of k will be                                                                                                               (2017 JEE Main)

(a) 10^-3 kg^-1                                               (b) 10^-4 kgm^-1

(c) 10^-1 kgm^-1s^-1                                        (d) 10^-3 kgm^-1

Solution: Given, force,   F=kv2

∴Acceleration,a= -kmv2  

or   dvdt= -kmv2  

⇒    dvv2= -km∙dt  

Now, with limits, we have

10vdvv2=-km0tdt     

⇒    -1v10v=-kmt  

⇒  1v=0.1+ktm  

⇒    v=10.1+ktm=10.1+1000k  

⇒ 12×m×v2=18×v02  

⇒v=v02=5  

⇒ 10.1+1000k=5  

Þ    1 = 0.5 + 5000 k

⇒  k=0.55000⇒k=10-4kg/m                                                                Answer (b)

Potential Energy: Scalar quantity may be +, - , Zero 

Here F= -dUdrHere U is potential energy  

F.dr=-dU  

Note: if we find slope (tanq) of u –r graph then we will find amount of force.

 

BY THE U –R GRAPH

Slope at point A is positive

Slope at point B is zero here force at B will be zero similarly slope at C is negative slope at point D is zero.

 

 

 

 

 

 

 

TYPES OF POTENTIAL ENERGY

(1) Elastic potential energy: Energy associated with state of Compression or expansion of an elastic object like spring. U = ½ kx²  Here x is the stretch of compression.

(2) Electric Potential energy: Energy associated with state of separation between charged particle

(3) Gravitational potential energy

U=-Gm1m2r  

Note: Attraction force the An U → negative always

U=mgh1+h/r  

If h <<<<< R

U = mgh                                                                                                                                                                                                         

 

 

Example: A chain is held on a frictionless table with (1/n)th of its length hanging over the edge if the chain has a length L and a mass M. how much work is required to pull the hanging part back on the table.

amgl8n2                                bmgl4n2

cmgl2n2                                dmgln2

Solution:   L length chain mass –M

I length chain mass –M/L

dy length chain mass MLdy  

W = DU

W=Mgl2n2  

du₁ = (dm) gy

du1=MLg0l/nydy  

U1=mLgY220Ln=MLgL22n2  

U1=MgL2n2,  

U₂ = 0 at table

 

CONSERVATIVE AND NON-CONSERVATIVE FORCE

Conservative forces: Work done by conservative forces are independent of path. It only depends on initial and final position.

Ex:- Gravitational forces, electrostatic forces etc.

Note: Work done by conservative forces in closed path is always zero.

W_path1 = W _path2 = W _path3

Non conservative force: Work done by non-conservative forces depends on path.

Ex:- frictional force, viscous force etc.

Power Work done per unit time P =F.V  

As we know W=∆K=∆U  

P=dwdt=w∆t=dkdt=dudt  

Unit. J/sec or watt

1 horse power = 746 watt.

 

CONSTANT POWER

Case I:- dependency of v on t.

Let an car of mass m accelerates staring from rest, while the engine supplies constant power P then how the V depends on t.

As we know P = Fv

P=mvdvdt  

Pdt=mvdv  

pt=mv22=V2=2ptm  

V=2Ptm=V∝t1/2  

Case II:- dependency of S on t

Let above result V=2ptm  `

dsdt=2Pmt  

0sds=2pmt dt  

S=2Pmt3/23/2   

S= 8P9m1/2 t3/2  

S∝t3/2   

Question: A uniform chain of length pr lies inside a smooth semicircular tube AB of radius r. assuming a slight disturbance to start the chain in motion, the velocity with which it will emerge from the end B of the tube will be

agr1+2π                                          b 2gr2π+π2

c gr  π+2                                         d πgr

Solution: dm=mπ dθ

∴  dU=dmgh= mπdθgrsinθ                 

  Ui= 0πdU=2mgrπ  

Now,  Ki + Ui = Kf + Uf

∴     0+2mgrπ=12mv2-mgπr2  

v=2gr 2π+π2                                                                                                          Answer (b)

NEWTON’S LAW OF GRAVITATION

 

Here G is universal gravitational Constant

G = 6.67 x 10^-11 Nmt²/kg²

Dimensional formula = [M^-1 L³ T^-2]

Note: This formula is applicable for point mass only.

PROPERTIES OF GRAVITATIONAL FORCE

(1) It is always only attraction type force never repel.

(2) It is independent of medium between particles

(3) It is mutual force

(4) It is applicable for very far distant objects like interplanetary distance as well as for very short distance like inter-atomic distance

(5) It is conservative force means work done by this force is independent of path.

Example: Two particles of equal masses move in a Circle of radius r under the action of mutual gravitational attraction force. Then the speed of each particle if the mass of each particle is m .

                                                        

                                              

Solutions:

Here gravitational force F is centripetal force provider

       Answer (c) 

GRAVITATIONAL FIELD

Space of influence surrounding a mass m in which its gravitational effects are effective is called gravitational field of given mass.

Intensity of gravitational field due to point mass

GRAVITATIONAL FIELD OF EARTH HERE WE ARE ASSUMING

(1) Shape of earth is perfect sphere

(2) Center of mass of earth is at its center

(3) Uniform mass distribution

                                                                                                        

 

 

Intensity of earth gravitational field at its surface

Now here E_g = g we will use another term specially for earth is acceleration due to gravity Now At earth’s Surface    Direction of E_g or g is always towards the Center of earth Also weight of any object at the surface of earth W = F = m E_g = mg Force acting towards the center of earth Properties of ‘g’

(1) It is independent of mass of any object

(2) It is not a universal constant its value depends on place, position and planet.

NOTE: For example value of g on the surface of moon is approximately g/6 i.e. 1/6 of the value on the earth

Variation of ‘g’

(1) Due to change of planet

As we know at the surface of earth

       

 

 

 

(2) Due to shape of earth

The earth is not perfectly spherical in shape but is on oblate sphere. It is bulged at the equator and flattened at the pole.

Equatorial radius is 21 km more than polar radius

 

g ® Minimum at equator

g ® Maximum at poles   

At the surface

g_poles > g_equator

(3)  DUE TO ROTATION OF EARTH

USING LAW OF PARALLELOGRAM

R² = P² + Q² + 2 PQ cos a

(mg^’)² = (mg)² + (mrw²)² + 2 mg mrw² cos (p - ϕ)

 

 

 

 

 

At the equator f= 0°

At the poles g^’= g - Rw²

f = 90°, g¹ = g – Rw² cos² 90° g^’ = g

4. At the depth h inside the earth

At the surface of earth

                       

At depth h

           

Here

 

 

    '  

 

 

At the Center of the earth

 

 g’ = 0 weight less

5.  At h height above the earth’s surface

At the surface of earth

           

At the h height above

 

 

 

 

 

If  h << R

Binomial expansion

(1+ x)ⁿ = 1+ nx

If x <<<<<<1

Field due to a Uniform Circular Ring at some Point on its Axis

Field strength at a point P on the axis of a circular ring of radius R and mass M is given by,

 

This is directed towards the center of the ring. It is zero at the center of the ring and maximum at   (can be obtained by putting ).  Thus, E-r graph is as shown in the maximum value is

 

 

KEPLER’S LAW

(1) Law of orbit: Each planet moves around the sun in an

elliptical orbit with the sun at one of the foci.

Here e is eccentricity of an ellipse

 

 

 

(2) LAW OF AREA

The line joining the sun and planet sweeps out equal area in equal interval of time.

 

Here L = angular momentum of planet about the sun

(3) LAW OF PERIODS

Square of time period µ cube of semi major axis of elliptical orbit.

                                                             

Final Touch Points

Polar and Geostationary Satellites

1. Satellites in low polar orbit pass over the poles. They orbit between 100 km and 200 km above the Earth’s surface, taking around 90 minutes to make each orbit. The earth spins beneath the satellites as it moves, so the satellite can scan the whole surface of the earth. Low orbit polar satellites have uses such as

• Monitoring the weather.
• Observing the earth’s surface.
• Military uses including spying.

2. Geostationary satellites have a different trajectory to polar satellites. They are in orbit above the equator from west to east. The height of their orbit-36,000 km is just the right distance so that it takes them one day (24 hours) to make each orbit. This means that they stay in a fixed position over the earth’s surface. A single geostationary satellite is on a line sight with about 40 percent of the earth’s surface. Three such satellites, each separated by 120 degrees of longitude, can provide coverage of the entire planet. Geostationary satellites have uses such as:

• Communications-including satellites phones
• Global positioning or GPS,

Geostationary satellites always appear in the same position when seen from the ground. This is why satellite television dishes can be bolted into one position and do not need to move.

3. Acceleration due to moon’s gravity on moon’s surface is

                                           

while acceleration due to earth’s gravity on moon’s surface is approximately  .  This because distance of moon from the earth’s centre is approximately equal to 60 times the radius of earth and

4. Total energy of a closed system is always negative. For example, energy of planet-sun, satellite-earth or electron-nucleus system is always negative..

5. If the law of force obeys the inverse square law

K=U2=E  

The same is true for electron-nucleus system because there also, the electrostatic force Fe∝1r2.

6. Trajectory of a body projected from point A in the direction AB with different initial velocities:

Let a body projected from point A with velocity v in the direction AB. For different values of v the paths are different. Here, are the possible cases.

2. If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.

 

 

 

 

 

(i) If times period of rotation of earth becomes 84.6 min, effective value of g on equator becomes zero or we feel weightlessness on equator.

(ii) Time period of a satellite close to earth’s surface is 84.6 min.

(iii) Time pendulum of a pendulum of infinite length is 84.6 min.

(iv) If a tunnel is dug along any chord of the earth and a particle is released from the surface of earth along this tunnel, them motion of this particle is simple harmonic and time period of this is also 84.6 min.

Note (a) Points (iii) and (iv) come in the chapter of simple harmonic motion.

(b)  is also the time period of small oscillations of a block inside a smooth spherical bowl of radius R.

But this is not 84.6 min because here R is the radius of bowl not the radius of earth.

This expression can be compared with the time period of a pendulum.

MOMENT OF INERTIA OF A SINGLE PARTICLE

I = mr²                  

 

MOMENT OF INERTIA OF A RIGID BODY

 

Theorem on Moment of Inertia

Theorem of Parallel Axis

I = I_cm +Mr²

Theorem of Perpendicular Axis

Moment of Inertia is independent of Mass depend on distribution of mass

Moment of inertia of some regular shaped rigid bodies

 

 

 

RADIUS OF GYRATION

Let mass of the rigid body = M                    

Now  I = S mr²

 

 

 

 

Here K is radius of gyration

RMS value of

Case I

A thin disc of mass M and radius R has mass per unit area  where r is the distance from its center. Its moment of inertia about an axis going through its center of mass and perpendicular to its plane is    

Given, surface mass density,  

So, mass of the disc can be calculated by considering small element

of area 2prdr on it and then integration it for complete disc, i.e.

 

 

 

                                    ….(i)

Moment of inertia about the axis of the disc,

 

 

 

Case II

Two identical spherical balls of mass M and radius R each are suck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the center of the rod is Key idea This problem will be solved by applying parallel axis theorem, which states that moment of inertia of a rigid body about any axis is equals to its moment of inertia about a parallel axis through its center of mass plus the product of the mass of the body the square of the perpendicular distance between the axis.

We know that moment of inertia (MI) about the principle axis of the sphere is given by

 

Using parallel axis theorem, moment of inertia about the given axis

 

 

Considering both sphere at equal distance form the axis, moment of inertia due to both sphere about this axis will be

 

Now, moment of inertia of rod about its perpendicular bisector axis is given by

 

Here, given that

L = 2R

 

So, total moment of inertia of the system is

 

 

Case III

Seven identical circular planner discs, each of mass M and radius R are welded symmetrically The moment of inertia of the arrangement about an axis normal to the plane and passing through the point P is                                                          

From theorem of parallel axis,

 

 

Case IV Cavity

Form a uniform circular disc of radius R and 9M, a small disc of radius  is removed The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through center of disc is

 

Case V

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. what is the ratio l/R such that the moment of inertia is minimum?

MI of a solid cylinder about its perpendicular bisector of length is

 

                                       

For I to be minimum,

 

 

 

 

 

 

  

Case:- Moment of Inertia

Rod → mass = m, Length = L

Disc → mass = m, Radius = R

Multiple rods

Each rod of mass m and length L

 

  

          

  

 

Torque of Force about the Axis of Rotation

 

 

 

 

 

 

Case 1

A cylinder uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see the figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle q with the vertical, is

As the rod rotates in vertical plane so a torque is acting on it, which is due to the vertical component of weight of rod.

Now, Torque t = force x perpendicular distance of line of action of force from axis of rotation

 

Again, Torque, t = Ia

Where, I = moment of inertia =  

[forced and Torque frequency along axis of rotation

passing through in end]

a = angular acceleration

               

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick as shown in the figure of 2N on the ring rolls it without slipping with an acceleration of 0.3 m/s². The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the rings is .

There is no slipping between ring and ground. Hence f₂ is not maximum. But there is slipping between ring and stick. Therefore f₁ is maximum. Now let us write the equations

 

 

 

 

 

Solving above four equations we get,