Mechanics Notes

UNIT & DIMENSION

1parsec=3.26 light years
1 barn= 10^-28 m²
1 shake = 10^-8 sec
1 light year= 9.45 x 10¹⁵ m
1 Astronomical unit= 1.5 X 10¹¹m
Dimension finding techniques
For Gradient – divide by L
F= [MLT^-2]
Use formulas and dimension of force to find other dimension.
V=[ML² T^-3 A^-1]
You can use this to find dimension of resisitance too.
As, V=IR
R=V/I put dimension of I and V you will get dimension of R.
 

is dimension less

           

has dimension of velocity.

Linear momentum- mass x velocity
Angular velocity- mass x velocity x distance
L/r = √LC =CR= [T]

Uses of Dimension

Testing correctness of a relation.
Identifying unknown quantity in a relation.
Derivation of relation for a physical quantity.
Conversion from one system to another system of unit

Dimensionless Quantities

Relative density ( ρ = ρ_m / ρ_w)
Plane angel ( θ = s / r)
Solid angle (ω = A / r² )
Strain ( = Δl / l)
Poission's ratio ( P_r = Δd/d ^ₓ  l/Δl )
Loudness ( L = log(I / I_o))
Mechanical equivalent of Heat ( J = W / H)
Emissivity (E = e / e_b)
Relative permitivity or dielectric constant (K = ϵ_r =  ϵ / ϵ_o )
Relative permeability (μ_r = μ / μ_o )
Refractive index (μ = c / v)
Magnification (m = v / u)
Dispersive power (ω = (δ_v - δ_r) / δ )

Quantities Having Identical Dimensions

Acceleration (a or g) and Gravitational field strength (E=mg/m) have dimension [M⁰L¹T^-2]
Force and Energy gradient (F*l/l = F) have dimension [M¹L¹T^-2]
Surface tension (=F/l) and Spring constant (k=F/x) have dimenion [M¹L⁰T^-2]
Angular velocity (ω=2πf), frequency (f), velocity gradient (v/l = l/tl = f), radioactive decay constant have dimension [M⁰L⁰T^-1]
Pressure (=F/A), stress (=F/A), elastic modulii (γ,Ƞ,k=stress/strain), energy density (Fl/Al) have dimension [M¹L^-1T^-2]
Work (=F*l), energy, torque (=r ₓ F) have dimension [M¹L²T^-2]

SCALAR & VECTOR

Scalar quantities: Distance, volume, mass, time, work energy power, charge, electric current, potential, electric flux, magnetic flux etc.
Vector quantities: Displacement, momentum, force, field strength, potential gradient, magnetizing force.
Note: Electric current possesses both magnitude & direction but is Scalar, because it doesn't obey the laws of vector addition.
Note: But moment of inertia is a tension quantity.

To be a vector

Necessary condition---------------> should have direction
Sufficient condition---------------> should obey law of vector addition

Gradient of scalars are always vector: eg. temperature, gradient, velocity gradient, pressure gradient, potential gradient, energy gradient etc.

 Field intensity is always vector: eg: Gravitational intensity, magnetic intensity, electric field intensity.

• Minimum no. of collinear vector giving zero resultant is 2.

• Minimum no. of collinear vectors of unequeal magnitude giving zero resultant is 3.

• Minimum no. of co-planar vectors giving zero resultant is 3.

• Minimum no. of non co-planar vectors giving zero resultant is 4.

Axial vectors: Angular displacement, torque
F.S is a scalar, r. F is a vector quantity.
If resultant of two force F1 & F2 is F, & the resultant is at right angle to the force F1, then the force F2 is equal to 2F1. If al = |6|= A, then |a + b= 2Acos

Note: (Angle between A +B and A x B is 90⁰)
A.( B x A) is always zero.
 If A +B+c =0 then
A.(BxC)= 0
If lal= |bl= A, then |a - b|=2Asin x/2

MOTION IN STRAIGHT LINE

If V1, V2 & V3 be the velocities of a body to cover equal distance then average velocity of the body along the source will be:

Hint: Av. Velocity = total displacement / total time

Similarly,

For velocity V1 & V2

Note:

• When a body thrown vertically upwards reaches highest point
  • velocity is zero but acceleration is still equal to g.
• To cross a river in least possible time start swimming at right angle to flowing river.
• To cross along shortest path = swim at angle 90°+ sin (u/v) were,
  • u = velocity of water
  • v= velocity of person in still water.

To cross the river in the shortest time it ( boat or swimmer) should try to move straight .
To cross the river in shortest distance boat or swimmer should cross making angle θ with the straight line opposite to the flow of water.

PROJECTILE MOTION

Path followed by a body is parabolic only if initial velocity & acceleration act on a body at angles other than 0° to 180°.
Projectile motion is parabolic in nature & path is called trajectory.
K.E. of projectile at maximum height = E0 cos2 Ө
Hint: K.E. = ½ mv² =1/2 m (u cos Ө)²        [v = u cos Ө]
= ½ mu² cos²Ө
= E0 cos² Ө
Potential energy= E0 sin² Ө

	When object Reaches to highest point	When object returns to ground
1. Change in speed	u(1-cos Ө) = 2u = sin2 Ө /2	Zero
2. Change in velocity	u sin Ө	2u sin Ө [Hint: Directions changes so add]
3. Change in direction	Ө	2 Ө
4. Av. velocity	(u√ 3cos2 Ө + 1)/2	ucos Ө
5. Change in momentum	mu sin Ө or P0 sin Ө	2mu sin Ө or 2P0 sin Ө

At highest point,                  

Velocity = ucosθ                                                   

P = mu cosθ                                                                       

L= mu cosθ x Hmax                                                                  

∆p = musin θ                                                          

Torque = mg x R/2                                    

Y=x tanθ -1/2*g*(x/ucosθ)²

At final point

velocity = u

P =mu

L = mu x Rsinθ

∆P = 2musin θ

torque = mg x R

👉If a passenger drops a body from a uniformly moving vehicle, the trajectory would appear a straight line for the passenger

o  but it would appear a parabola to a person on ground.

👉When friction is present, time of ascend < time of descend.

Q)A body released from height h reaches ground in time t then,

1)h/2 distance in time t/√2

2) in t/2 time it travels 3h/4 distance.

👉Ball crosses top of tower in t1 and t2 time

 t1 + t2 =larger time

t1 - t2 = smaller time

height =1/2 gt1t2

👉Maximum height and range is related as

R = 4Hcotθ

👉If h1 and h2 are maximum height for same range then R = 4 √(h1h2) and one height is triple of another.

T=2usin(α-β) /gcosβ
R= 2u²sin(α-β)cosα/gcos²β
H= u²sin2(α-β)/2gcosβ
Range is maximum for θ= α-β = π/4 – α/2

NEWTON'S LAWS OF MOTION

(a) 1 law (law of inertia): Gives qualitative definition of force
b) 2nd law: gives measurement (quantitative definition) of force
(c) 3 law: gives property of force.
When lift moves with constant acceleration

• Upwards: R-mg = ma
• So, R m(g +a)
• i.e. Apparent wt. > real wt.
• Tensions in rope> wt. of the system
• Downwards: mg - R= ma
• So, R = M (g-a)
• If a>g, R (apparent wt) will be negative. So man will be exerting his weight upwards.

In case of gun (mass 'M’) firing bullet (mass 'm') with velocity u, recoil velocity V= mu/M & centre of mass remains stationary.
Fig .
T₁=(m₂ +m₃)/(m₁+m₂ +m₃) *F
Mass behind the point where torque is to be measured is added at numerator and total mass in the denominator.
Fig.
T= L-x /L x F + m/L Xg
Fig:
 Acceleration (a)=(m₁gsinα –m₂gsinβ )/(m₁+m₂)
Torque (T)= m₁m₂/(m₁+m₂) x g (sinα +sinβ)
Fig:
Acceleration (a)= (m₁gsin90⁰ – m₂gsin0)/m₁ +m₂ = m₁/(m₁ +m₂) x g
Torque (T)= (2m₁m₂/(m₁ +m₂)) x g
Thrust on pulley = √2 T

FRICTION

Friction force:
(i) Acts in opposite direction to motion
(ii) Acts parallel to the surface in contact
Rolling friction< kinetic friction< static friction or, μ_r< μ_k<μ_s
Stopping distance, S = V²/2μg α 1/μ  when a body moving with initial Speed v on a surface comes to rest after travelling a certain distance 's’ in time 't’.
When an iron chain of length L lies on a rough table of coefficient of friction μ,then the fraction of chain that can just overhang the table is {μ/(μ+1)}L.
Hint: for just overhanging, frictional force ≥ wt. of hanging part x g
So, μ  x mass per unit length x wt. of part on table x g ≥ μ x mass per unit length x wt. of part over hanging x g
Or, μ x M( L-x)g/L   ≥ Mg/L
or, μ (L-x) ≥ x
 or, μL-μx  ≥ x
or, μL ≥ x + μx
or, x ≤ μL/(μ+1)
x/L =μ / (μ+1)
stopping acceleration = μg
stopping distance = v²/2μg
time = v/μg
change in pd after rotating by an angle θ =mgl/2 (1- cosθ)
climb up ,                                                                  climb down,
a = g (sinθ + μcosθ )(rough surface)                 a= g (sinθ –μcosθ) (rough surface)
                                                                                    a= gsinθ (smooth)
μ = (1-1/n²)tanθ
μ = total no. of equal part /total rough part  xtanθ
μ = (n² -1)/(n²+1) --------à climbing up and down
pulling force = mgsinθ/cos(α-θ)
minimum force to slide the body on surface = μmg/√(μ²+1)
minimum force to held a body against vertical wall = mgcotθ or, mg/μ

CIRCULAR MOTION

a_c = v²/r         V _A =√(rg)
                        V_B = √(3rg)
                        V_C= √(5rg)
Body slips tangentially at h=R/3 from centre of circle.
Leveled road,
F₁ +F₂ = mv²/r                                              v=√(μrg)-----à (max safe velocity)
R₁ + R₂ = mg
Banked road,
Rsinθ = mv²/r                                              v=√(rgtanθ) ----------à ( max safe velocity)
R cosθ = mg
Tanθ = v²/rg
Neutron striking α particle (  ₂He⁴ )= 4x/(1+x)² = 4 x4 /(1+4)²
Uniform circular motion:
has constant speed, variable velocity & acceleration.

• is only possible in horizontal plane

• is periodic

A. Unchanged	B. Changed in direction, not in magnitude
1. Speed 2. Angular speed 3. Kinetic energy 4. Angular momentum	1. Velocity  2. Linear momentum  3. Acceleration 4. Displacement
C. Remaining zero	D. Acting radially towards centre
1. Work done 2. Angular acceleration	1. Force so called centripetal 2. Acceleration
3. Moment of force about axis
E. Acting tangentially 1. Velocity 2. Momentum Note: Angular velocity is perpendicular to plane of circular moion as given by right hand screw rule. So, a ┴v┴ω.

• To be in a vertical circular motion:

(a) Minimum tension at highest point ≥ 0 (zero)
(b) Minimum velocity at highest point ≥√gr
 (c) Min velocity at the bottom ≥√5gr
(d) When a mass 'm is attached to one end of a rod of length the other end is hinged at centre of a vertical circular path then minimum velocity that must be imparted at the bottom to complete the circle will be  ≥√4gr.
Hint: As the mass is attached to the end of a rod which doesn't slacken, therefore, taking v= 0 at the highest point from:
v²-u² = 2as
0 – u² = 2as
0 – u²= 2(-g)2l
Therefore, u = √4gl

• When a body moves from the lowest point to highest point in circular motion.

(a) Vel _{lowest point}  - Vel _{highest point} = √4gr
Hint: Change in K.E. = change in P.E.
½ mv² _{lowest point}   - ½ mv² _{lowest point}  = mg x 2r
(since difference in highest and lowest point is 2r)
(b) Difference in tension =  6mg
When a block resting on the top of sphere of radius R is released, it slipst tangentially after falling through a height of R/3 or a height o of 5R/3 from the ground.
Hint: At any point, in a sphere,
mg cosθ -normal reaction (N)= mv²/r
N = mgcosθ – mv²/r                                               
Block slips tangentially if N= 0
or, 0 = mgcosθ- mv²/r

Also,
or, cosθ = v²/ gR                                                                                    
v²= u²+2gh
so, cosθ= 2h/R
But cosθ =( R-h)/R [From fig]
(R-h)/R = 2h/R
H = R/3 [from top]
or, H = 5R/3 [from bottom]

ROTATIONAL MOTION

• When a body rolls on the ground without slipping:

(a)Rotational K.E. =½ IW² =½ K² MV² / R²
(b) Translational K.E =½ MV²
(c)Total energy E=1/2IW² + 1/2MV²  = 1/2MV²(K²/ R² + 1)
(d) E_rotational : E _{translational} : E _total = K²: R²: (K²+R²)
Velocity of body rolling down an inclined plane of inclination θ  without slipping from rest has: velocity at bottom (V)= √ 2gh/1+k²/r²
and Acceleration a = gsin θ/ 1 + k²/r²

• In equilibrium, acceleration & angular acceleration is zero.

Note: A body with minimum potential energy in stable equilibrium. A body with maximum potential energy is in unstable equilibrium.
 If 2 solid spheres of same material have radii in the ratio 1:2, then their moment of inertia will be the ratio of 1:32 [Hint: M.I for solid sphere = 2/5MR²]
Since, both the spheres are or same material; their densities are equal. So,
I = 2/5 x ρ x V x R²
I = 2/5 x ρ x 4/3 x π x R³ x R²
I = 2/5 x ρ x 4/3 x π x R⁵
I₁/I₂ ∝ R₁⁵/ R₂⁵
I₁/I₂ = R₁⁵/ R₂⁵ = (1/2)⁵= 1:32
Note:

If an object moves down a smooth inclined plane of height h	If it rolls from rest without slipping
.vel. at bottom V0 =√2gh	V  = √2h/g( 1 + K2/R2)
. acceleration at bottom a= gsin θ	A= gsin θ/(1 + K2/R2) = a0/(1 + K2/R2)
. time of fall t θ = √2h/g cosecθ	t  = √2h/g( 1 + K2/R2) cosecθ        t0=√ ( 1 + K2/R2)

WORK, ENERGY, POWER & COLLISION

Work done = Mgl/2n²
Power= (blood pressure  x volume )/ time taken
A bullet losses 1/n velocity after penetrating x distance then total distance travelled= nx²/(2n-1)
Bullet losses 1/n K.E after penetrating x distance then total distance travelled = nx
No. of planks required to stop the bullet ,
N = (n+1) /2 (odd)
N= (n+2)/2 (even)
h_n=e²ⁿ h₀
v_n=eⁿ h₀                     e= v/u = √(h/h₀)

• 1 H.P. =746 watts

• K.E. = ½ mv² = P²/2m is never negative.

• When two bodies of masses m & 2m are moving with same K.E., the ratio of their linear momentum is 1: √2.
• If you have a heater of 1000 W, then energy consumed by heater in 2 hours is 2 KWh.

• Change in potential energy when an object is taken from earth’s surface to height 'h'.

• Δ U = mgh[R/(R + h)]
  • [Hint: U=-GMm/(R+h)      = mgR²/(R+h)]

Note:
(a) When h<<<R, R/ (R+h) =1
So, Δ U=mgh
b)When h>>>R, h/ (h+R) =1
So, Δ U =mgR

• Elastic potential energy stored in wire
  • U = ½ kx²
  • Where x= extension, k = force constant

Conservation forces:

• Eg. Gravitational force, Electrostatic force, Magnetostatic force.

Non conservative forces

• eg: Frictional force, viscous force
• Hint: All conservative forces are central forces but vice versa may not be]

In collision
e (coefficient of restituton)= velocity of separation / velocity of approach = (V₂-V₁)/(u₁-u₂)
 For elastic collision, e=1
 For perfectly inelastic collision, e =0
For collisions in practice, 0 <e<1

	Elastic collision	Inelastic collision
1. Linear momentum	Conserved	Conserved
2. Total energy	Conserved	Conserved
3. K.E	Conserved	Not conserved
4. Forces involved	Must be conserved	Any
5. Mechanical energy	Conserved	Depends on forces involved
6. Value of 'e’	1	<1

• When earth's a body is dropped from a height h₀, gets in elastic collision with earth's surface and the body rebounds to height h1 then, height raised after bounce.
  • h_n =e²h₀

• If it follows successive collision and finally comes to rest then, height raised after nth bounce
  • h₀ = (e²) h₀

In an elastic collision:
Velocity of approach = Velocity of separation
i.e. u₁-u₂=  v₂-v₁ & e = (v₂-v₁)/( u₁-u₂)
 Note:
(i) When two bodies of equal masses collide each other in an elastic collision, they interchange their velocities i.e. v₁ = u₂ and v₂ = u₁
 (ii) When one is stationary, the stationary one will take the velocity of moving one i.e. v₁ = 0, v₂ = u₁, when m₁=m₂
When mass of one particle (A) <<< other particle (B) & B is at rest i.e. m1 << m2 & u2 = 0, then particle A will return back with same speed i.e. v1=-u1 & v2= u2
When body A is huge & B is at rest, then A will continue its initial velocity &B will move with double the velocity to that of A
i.e. v₁  ≈u₁ & v₂  ≈2u₁

GRAVITATION

Dimension of G [M^-1 L³ T^-2 ]

• Variation in 'g' due to shape i.e. g _poles - g _equator = 18cm/s² .
•  Variation in 'g' due to rotation:
  •  g'= g ω² R cos²θ where θ= 7.29 * 10^-5 rad/s is angular velocity.

At equator,
g'= g –  ω² R
∆g=g-g' = ω² R = 0.034m/s² [At equator]
[But no effect on 'g' at poles]
g _poles > g _equator by 0.052 m/s²
Note:
(a) If earth stops rotating, value of 'g' will increase by 0.034 m/s² because it is decreased by equal amount due to rotation.
(b) For the objects on the earth's surface to start flying, earth should rotate with an angular velocity 17 times more than the present. Then length of day will be of 84 minutes.

• 'g will be half at height (√2-1) R height from earth's surface.

Similarly, it will be 1/3rd at height (√3-1) R from earth's surface.
[Hint:'g' will be 1/nth  time at height (√n-1) R]

• Escape velocity (V_e)& orbital velocity (V₀) are in the ratio of √2:1
• If a body is projected thrown from the earth's surface at velocity n times the escape velocity (V_e ), its velocity on the free space will be√ n² -1) V_e.  
• So, When projected with twice the Ve, it will be√3 V_e. in free space.

• Total energy & potential energy of a satellite are always negative & K.E. is positive.

•  K.E= -Total energy

• P.E.= -2KE

• P.E = 2 Total energy

Rocket launched with	follows
Escape velocity	Parabolic motion
V < Vesc.	Spiral path falls back to earth
Vorb <V < Vesc	Elliptical path
V> Ve	Hyperbolic path

SIMPLE HARMONIC MOTION

K.E = ½ mw² (a²-y²) [P.E if only y² and T.E if only a²]
Time period of oscillation in earth’s hole = 84.6 min
Frequency if scaler = 2f
                         If vector=f
Total mechanical energy = 0
1 oscillation = 4 amplitude
F=kx   k=YA/L
FIG-----

ELASTICITY

SURFACE TENSION

HYDROSTATICS