Projectile Motion

When any object is thrown from horizontal at an angle θ except 90°, then the path followed by it is called trajectory, the object is called projectile and its motion is called projectile motion.

If any object is thrown with velocity u, making an angle θ, from horizontal, then

• Horizontal component of initial velocity = u cos θ.
• Vertical component of initial velocity = u sin θ.
• Horizontal component of velocity (u cos θ) remains same during the whole journey as no acceleration is acting horizontally.
• Vertical component of velocity (u sin θ) decreases gradually and becomes zero at highest point of the path.
• At highest point, the velocity of the body is u cos θ in horizontal direction and the angle between the velocity and acceleration is 90°.

Important Points & Formulae of Projectile Motion

1. At highest point, the linear momentum is mu cos θ and the kinetic energy is (1/2)m(u cos θ)².
2. The horizontal displacement of the projectile after t seconds
   x = (u cos θ)t
3. The vertical displacement of the projectile after t seconds
   y = (u sin θ) t — (1/2)gt²
4. Equation of the path of projectile
   
5. The path of a projectile is parabolic.
6. Kinetic energy at lowest point = (1/2) mu²
7. Linear momentum at lowest point = mu
8. Acceleration of projectile is constant throughout the motion and it acts vertically downwards being equal to g.
9. Angular momentum of projectile = mu cos θ x h, where h denotes the height.
10. In case of angular projection, the angle between velocity and acceleration varies from 0° < θ < 180°.
11. The maximum height occurs when the projectile covers a horizontal distance equal to half of the horizontal range, i.e., R/2.
12. When the maximum range of projectile is R, then its maximum height is R/4.

Lets consider a body is acted by the variable force

a ns w  e r
Energy is involved in a chemical change because of formation and breakdown of bonds. In exothermic reactions energy is released and in endothermic reactions energy is taken in.

Weak Nuclear Force: This force appears only in certain nuclear processes such as the β-decay of a nucleus. In β-decay, the nucleus emits an electron and an uncharged particle called neutrino.This particle was first predicted by Wolfgang Pauli in 1931.

Density = Mass/Volume  = 5.74/1.2 = 4.783 g/cm³ 
Here least significant figure is 2,
so density = 4.8 g/cm³ 

it is based on Young's modulus concepts .

we know, Young's modulus = stress/strain
stress = F/A , strain = ∆L/L

so, Young's modulus, Y = FL/A∆L
so, F = YA∆L/L
hence, it is clear that tension in rope is directly proportional to area of cross section of rope.

area of cross section of rope , A = πd²/4

so, 
we can write , 

so, 



hence, maximum tension that may be given to a similar role of diameter 2cm is 2000N

(4)

not accelerating .

Reason:

Forces that are equal in size but opposite in direction are called balanced forces. Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force

Acceleration of a body moving with a uniform velocity = 0

 because acceleration = change in velocity​/time

 change in velocity= 0

SO, acceleration = 0

Earth attracts all things towards it through an unseen force of attraction. This force of attraction is called as gravitation or gravitational pull. You must have noticed that every time you throw an object upwards, it reaches a certain height and then falls down on the earth's surface

Using S = ut + (1/2)at²

S = 500 m

u = 0   Start from rest

a = ?

t = 30 sec

=> 500 = 0 + (1/2)a(30)²

=> a = 1000/900

=> a = 10/9  m/s²

V = u + at

=> V = 0 + (10/9) * 30

=> V = 100/3 m/s

velocity of the aeroplane at the take off​ = 100/3 m/s

We know that, distance = speed×time

So, d = 50×40 = 2000m

d = 2×3.14×r (circumference of the circle)

=> 2000=2×3.14×r

On calculating, r = 318.47 m

In this case, required acceleration will be centripetal acceleration.

So, a = (v×v)/r

        a = (50×50)/318.47

        a = 7.85m/s^2

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the center of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.

In a Projectile Motion, there are two simultaneous independent rectilinear motions:

1. Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
2. Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.

The expression for rigid body’s “rotational kinetic energy” is 

SOLUTION:

A body undergoing a rotational dynamics with angular velocity ω will also poses a translational motion. Let the velocity for translation motion be ‘’v’’.

We know for translational motion the energy poses by body is kinetic and given as

Also we know that for a body obeying rotation  

v=rω

Substituting in translation

Also there is a relation for moment of inertia i.e 

Substituting we get,

Motion of a car on a banked road

In the vertical direction (Y axis)

Ncosϴ = fsinϴ + mg --------------------(i)

In horizontal direction (X axis)

fcosϴ + Nsinϴ = mv²/r  ----------------(ii)

Since we know that f      μ_sN   

For maximum velocity, f = μ_sN

(i)becomes:

       Ncosϴ = μ_sNsinϴ + mg

Or, Ncosϴ - μ_sNsinϴ = mg

Or, N = mg/(cosϴ- μ_ssinϴ)

Put the above value of N in (ii)

μ_sNcosϴ + Nsinϴ = mv²/r 

μ_smgcosϴ/(cosϴ- μ_ssinϴ) + mgsinϴ/(cosϴ- μ_ssinϴ) = mv²/r 

mg (sinϴ + μ_scosϴ)/ (cosϴ - μ_ssinϴ) = mv²/r 

Divide the Numerator & Denominator by cosϴ, we get

v² = Rg (tanϴ +μ_s) /(1- μ_s tanϴ)

v = √ Rg (tanϴ +μ_s) /(1- μ_s tanϴ)

This is the miximum speed of a car on a banked road.

Special case:

When the velocity of the car = v₀ ,

• No f is needed to provide the centripetal force. (μ_s =0)
• Little wear & tear of tyres take place.

             v_o = √ Rg (tanϴ)

 Problem: A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the

 (a) optimum speed of the racecar to avoid wear and tear on its tyres, and

(b) maximum permissible speed to avoid slipping ?

 Solution.

R = 300m

ϴ = 15^o

μ_s  = 0.2

• v_o = √ Rg tanϴ

      = √300 * 9.8 * tan 15^o

      = 28.1 m/s

• v_max = √ Rg (tanϴ +μ_s) /(1- μ_s tanϴ)

             = √ 300 * 9.8 * (0.2 + tan 15^o)/(1-0.2tan15^o)

             = 38.1 m/s

Newton’s second law of motion states that the force exerted by a body is directly proportional to the rate of change of its momentum. For a body of mass ‘m’, whose velocity changes from u to v in time t, when force ‘F’ is applied.

Relation between linear velocity and angular velocity

Let us consider the randomly shaped body undergoing a rotational motion as shown in the figure below. The linear velocity of the particle is related to the angular velocity. While considering the rotational motion of a rigid body on a fixed axis, the extended body is considered as a system of particles moving in a circle lying on a plane that is perpendicular to the axis, such as the center of rotation lies on the axis.

In this figure, the particle P has been shown to rotate over a fixed axis passing through O. Here, the particle represents a circle on the axis. The radius of the circle is the perpendicular distance between point P and the axis. The angle indicates the angular displacement Δθ of the given particle at time Δt. The average angular velocity in the time Δt is Δθ/Δt. Since Δt tends to zero, the ratio Δθ/Δt reaches a limit which is known as the instantaneous angular velocity dθ/dt. The instantaneous angular velocity is denoted by ω.

From the knowledge of circular motion, we can say that the magnitude of the linear velocity of a particle traveling in a circle relates to the angular velocity of the particle ω by the relation υ/ω= r, where r denotes the radius. At any instant, the relation v/ r = ω applies to every particle that has a rigid body.

If the perpendicular distance of a particle from a fixed axis is r_i, the linear velocity at a given instant v is given by the relation,

V_i = ωr_i

Similarly, we can write the expression for the linear velocity for n different particles comprising the system. From the expression, we can say that for particles lying on the axis, the tangential velocity is zero as the radius is zero. Also, the angular velocity ω is a vector quantity which is constant for all the particles comprising the motion.

Let us consider the randomly shaped body undergoing a rotational motion as shown in the figure below. The linear velocity of the particle is related to the angular velocity. While considering the rotational motion of a rigid body on a fixed axis, the extended body is considered as a system of particles moving in a circle lying on a plane that is perpendicular to the axis, such as the center of rotation lies on the axis.

In this figure, the particle P has been shown to rotate over a fixed axis passing through O. Here, the particle represents a circle on the axis. The radius of the circle is the perpendicular distance between point P and the axis. The angle indicates the angular displacement Δθ of the given particle at time Δt. The average angular velocity in the time Δt is Δθ/Δt. Since Δt tends to zero, the ratio Δθ/Δt reaches a limit which is known as the instantaneous angular velocity dθ/dt. The instantaneous angular velocity is denoted by ω.

From the knowledge of circular motion, we can say that the magnitude of the linear velocity of a particle traveling in a circle relates to the angular velocity of the particle ω by the relation υ/ω= r, where r denotes the radius. At any instant, the relation v/ r = ω applies to every particle that has a rigid body.

If the perpendicular distance of a particle from a fixed axis is r_i, the linear velocity at a given instant v is given by the relation,

V_i = ωr_i

Similarly, we can write the expression for the linear velocity for n different particles comprising the system. From the expression, we can say that for particles lying on the axis, the tangential velocity is zero as the radius is zero. Also, the angular velocity ω is a vector quantity which is constant for all the particles comprising the motion.