Motion in Plane

Projectile Motion

When an object is in flight after being thrown or projected, the object is called projectile and its motion is called projectile motion. The motion of a projectile may be considered as the result of two separate but simultaneously occurring components of motion. One component is along a horizontal direction without acceleration, and other component is in vertical direction with constant acceleration due to force of gravity.

Let us assume there is negligible effect of air resistance on the motion of projectile. Let us assume that the projectile is launched with velocity v₀ that makes an angle θ₀ with x-axis.

Once the object has been projected, acceleration due to gravity is directed vertically downward and is acting on it.

`a=-gj`

Or, `a_x=0`, and `a_y=-g`

The components of initial velocity are:

`v_(0x)=v_0co\s\θ_0`

`v_(0y)=v_0si\n\θ_0`

If we take initial position as origin, then x₀ = 0 and y₀ = 0

`x=v_(0x)t=(v_0co\s\θ_0)t`

`y=(v_0si\n\θ_0)t-1/2g\t^2`

These equations give x- and y-coordinates of position of a projectile at time t in terms of two parameters, viz. initial speed v₀ and projection angle θ₀. Here, the choice of mutually perpendicular axes for the analysis of projectile motion has resulted in a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y-component changes, the way it happens in case of free fall. So, at the point of maximum height v_y = 0 and hence,

`θ=ta\n^(-1)(v_y)/(v_x)=0`

Equation of path of a projectile

It can be seen by eliminating the time between the expressions for x and y as follows:

`y=((ta\n\θ_0)x-g/(2(v_0co\s\θ))^2)x^2`

Since g, θ₀ and v₀ are constants, this equation is in the form of a quadratic equation. We know that quadratic equation is the equation of a parabola. Hence, it is clear that path of the projectile is a parabola.

Time of Maximum Height

Let us assume the time taken to reach maximum height is t_m. At this point v_y = 0, so we get following equation:

`v_y=v_0\si\n\θ_0-g\t_m=0`

Or, `t_m=(v_0\si\n\θ_0)/g`

Total time of flight of projectile can be obtained putting y = 0. We get following equation for total time:

`T_f=2(v\si\n\θ_0)/g`

Maximum height of projectile

Maximum height can be calculated as follows:

`h_m=((v_0\si\n\θ_0)^2)/(2g)`

Horizontal Range of Projectile:

`R=(v_0^2si\n2θ_0)/g`

This equation shows that R is maximum when sin 2θ₀ is maximum, i.e. when θ₀ = 45°

So, the maximum horizontal range `=R_n=(v_0^2)/g`

Uniform Circular Motion

Let us assume an object is moving with uniform speed v in a circle of radius R. The object undergoes acceleration because its velocity is continuously changing direction.

Let us assume that r and r' are position vectors, and v and v' are velocities of object when it is at point P and P'. We know that in case of circular motion, velocity at a point is along the tangent at that point in the direction of motion.

In this figure, velocity vectors v and v' are shown as tangents at points P and P'.

Using the triangle law of addition, we get Δv (as shown in smaller figure).

As the path is circular, v is perpendicular to r, and similarly v' is perpendicular to r'

We know that average acceleration along Δv is as follows:

`a_(av)=(Δv)/(Δt)`

So, average acceleration is perpendicular to Δr.

If Δv is placed on the line that bisects the angle between r and r' it is clear that it is directed towards the centre of the circle. Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre.

Centripetal acceleration is given by following equation:

`a_c=(v/R)v=(v^2)/R`

Another way of describing velocity and acceleration of object in uniform circular motion:

It can be defined in terms of angular speed (ω) which is the time rate of change of angular displacement.

`ω=(Δθ)/(Δt)`

If the distance travelled by the object during time Δt is Δs then

`v=(Δs)/(Δt)`

Since Δs = R Δθ, hence:

`v=R(Δθ)/(Δt)=Rω`

Or, `v= Rω`

Centripetal acceleration a_c can be expressed in terms of angular speed as follows:

`a_c=(v^2)/R=(ω^2R^2)/R=ω^2R`

Or, `a_c =ω^2R`

In terms of frequency ν we get following equations for circular motion:

`ω=2πν`

`v=2πRν`

`a_c=4π^2ν^2R`