Question

2-16. Consider a body falling freely from a height xo according to Figure 2.9a. If we neglect ir resistance or viscous drag, the only force acting upon the body is the gravitational force mg. Using the coordinates in Figure 2.9a, mg acts in the same direction as x and so the differential equation corresponding to Newton's second law is dt Show that where x0 and uo are the initial values of x and u. According to Figure 2.9a, xo-0 and so If the particle is just dropped, then v 0 and so Discuss this solution. Now do the same problem using Figure 2.9b as the definition of the various quantities involved, and show that although the equations may look different from those above, they say exactly the same thing because the picture we draw to define the direction of x, vo, and mg does not affect the falling body. 0 mg m g 7 0 FIGURE 2.9 )A coordinate system for a body falling from a height o and (b) a different coordinate system for a body falling from a height

Answer 1

From figure 2.9(a)

Differential equation of motion is

$m \frac{d^2x}{dt^2}=mg$

Or

$m\frac{dv}{dt}=mg$

Or

$dv=gdt$

Integrating both sides we get

$\int dv=g \int dt$

$v =gt+C_1$ [ where C₁ is an integration constant ]

From the initial condition , at t=0, , v=v₀ . So we find $C_1=v_0$.

So  $v =gt+v_0$

Again

$v=\frac{dx}{dt}=gt+v_0$

or

$dx=[gt+v_0]dt$

Integrating both sides we get

$\int dx=\int [gt+v_0]dt$

Or

$x=\frac{gt^2}{2}+v_0t+C_2$   [ where C₂ is an integration constant ]

From the initial condition , at t=0, , x=x₀ . So we find $C_2=x_0$

Hence

$x=\frac{gt^2}{2}+v_0t+x_0$

If x₀=0, then we get

$x=\frac{gt^2}{2}+v_0t+0=\frac{gt^2}{2}+v_0t$

If the particle is just dropped, then

$x=\frac{gt^2}{2}+0=\frac{gt^2}{2}$

Hence

.$x=\frac{gt^2}{2}+v_0t+x_0$.............................(i)

From figure 2.9(b)

Differential equation of motion is

$m \frac{d^2x}{dt^2}=-mg$

Or

$m\frac{dv}{dt}=-mg$

Or

$dv=-gdt$

Integrating both sides we get

$\int dv=-g \int dt$

$v =-gt+C_3$ [ where C₃ is an integration constant ]

From the initial condition , at t=0, , v=-v_{0​​​​​​​} . So we find $C_3=-v_0$.

So  $v =-gt-v_0$

Again

$v=\frac{dx}{dt}=-gt-v_0$

or

$dx=-[gt+v_0]dt$

Integrating both sides we get

$\int dx=-\int [gt+v_0]dt$

Or

$x=-\frac{gt^2}{2}-v_0t+C_4$   [ where C₄ is an integration constant ]

From the initial condition , at t=0, , x=-x_{0​​​​​​​} . So we find $C_4=-x_0$

Hence

$x=-\frac{gt^2}{2}-v_0t-x_0$............................................(ii)

If x₀=0, then we get

$x=-\frac{gt^2}{2}-v_0t-0=-\frac{gt^2}{2}-v_0t$

If the particle is just dropped, then

$x=-\frac{gt^2}{2}-0=-\frac{gt^2}{2}$

Equation (i) and (ii) both are same

Eqn(i) tells us that the acceleration of the body is along downward direction (toward +ve x- axis, as from figure 2.9(a))

Eqn(ii) tells us that the acceleration of the body is along downward direction (toward -ve x- axis, as from figure 2.9(b))