Faiq said:
The equations aren't resulting in any contradiction but I want to understand the reason.
The problem is whenever the velocity is horizontal and the acceleration is perpendicular to it, there's no component of acceleration to reduce the horizontal component but there's a component of acceleration to increase the vertical component. So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component
The problem is,
you're not using a sufficiently accurate finite difference approximation.
Consider the times t = 0 and t = Δt.
At t = 0,
$$x(0)=r$$
$$y(0)=0$$
$$v_x(0)=0$$
$$v_y(0)=v$$
$$T_x(0)=-T=-m\frac{v^2}{r}$$
$$T_y(0)=0$$
At t = Δt,
$$x(\Delta t)=r\cos{\left(\frac{v}{r}\Delta t\right)}\approx r\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$y(\Delta t)=r\sin{\left(\frac{v}{r}\Delta t\right)}\approx v\Delta t$$
$$v_x(\Delta t)=-v\sin{\left(\frac{v}{r}\Delta t\right)}\approx -\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=+v\cos{\left(\frac{v}{r}\Delta t\right)}\approx v\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$T_x(\Delta t)=-m\frac{v^2}{r}\cos{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^2}{r}\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$T_y(\Delta t)=-m\frac{v^2}{r}\sin{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^3}{r^2}\Delta t$$
The differential equations for the force balances are: $$m\frac{dv_x}{dt}=T_x$$
$$m\frac{dv_y}{dt}=T_y$$
The
first order (in Δt) forward finite difference approximations to these differential equations over the time interval Δt are:
$$m\frac{v_x(\Delta t)-v_x(0)}{\Delta t}=T_x(0)=-m\frac{v^2}{r}$$
$$m\frac{v_y(\Delta t)-v_y(0)}{\Delta t}=T_y(0)=0$$
The solution to these equations for the velocities at time Δt are:
$$v_x(\Delta t)=-\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=v$$
By comparing with the values of ##v_x(\Delta t)## and ##v_y(\Delta t)## with the second order accurate values above, we see that the x velocity is accurate to terms of second order in ##\Delta t##, but the y velocity is not. Furthermore, for this approximation, if we take the sum of the squares of the velocity components, we obtain:$$(v^2_x+v^2_y)_{(t=\Delta t)}=v^2\left(1+\frac{v^2}{r^2}(\Delta t)^2\right)$$
So, to this level of approximation, the sum of the squares of the velocity components has increased by a term proportional to ##(\Delta t)^2##. However, even here, as the time interval Δt becomes smaller, the increase becomes less and less.
Now let's consider the second order finite difference approximation. In this approximation, we use the trapazoidal rule, and write:
$$m\frac{v_x(\Delta t)-v_x(0)}{\Delta t}=\frac{(T_x(0)+T_x(\Delta t))}{2}=-m\frac{v^2}{r}$$
$$m\frac{v_y(\Delta t)-v_y(0)}{\Delta t}=\frac{(T_y(0)+T_y(\Delta t))}{2}=-m\frac{v^3}{2r^2}\Delta t$$
The solution to these equations for the velocities at time Δt are:
$$v_x(\Delta t)=-\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=v-\frac{v^3}{2r^2}(\Delta t)^2$$
These are in total agreement with the equations above for the 2nd order approximations to the velocity components. Furthermore, for this approximation, if we take the sum of the squares of the velocity components, we obtain: $$(v^2_x+v^2_y)_{(t=\Delta t)}=v^2(1+terms \ of \ order \ (\Delta t)^4)$$
So, by using a more accurate finite difference approximation, we have come much closer to numerically satisfying the condition that the speed of the mass is constant.
