91Ó°ÊÓ

The centre of the car moves along a circular arc of radius R.

A force is a push or pull on an object is because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them .The acted force may be of attraction or of repulsion .The two items no longer feel the force after the interaction ends

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum$\frac{d\stackrel{â‡\P Ä}{p}}{dt}$and the perpendicular rate of change of momentum$\frac{d\stackrel{â‡\P Ä}{p}}{dt}$are the two elements that we are concerned with.

$$\begin{gathered} F_{net}=\frac{d\stackrel{â‡\P Ä}{p}}{dt} \\ =\frac{d\stackrel{â‡\P Ä}{p}}{d{t}_{ll}}+\frac{d\stackrel{â‡\P Ä}{p}}{dt} \end{gathered}$$

In this case, the force acting on the object is in +y or -y direction, therefore, there is no parallel force and equals zero.

$\frac{d\stackrel{â‡\P Ä}{p}}{dt}=0$

As a result, the rate change is the direction change owing to the perpendicular rate of change.

The net force exerted on the object equals the rate change of the momentum and the magnitude of the perpendicular rate change at speeds much less than the speed of light is given by

$$\begin{gathered} F_{net}=\frac{d\stackrel{â‡\P Ä}{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

Where R the radius of the circle path is, v is the speed of the roller car and m is the mass of the car.

Also, the car is under two forces, the centrifugal force$F_c$ and its weight$mg$, and at the top of the path, both forces are in opposite direction to each other so the net force of both forces is

$F_{net}=F_c-mg$

To make the roller car just barely lose contact with the track at the top of the loop, the net force on the truck must be zero.

$$\begin{gathered} F_{net}=F_c-mg \\ 0=\frac{m{v}^2}{R}-mg \\ \frac{v^2}{R}=g \\ v=\sqrt{Rg} \end{gathered}$$

The speed, in this case, depends on the radius of the circle path.