91Ó°ÊÓ

A car of mass is moving over the rounded top of a hill of radius R At the instant when the car is at the very top of the hill, the car has a speed v.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum $\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\left|}\mathbf{\right|}}}$and the perpendicular rate of change of momentum$\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{âŠ\yen }}}$are the two elements that we are concerned with. As a result, the object's net force ${\mathbf{F}}_{\mathbf{net}}$is given by

$$\begin{gathered} {\mathbf{F}}_{\mathbf{net}}\mathbf{=}\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{p}}}{\mathbf{dt}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{p}}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{p}}}{\mathbf{dt}} \end{gathered}$$

The automobile is subjected to three forces (the system). The first force is the friction force f between the car and the hill, which causes the car to slow down; the second force is the hill's upward force${\mathrm{F}}_{\mathrm{N}}$; and the third force is the weight of the car owing to gravitational force.

The non-negligible force on the system is friction force ,the upward force is and the gravitational force is mg

Three forces are applied to the car in the free-body illustration below.

The graph represents the car is at the very top of the hill

The car's speed is constant, the change in magnitude of the momentum rate is zero, and it equals the parallel component of the momentum.

$\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{\left|\right|}}=0$

The upward force${\mathrm{F}}_{\mathrm{N}}$equals the rate change, which is the change in direction owing to the perpendicular rate of change. At speeds significantly slower than the speed of light, the magnitude of the perpendicular rate change is given by

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}=\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{\mathrm{dt}} \\ =\frac{{\mathrm{Mv}}^2}{\mathrm{R}} \end{gathered}$$

Where R is the circle path's radius, v is the car's speed, and is the car's mass. You should also be aware that the upward force is the perpendicular force that keeps the car spinning.

The force exerted by the road on the car is$\frac{{\mathrm{Mv}}^2}{\mathrm{R}}$

It's worth noting that the normal force (N) is equal to zero for a given speed (v).In the vertical direction, we write Newton's second law for the automobile. Also, the centripetal acceleration $\left({\mathrm{a}}_{\mathrm{c}}\right)$is equal to the sum of vertical forces (in which the downhill direction is positive). In order to determine the speed at which the road ceases to exert an upward force on the car, we set (N) to zero in this equation. Finally, determine

$$\begin{gathered} \underset{}{∑{\mathrm{F}}_{\mathrm{y}}=\mathrm{mg}-\mathrm{N}} \\ ={\mathrm{ma}}_{\mathrm{c}} \\ \mathrm{N}=0 \\ \cancel{\mathrm{m}}\mathrm{g}=\cancel{\mathrm{m}}\frac{{\mathrm{v}}^2}{\mathrm{R}} \\ \mathrm{v}=\sqrt{\mathrm{gR}} \end{gathered}$$

The force exerted by the road on the car be zero is $\sqrt{\mathrm{gR}}$.