Question

A sports car (and its occupants) of massis moving over the rounded top of a hill of radius RAt the instant when the car is at the very top of the hill, the car has a speed v. You can safely neglect air resistance.

(a) Taking the sports car as the system of interest, what object(s) exert non negligible forces on this system?

(b) At the instant when the car is at the very top of the hill, draw a diagram showing the system as a dot, with force vectors whose tails are at the location of the dot. Label the force vectors (that is, give them algebraic names). Try to make the lengths of the force vectors be proportional to the magnitudes of the forces.

(c) Starting from the Momentum Principle calculates the force exerted by the road on the car.

(d) Under what conditions will the force exerted by the road on the car be zero? Explain.

Short answer

(a)The non negligible force on the system is friction force f ,the upward force is${\mathrm{F}}_{\mathrm{N}}$and the gravitational force is

(b)The graph of step 2 represents the car is at the very top of the hill.

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

(d)The force exerted by the road on the car be zero is localid="1656916870657" $\sqrt{\mathrm{gR}}$

Step-by-step solution

Given

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.

Define net force

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 $\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\left|}\mathbf{\right|}}}$and the perpendicular rate of change of momentum$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\perp }}}$are the two elements that we are concerned with. As a result, the object's net force ${\mathbf{F}}_{\mathbf{net}}$is given by

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

(a) The non-negligible force of the system

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

(b) Represent diagrammatically the car and label force vectors

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

(c) The force exerted by the road on the car

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.

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

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

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

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$\frac{{\mathrm{Mv}}^2}{\mathrm{R}}$

(d) Determine the condition for the force exerted by the road on the car be zero

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

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

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