A car in an amusement park ride rolls without friction around a track (Fig. P7.42).The car starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)? (b) If $\mathit{h}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathit{R}$and $\mathit{R}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$, compute the speed, radial acceleration, and tangential acceleration of the passengers when the car is at point C, which is at the end of a horizontal diameter. Show these acceleration components in a diagram, approximately to scale.

The given data can be listed below,

• The height of the track is,$h=3.50R$
• The radius of the car is, $R=14\mathrm{m}$

The part of acceleration that is perpendicular to an object's motion is called tangential acceleration. When the direction of motion is not always the same as the tangential acceleration, tangential acceleration is typically utilised.

The velocity of the car at point B is given by,

$v_B=\sqrt{gR}$

Here,g is the acceleration due to gravity, and R is the radius of the loop.

From the conservation lay of energy the height of the track is given by,

role="math" localid="1667959710199" $$\begin{gathered} E=K+U_g \\ mgh=\frac{1}{2}m{v}^2+mgy \\ h=\frac{{\displaystyle \frac{1}{2}}m{v}^2+2mgR}{mg} \end{gathered}$$

Here, m is the mass of the car, v is the velocity of the car at point B,g is the acceleration due to gravity, and R is the radius of the loop.

Substitute all the values in the above,

$$\begin{gathered} h=\frac{1}{2}\frac{v_B^2}{g}+2R \\ =\frac{1}{2}\frac{vR}{g}+2R \\ =\frac{5}{2}R \end{gathered}$$

Thus, the minimum height of the track in terms of R is $\frac{5}{2}R$.

The figure to show the motion of the car is shown below as,

The speed of the passenger in the car at point C can be calculated by the use of conservation of energy at the point which can be given by,

$$\begin{gathered} E=mgh \\ =\frac{1}{2}m{v}_C^2+mgR \\ {v}_C=\sqrt{2g\left(h-R\right)} \end{gathered}$$

Here, h is the minimum height of the track, R is the radius of the loop, and g is the acceleration due to gravity.

Substitute all the values in the above,

$$\begin{gathered} v_C=\sqrt{2\mathrm{g}\left(\frac{5\mathrm{R}}{2}\right)} \\ =\sqrt{5\mathrm{gR}} \\ =\sqrt{5\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(14\mathrm{m}\right)} \\ =26.2\mathrm{m}/\mathrm{s} \end{gathered}$$

The tangential acceleration of the passenger in the car when the Gis the force acting in tangential direction is given by,

$a_t=\frac{F}{m}$

Here, Fis the force in the tangential direction and mis the mass of the car.

Substitute all the values in the above,

$$\begin{gathered} a_t=\frac{G}{m} \\ =g \end{gathered}$$

The radial acceleration of the passenger sitting inside the car at point C is given by,

$a_R=\frac{v_C^2}{R}$

Here. is the velocity of the car at point C and Ris the radius of the loop.

Substitute all the values in the above,

$$\begin{gathered} a_R=\frac{\left(5gR\right)}{R} \\ =5g \end{gathered}$$

Thus, the velocity, tangential and radial acceleration of the passenger in the car are 26.2 m/s, g and 5g respectively.