Question

A roller-coaster car at an amusement park has a mass of${}^{\mathbf{1200}\mathbf{kg}}$when fully loaded with passengers. As the car passes over the top of a circular hill of radius${}^{\mathbf{18}\mathbf{}\mathbf{m}}$, assume that its speed is not changing. At the top of the hill, what are the (a) magnitude${}^{{\mathbf{F}}_{\mathbf{N}}}$and (b) direction (up or down) of the normal force on the car from the track if the car’s speed is${}^{\mathbf{V}\mathbf{=}\mathbf{11}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$? What are (c)${}^{{\mathbf{F}}_{\mathbf{N}}}$and (d) the direction if${}^{\mathbf{V}\mathbf{=}\mathbf{14}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$?

Short answer

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{roller}-\mathrm{coaster}\mathrm{car}=1200\mathrm{kg}, \\ \mathrm{Radius}\mathrm{of}\mathrm{cirular}\mathrm{hill}=18\mathrm{m}, \\ \mathrm{Velocity}=11\mathrm{m}/\mathrm{s}\mathrm{and}14\mathrm{m}/\mathrm{s} \end{gathered}$$

Step-by-step solution

Given

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{roller}-\mathrm{coaster}\mathrm{car}=1200\mathrm{kg}, \\ \mathrm{Radius}\mathrm{of}\mathrm{cirular}\mathrm{hill}=18\mathrm{m}, \\ \mathrm{Velocity}=11\mathrm{m}/\mathrm{s}\mathrm{and}14\mathrm{m}/\mathrm{s} \end{gathered}$$

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves Newton’s second law of motion.

Formula:

The velocity in uniform circular motion is given by,

$\mathrm{v}=\frac{2\mathrm{\pi R}}{\mathrm{T}}$

where, v is the velocity, R is the radius and T is the time period

According to Newton’s second law of motion

${\text{F}}_N-mg=m{a}_c$

where,${}^{{\mathrm{a}}_{\mathrm{c}}}$is an acceleration, g is an acceleration due to gravity, m is mass,${}^{F_{\mathrm{N}}}$is the normal force and R is the radius.

(a) Determining the magnitude of force

In this case consider the normal force pointing upward direction and gravity pointing downwards. Also, the direction to the center of the circle (the direction of centripetal acceleration) is down.

The centripetal acceleration then will be,

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$

Thus, equation (ii) leads to,

${\text{F}}_{\mathrm{N}}-\mathrm{mg}=\mathrm{m}\left(-\frac{{\mathrm{v}}^2}{\mathrm{r}}\right)$

${\mathrm{F}}_{\mathrm{N}}=1200\mathrm{g}-1200\left(\frac{{\left(11\right)}^2}{18}\right)$

$=3.7\times {10}^3\mathrm{N}$

Thus, when${}^{v=11\mathrm{m}/\mathrm{s}}$, the magnitude of${}^{{\mathrm{F}}_{\mathrm{N}}=3.7\times {10}^3\mathrm{N}}$.

(b) Determining the direction of the normal force on the car from the trace

From above, the value of${}^{{\mathrm{F}}_{\mathrm{N}}}$is positive.

Thus,${}^{{\mathrm{F}}_{\mathrm{N}}}$points upward.

(c) Determining the magnitude of force if v= 14 m/s

When $$\begin{gathered} {}^{\mathrm{v}=14\mathrm{m}/\mathrm{s}} \\ {\mathrm{F}}_{\mathrm{N}}=-1.3\times {10}^3\mathrm{N},\mathrm{or}\left|{\mathrm{F}}_{\mathrm{N}}\right|=1.3\times {10}^3\mathrm{N} \end{gathered}$$

Thus, the magnitude of${\mathrm{F}}_{\mathrm{N}}$is$1.3\mathrm{kN}$

(d) Determining the direction of force if v=14 m/s

The fact that this answer is negative means that${}^{{\mathrm{F}}_{\mathrm{N}}}$points opposite to what was assumed.

Thus, its direction is down.