Chapter 6, Problem 16P

(a)

To determine

The forceexerted by the track on the fully loaded roller-coaster at point $A$.

Answer to Problem 16P

The forceexerted by the track on the fully loaded roller-coaster at point $A$ is $\underset{\_}{2.49\times {10}^4\text{N}}$

Explanation of Solution

Consider the fully loaded roller-coaster car is moving on a curved path which has a radius of curvature. If the roller-coaster car rotates on a circular path, then it experiences a force acts towards the center of the circle which is known as centripetal force.

The magnitude of centripetal force $\left(F_c\right)$ depends on the following parameters.

1. 1. The speed of the roller-coaster car.
2. 2. The mass of the roller-coaster car.
3. 3. The radius of curvature of the curve path.

The forces act on the fully loaded roller-coaster car when it is at point $A$ as shown in figure below.

Write the expression for the fully loaded roller-coaster car corresponding to Newton’s second law in $y$-direction as.

  $n-mg=m{a}_c$                                                                                               (I)

Here, $n$ is the force exerted by the track on the fully loaded roller-coaster at point $A$, $m$ is mass of the fully loaded roller-coaster car, $a_c$ is centripetal acceleration of the car at point $A$ and $g$ is gravitational acceleration.

Write the expression for centripetal acceleration as.

  $a_c=\frac{v^2}{r_1}$                                                                                                        (II)

Here, $v$ is linear speed of the fully loaded roller-coaster at point $A$ and $r_1$ is radius of radius of curvature of the curve path at point $A$.

Substitute $\frac{v^2}{r_1}$ for $a_c$ in equation (I).

  $n-mg=m\left(\frac{v^2}{r_1}\right)$

Simplify the above expression for $n$ as.

  $n=mg+\frac{m{v}^2}{r_1}$                                                                                            (III)

Conclusion:

Substitute $500\text{kg}$ for $m$, $20.0\text{m}/\text{s}$ for $v$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10.0\text{m}$ for $r_1$ in equation (III).

  $\begin{array}{c}n=\left(500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)+\frac{\left(500\text{kg}\right){\left(20.0\text{m}/\text{s}\right)}^2}{\left(10.0\text{m}\right)}\\ =4900\text{N+}20000\text{N}\\ \text{=}\left(24900\text{N}\right)\\ =2.49\times {10}^4\text{N}\end{array}$

Thus, the force exerted by the track on the fully loaded roller-coaster at point $A$ is $\underset{\_}{2.49\times {10}^4\text{N}}$.

(b)

To determine

The maximum speed the vehicle can have at point $B$ and still remain on the track.

Answer to Problem 16P

The maximum speed the vehicle can have at point $B$ and still remain on the track is $\underset{\_}{12.1\text{m}/\text{s}}$.

Explanation of Solution

The direction of the centrifugal force and the normal force exerted by the track on the fully loaded roller-coaster is in same direction. As the speed of the car increases, the centripetal force increase which results decrease in normal reaction force.

The car keeps on track because of normal force reaction exerted by the track on the car.

At the verge to keep the car on the track, the normal force exerted by the track on the car is equal to zero. If the normal reaction force is equal to zero, then weight of the car will be equal to the centripetal force.

Write the expression for the condition of a car to keep on the track as.

  ${\left(F_c\right)}_B=mg$                                                                                                (IV)

Here, ${\left(F_c\right)}_B$ is centrifugal force acts on the car at point $B$.

Write the expression for centrifugal force acts on the car at point $B$ as.

  ${\left(F_c\right)}_B=\frac{m{v}^2{}_{\max }}{r_2}$                                                                                           (V)

Here, $r_2$ is radius of curvature of the curve path at point $B$ and $v_{\max }$ is maximum speed of the car at point $B$.

Substitute $mg$ for ${\left(F_c\right)}_B$ in equation (V).

  $\frac{m{v}^2{}_{\max }}{r_2}=mg$

Re-arrange the terms

  $v^2{}_{\max }=g{r}_2$

Simplify the above expression for $v_{\max }$ as.

  $v_{\max }=\sqrt{g{r}_2}$                                                                                               (VI)

Conclusion:

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $15.0\text{m}$ for $r_2$ in equation (VI).

  $\begin{array}{c}{v}_{\max }=\sqrt{\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(15.0\text{m}\right)}\\ =12.124\text{m}/\text{s}\\ \approx 12.1\text{m}/\text{s}\end{array}$

Thus, the maximum speed the vehicle can have at point $B$ and still remain on the track is $\underset{\_}{12.1\text{m}/\text{s}}$.