Chapter 13.1, Problem 13.43P

In Prob. 13.42. determine the range of values of h for which the roller coaster will not leave the track at D or E, knowing that the radius of curvature at E is $p=75$ ft. Assume no energy loss due to friction.

To determine

The range of values of h for which the roller coaster will not leave the track.

Answer to Problem 13.43P

The value of range of h is 50ft

Explanation of Solution

Given information:

Radius of curvature at point E is=75ft.

Consider free body diagram of the roller coaster at point D.

${\left(a_D\right)}_n=$ Resultant acceleration acting on the coaster at the point D downwards.

$N_D=$ The normal force acting on the coaster at the point D.

$m_c=$ Mass of the coaster.

Resolve the forces acting on the rider by Newton’s second law.

$\begin{array}{l}{N}_D+m_{c}g=m_c{\left(a_D\right)}_n\\ N_D=m_c{\left(a_D\right)}_n-m_{c}g---(1)\end{array}$

Apply the principle of work and energy to the roller coaster at point A and point D.

Find the resultant acceleration acting on the coaster at the point B is ${\left(a_D\right)}_n.$

$T_A+U_{\left(A-D\right)}=T_D-----(2)$

$T_A=$ Kinetic energy of the coaster at point A.

$T_D=$ Kinetic energy of the coaster at point D.

$U_{\left(A-D\right)}=$ Work done by the coaster in reaching position D from position A.

The coaster is at rest at position A,

$T_A=0$

Kinetic energy of the rider at point D is,

$T_D=\frac{1}{2}{m}_c{v}_D^2$

$v_D=$ Velocity of the rider at point D.

Work done by the rider in reaching position D from position A,

$U_{\left(A-D\right)}=m_{c}g{h}_{AD}$

Substitute above values we get,

$\begin{array}{l}{T}_A+U_{\left(A-D\right)}=T_D\\ 0+m_{c}g{h}_{AD}=\frac{1}{2}{m}_c{v}_D^2\\ v_D^2=2g{h}_{AD}\\ v_D^2=2g\left(h-2r\right)\end{array}$

Resultant acceleration acting on the rider at point D travelling in a circular track with radius of curvature r.

$\begin{array}{l}{\left(a_D\right)}_n=\frac{v_D^2}{r}\\ {\left(a_D\right)}_n=\frac{2g\left(h-2r\right)}{r}\\ N_D=m_c{\left(a_D\right)}_n-m_{c}g\\ N_D=m_c\left[\frac{2g\left(h-2r\right)}{r}\right]-m_{c}g\\ N_D=m_{c}g\left(\frac{2h-5r}{r}\right)\end{array}$

The normal force acting on the coaster at point D should be greater than zero.

For the roller coaster to not leave the track,

$\begin{array}{l}{N}_D>0\\ m_{c}g\left(\frac{2h-5r}{r}\right)>0\\ 2h-5r>0\\ h>\frac{5r}{2}\\ h>\frac{5}{2}\left(20ft\right)\\ h>50ft-----(3)\end{array}$

Consider free body diagram roller coaster at point E as shown below:

Resultant acceleration acting on the roller coaster at the point E downwards is ${\left(a_E\right)}_n.$

Resolve the forces acting on the coaster by Newton’s second law at point E.

$\begin{array}{l}{N}_E-m_{c}g=-m_c{\left(a_E\right)}_n\\ N_E=m_{c}g-m_c{\left(a_E\right)}_n-----(4)\end{array}$

Apply the principle of work and energy to the roller coaster at point A and point E.

Find the resultant acceleration acting on the coaster at the point E is ${\left(a_E\right)}_n.$

$T_A+U_{\left(A-E\right)}=T_E-----(5)$

$T_E=$ Kinetic energy of the coaster at point E.

$U_{\left(A-E\right)}=$ Work done by the coaster in reaching position E from position A.

The coaster is at rest at position A.

$T_A=0$

Kinetic energy of the rider at point E is,

$T_E=\frac{1}{2}{m}_c{v}_E^2$

$v_E=$ Velocity of the rider at point E.

Work done by the rider in reaching position E from position A.

$U_{\left(A-E\right)}=m_{c}g{h}_{AE}$

Substituting the above values we get,

$\begin{array}{l}{T}_A+U_{\left(A-E\right)}=T_E\\ 0+m_{c}g{h}_{AE}=\frac{1}{2}{m}_c{v}_E^2\\ v_E^2=2g{h}_{AE}\\ v_E^2=2g\left(h-r\right)\end{array}$

Resultant acceleration acting on the rider at point E travelling in a circular track with radius of curvature $\rho .$

$\begin{array}{l}{\left(a_E\right)}_n=\frac{v_E^2}{\rho }\\ {\left(a_E\right)}_n=\frac{2g\left(h-r\right)}{\rho }\\ N_E=m_{c}g-m_c\left[\frac{2g\left(h-r\right)}{\rho }\right]\\ N_E=m_{c}g\left(\frac{\rho -2h+2r}{\rho }\right)\end{array}$

The normal force acting on the coaster at point E should be greater than zero.

For the roller coaster to not leave the track,

$N_E>0$

Substituting the above values we get,

$\begin{array}{l}{m}_{c}g\left(\frac{\rho -2h+2r}{\rho }\right)>0\\ \rho -2h+2r>0\\ 75ft-2h+2\left(20ft\right)>0\\ 115-2h>0\\ 2h<115\\ h<57.5ft------(6)\end{array}$

The range of values for h is $50ft<h<57.5ft.$