Chapter 13.1, Problem 13.42P

A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past point E. Knowing that $h=60$ ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point.

To determine

(a)

Estimate the force exerted by the seat on the rider by using given information.

Answer to Problem 13.42P

The value of force at point B is N_B=1120Ib and at point D is N_D=160Ib.

Explanation of Solution

Given information:

Diameter=40ft

Height=60ft

The force exerted by the seat on the rider at point B and D.

Consider the free body diagram we get,

Mass of the rider is $=m.$

Acceleration due to gravity is $=g.$

Resultant acceleration acting on the rider at the point B is $={\left(a_B\right)}_n.$

Normal force exerted on the rider by the seat at the point B is $N_B.$

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

$\begin{array}{l}{N}_B-mg=m{\left(a_B\right)}_n\\ N_B=m{\left(a_B\right)}_n+mg-----(1)\end{array}$

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

Acceleration acting on the rider at the point B

$T_A+U_{\left(A-B\right)}=T_B------(2)$

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

$T_B=$ Kinetic energy of the rider at point B.

Work done by the rider in reaching position B from position A is $U_{\left(A-B\right)}.$

The rider is at rest at position A is,

$T_A=0$

Then the kinetic energy of the rider at point B is,

$T_B=\frac{1}{2}m{v}_B^2$

Then the work done by the rider in reaching position B from position A,

$U_{\left(A-B\right)}=mgh$

Substitute above values we get,

$\begin{array}{l}{T}_A+U_{\left(A-B\right)}=T_B\\ 0+mgh=\frac{1}{2}m{v}_B^2\\ v_B^2=2gh\end{array}$

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

$\begin{array}{l}{\left(a_B\right)}_n=\frac{v_B^2}{r}\\ {\left(a_B\right)}_n=\frac{2gh}{r}\end{array}$

Substituting required values in the above equation we get,

$\begin{array}{l}{\left(a_B\right)}_n=\frac{2g\left(60ft\right)}{\left(20ft\right)}\\ =6g\end{array}$

Substitute ${\left(a_B\right)}_n=6g$ in equation 1 we get,

$\begin{array}{l}{N}_B=m{\left(a_B\right)}_n+mg\\ N_B=m\left(6g\right)+mg\\ N_B=7mg\end{array}$

Now we get,

$\begin{array}{l}{N}_B=7\left(160Ib\right)\\ =1120Ib\end{array}$

Forces acting on the riders by Newton’s second law

$\begin{array}{l}{N}_D+mg=m{\left(a_D\right)}_n\\ N_D=m{\left(a_D\right)}_n-mg-----(3)\end{array}$

Apply the principle of work and energy to roller coaster at point A and point D. To determine the resultant acceleration acting on the rider at the point B is,

$T_A+U_{\left(A-D\right)}=T_D------(4)$

$T_D=$ Kinetic energy of the rider at point D

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

The rider is at rest at position A is,

$T_A=0$

Kinetic energy of the rider at point D is,

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

$v_D=$ Velocity of the rider at point D

Work done by the rider in reaching position D from position A is $U_{\left(A-D\right)}=mg{h}_{AD}.$

Substituting required values in equation 4 we get,

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

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

$\begin{array}{l}{\left(a_D\right)}_n=\frac{v_D^2}{r}\\ {\left(a_D\right)}_n=\frac{2g\left(h-2r\right)}{r}\end{array}$

Substitute h and r values we get,

$\begin{array}{l}{\left(a_D\right)}_n=\frac{2g\left(60ft-40ft\right)}{\left(20ft\right)}\\ =\frac{2g\left(20\right)}{20}\\ =2g\end{array}$

Substitute required values we get,

$\begin{array}{l}{N}_D=m{\left(a_D\right)}_n-mg\\ =m\left(2g\right)-mg\\ =mg\end{array}$

Then,

$N_D=\left(160Ib\right)$

To determine

(b)

The minimum value of radius of curvature by using given information.

Answer to Problem 13.42P

The value of radius of curvature at point E is=80ft determined below.

Explanation of Solution

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

$N_E=$ The normal force acting on the coaster at the point E.

$m_c=$ Mass of the coaster.

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

$N_E-m_{c}g=-m_c{\left(a_E\right)}_n$

Roller coaster not to leave the tract at point E we get,

$m_{c}g=m_c{\left(a_E\right)}_n-----(5)$

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

Find the resultant acceleration acting on the coaster at the point E is,

$T_A+U_{\left(A-E\right)}=T_E----(6)$

The rider is at rest at position A.

$T_A=0$

Kinetic energy of the coaster at point E is,

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

Equating these two we get,

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

Substitute required values in above equation 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 coaster at point E travelling in a circular track with radius of curvature is $\rho .$

$\begin{array}{l}{\left(a_D\right)}_n=\frac{v_E^2}{\rho }\\ {\left(a_E\right)}_n=\frac{2g\left(h_{AB}-r\right)}{\rho }\end{array}$

$\begin{array}{l}{\left(a_E\right)}_n=\frac{2g\left(60ft-20ft\right)}{\rho }\\ =\frac{2g\left(40\right)}{\rho }\\ =\frac{80g}{\rho }\end{array}$

Substitute above values we get,

$\begin{array}{l}{m}_{c}g=m_c{\left(a_E\right)}_n\\ m_{c}g=m_c\left(\frac{80g}{\rho }\right)\\ \rho =80ft\end{array}$