Question 7.8: RIDING THE TRACKS GOAL Combine centripetal force with conser...

(a) Find the speed at the top of the loop.

Write Newton’s second law for the car:

(1)      m \overrightarrow{\mathbf{a}}_c=\overrightarrow{\mathbf{n}}+m \overrightarrow{\mathbf{g}}

At the top of the loop, set n=0. The force of gravity acts toward the center and provides the centripetal acceleration a_c=-v^2 / R

-m \frac{v_{\text {top }}^2}{R}=-m g

Solve the foregoing equation for v_{\text {top }} :

v_{\text {top }}=\sqrt{g R}

(b) Find the speed at the bottom of the loop.

Apply conservation of mechanical energy to find the total mechanical energy at the top of the loop:

E_{\text {top }}=\frac{1}{2} m v_{\text {top }}^2+m g h=\frac{1}{2} m g R+m g(2 R)=2.5 m g R

Find the total mechanical energy at the bottom of the loop:

\quad E_{\text {bot }}=\frac{1}{2} m v_{\text {bot }}^2

Energy is conserved, so these two energies may be equated and solved for v_{\text {bot: }} :

\begin{aligned}\frac{1}{2} m v_{\text {bot }}^2 &=2.5 m g R \\v_{\text {bot }} &=\sqrt{5 g R}\end{aligned}

(c) Find the normal force on a passenger at the bottom.

(This is the passenger’s perceived weight.)

Use Equation (1). The net centripetal force is m g-n .

-m \frac{v_{\text {bot }}^2}{R}=m g-n

(The normal force acts toward the center of the circle, in the negative radial direction. The car’s weight acts away from the center, in the positive radial direction.)

Solve for n :

n=m g+m \frac{v_{\text {bot }}^2}{R}=m g+m \frac{5 g R}{R}=6 m g

REMARKS The final answer for n shows that the rider experiences a force six times normal weight at the bottom of the loop! Astronauts experience a similar force during space launches.