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Chapter 7

Q. 7.8

RIDING THE TRACKS

GOAL Combine centripetal force with conservation of energy. Derive results symbolically.

PROBLEM Figure 7.13a shows a roller – coaster car moving around a circular loop of radius R. (a) What speed must the car have at the top of the loop so that it will just make it over the top without any assistance from the track? (b) What speed will the car subsequently have at the bottom of the loop? (c) What will be the normal force on a passenger at the bottom of the loop if the loop has a radius of 10.0 m?

STRATEGY This problem requires Newton’s second law and centripetal acceleration to find an expression for the car’s speed at the top of the loop, followed by conservation of energy to find its speed at the bottom. If the car just makes it over the top, the force \overrightarrow{\mathbf{n}} must become zero there, so the only force exerted on the car at that point is the force of gravity, m \overrightarrow{\mathbf{g}}. At the bottom of the loop, the normal force acts up toward the center and the gravity force acts down, away from the center. The difference of these two is the centripetal force. The normal force can then be calculated from Newton’s second law.

7.13

Step-by-Step

Verified Solution

(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.