Chapter 8, Problem 65PQ

A 50.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.00R.

a. What is the speed of the car at the top of the vertical loop?

b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?

To determine

The speed of the car at the top of the vertical loop.

Answer to Problem 65PQ

The speed of the car at the top of the vertical loop is $2\sqrt{gR}$ .

Explanation of Solution

The diagram of the motion of the car is shown in figure 1.

Write the expression for the conservation of energy as the car moves from the initial position to the top of the vertical loop.

  $U_i+K_i=U_f+K_f$                                                                                             (I)

Here, $U_i$ is the initial potential energy of the car, $K_i$ is the initial kinetic energy of the car, $U_f$ is the final potential energy of the car and $K_f$ is the final kinetic energy of the car.

Write the equation for the initial potential energy.

  $U_i=mgh$                                                                                                             (II)

Here, $m$ is the mass of the car, $g$ is the acceleration due to gravity and $h$ is the initial height of the car.

The car is initially at rest so that its initial kinetic energy must be zero.

Write the expression for $K_i$ .

`    $K_i=0$                                                                                                                     (III)

Write the equation for the potential energy of the car at the top of the vertical loop.

  $U_f=mg\left(2R\right)$                                                                                                  (IV)

Here, $R$ is the radius of the vertical loop.

Write the equation for $K_f$ .

  $K_f=\frac{1}{2}m{v}^2$                                                                                                            (V)

Here, $v$ is the speed of the car at the top of the vertical loop.

Put equations (II) to (V) in equation (I) and rewrite it for $v$ .

  $\begin{array}{c}mgh+0=mg\left(2R\right)+\frac{1}{2}m{v}^2\\ gh=2gR+\frac{1}{2}{v}^2\\ v^2=2g\left(h-2R\right)\\ v=\sqrt{2g\left(h-2R\right)}\end{array}$                                                                                   (VI)

Conclusion:

Given that the initial height is $h=4.00R$ .

Substitute $4.00R$ for $h$ in equation (VI) to find $v$ .

  $\begin{array}{c}v=\sqrt{2g\left(4.00R-2R\right)}\\ =\sqrt{4gR}\\ =2\sqrt{gR}\end{array}$

Therefore, the speed of the car at the top of the vertical loop is $2\sqrt{gR}$ .

To determine

The magnitude of the normal force acting on the car at the top of the vertical loop.

Answer to Problem 65PQ

The magnitude of the normal force acting on the car at the top of the vertical loop is $1.47\text{ N}$ .

Explanation of Solution

The forces acting on the car are the normal force and the gravitational force. Both these force act in the downward direction and the vertical sum of these forces provides the centripetal acceleration required for the car to move in the circular track.

Write the expression for the Newton’s second law.

  $\Sigma F=ma$                                                                                                                (VII)

Here, $\Sigma F$ is the net force acting on the car, $m$ is the mass of the car and $a$ is the acceleration of the car.

Write the expression for $\Sigma F$ .

  $\Sigma F=F_N+mg$

Here, $F_N$ is the normal force and $g$ is the acceleration due to gravity.

The acceleration of the car is centripetal acceleration.

Write the expression for $a$ .

  $a=\frac{v^2}{R}$

Put the above two equations in equation (VII) and rewrite it for $F_N$ .

  $\begin{array}{c}{F}_N+mg=m\frac{v^2}{R}\\ F_N=m\left(\frac{v^2}{R}-g\right)\end{array}$

Substitute $2\sqrt{gR}$ for $v$ in the above equation.

  $\begin{array}{c}{F}_N=m\left(\frac{{\left(2\sqrt{gR}\right)}^2}{R}-g\right)\\ =m\left(\frac{4gR}{R}-g\right)\\ =3.00mg\end{array}$                                                                                        (VIII)

Conclusion:

Given that the mass of the car is $50.0\text{ g}$ . The value of acceleration due to gravity is $9.81{\text{ m/s}}^2$ .

Substitute $50.0\text{ g}$ for $m$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (VIII) to find $F_N$ .

  $\begin{array}{c}{F}_N=3\left(50.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}\right)\left(9.81{\text{ m/s}}^2\right)\\ =1.47\text{ N}\end{array}$

Therefore, the magnitude of the normal force acting on the car at the top of the vertical loop is $1.47\text{ N}$ .