Chapter 6, Problem 92P

(a)

To determine

The speed at which the car reach the top of the loop.

Answer to Problem 92P

The speed at which the car reach the top of the loop is $19.8\text{m/s}$.

Explanation of Solution

The mass of the roller coaster car is $988\text{kg}$. The diameter of the circular loop is $20.0\text{m}$. The car is at rest initially at a height of $40.0\text{m}$.

Write the formula for the conservation of energy for the car.

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

Here, $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy, $U_f$ is the final potential energy

Consider the initial state to be at the top at rest and final stage to be at the top of the loop. The initial kinetic energy is thus zero.

Re-write equation (I).

$0+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f$

Here, $m$ is the mass of the car, $g$ is the acceleration due to gravity, $h_i$ is the initial height of the car, $v_f$ is the final velocity of the car, $h_f$ is the final height of the car.

Re-write the above equation to get an expression for $v_f$.

$\begin{array}{c}{v}_f=\sqrt{\frac{2mg\left(h_i-h_f\right)}{m}}\\ =\sqrt{2g\left(h_i-h_f\right)}\end{array}$ (II)

Conclusion:

Substitute $988\text{kg}$ for $m$, $40.0\text{m}$ for $h_i$, $20.0\text{m}$ for $h_f$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ in equation (III).

$\begin{array}{c}{v}_f=\sqrt{2\left(9.80{\text{m/s}}^{\text{2}}\right)\left(40.0\text{m}-20.0\text{m}\right)}\\ =19.8\text{m/s}\end{array}$

The speed at which the car reach the top of the loop is $19.8\text{m/s}$.

(b)

To determine

The force exerted by the track on the car at the top of the loop.

Answer to Problem 92P

The force exerted by the track on the car at the top of the loop is $29\text{kN}$.

Explanation of Solution

The mass of the roller coaster car is $988\text{kg}$. The diameter of the circular loop is $20.0\text{m}$. The car is at rest initially at a height of $40.0\text{m}$.

At the top of the loop, the normal force and the weight of the car points downward and provide the centripetal force.

Write the equation for forces at the top of the loop in vertical direction.

$\frac{m{v}^2}{r}=N+mg$

Here, $m$ is the mass of the car, $g$ is the acceleration due to gravity, $v$ is the velocity of the car at the top of the loop, $N$ is the normal force by the track on the car, $r$ is the radius of the loop.

Re-write the above equation to get an expression for $N$.

$N=m\left(\frac{v^2}{r}-g\right)$

Conclusion:

Substitute $988\text{kg}$ for $m$, $19.8\text{m/s}$ for $v$, $\frac{20.0\text{m}}{2}$ for $r$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ in equation (III).

$\begin{array}{c}N=\left(988\text{kg}\right)\left(\frac{{\left(19.8\text{m/s}\right)}^2}{20.0\text{m}/2}-9.80{\text{m/s}}^{\text{2}}\right)\\ =29051\text{N}\\ \simeq 29\text{kN}\end{array}$

The force exerted by the track on the car at the top of the loop is $29\text{kN}$.

(c)

To determine

The minimum height for the release of the car so that the car does not lose contact with the track.

Answer to Problem 92P

The force exerted by the track on the car at the top of the loop is $29.0\text{kN}$.

Explanation of Solution

The mass of the roller coaster car is $988\text{kg}$. The diameter of the circular loop is $20.0\text{m}$. The car is at rest initially at a height of $40.0\text{m}$.

At the top of the loop, the normal force and the weight of the car points downward and provide the centripetal force.

Write the equation for forces at the top of the loop in vertical direction.

$\frac{m{v}^2}{r}=N+mg$

Here, $m$ is the mass of the car, $g$ is the acceleration due to gravity, $v$ is the velocity of the car at the top of the loop, $N$ is the normal force by the track on the car, $r$ is the radius of the loop.

At the minimum height for the release of the car so that the car does not lose contact with the track, the normal force is zero.

Re-write the above equation.

$\frac{m{v}^2}{r}=N+mg$ (I)

From section (a) write the formula for the velocity at the top of the loop.

$v=\sqrt{2g\left(h_i-h_f\right)}$ (II)

Here, $h_i$ is the initial height of the car, $h_f$ is the final height of the car.

Substitute equation (II) in equation (I).

$\frac{m\left(2g\left(h_i-h_f\right)\right)}{r}=mg$

Re-write the above equation to get an expression for $h_i$.

$h_i=\frac{r+2h_f}{2}$

Conclusion:

Substitute, $20\text{m}$ for $h_f$, $\frac{20.0\text{m}}{2}$ for $r$ in equation (III).

$\begin{array}{c}{h}_i=\frac{20\text{m}/2+2\left(20\text{m}\right)}{2}\\ =\frac{10\text{m}+40\text{m}}{2}\\ =25\text{m}\end{array}$

The minimum height for the release of the car so that the car does not lose contact with the track is $25\text{m}$.