Chapter 6, Problem 92P

(a)

To determine

The speed of the car that has to reach the top of the loop.

Answer to Problem 92P

The speed of the car that has to reach the top of the loop is $\underset{\_}{19.8\text{m}/\text{s}}$.

Explanation of Solution

At the top of the ramp the car has only the potential energy given by

  $TE=mg{h}^{\prime }$        (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity

At the top of the circular loop the car has both kinetic and potential energy which is given by,

    $TE=mgh+\frac{1}{2}m{v}^2$        (II)

Here, $v$ is the velocity, $TE$ is the total energy.

Equating equation (I) and (II), and solve for $v$.

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

Conclusion:

Substitute $20\text{m}$ for $h$ and $40\text{m}$ for $h^{\prime }$ in equation (III) to find $v$,

    $\begin{array}{c}v=\sqrt{2\left(9.8\text{m}/{\text{s}}^2\right)\left(40\text{m}-20\text{m}\right)}\\ =19.8\text{m}/\text{s}\end{array}$

Therefore, the speed of the car that has to reach the top of the loop is $\underset{\_}{19.8\text{m}/\text{s}}$.

(b)

To determine

The force exerted at the top of the loop.

Answer to Problem 92P

The force exerted at the top of the loop is $\underset{\_}{29051.152\text{N}}$.

Explanation of Solution

Write the expression for the force exerted at the top of the loop.

    $F=\frac{m{v}^2}{r}-mg$        (IV)

here, $r$ is the radius, $v$ is the speed.

From the above diagram it is shown that

  $\begin{array}{c}r=\frac{h}{2}\\ =\frac{20\text{m}}{2}\\ =10\text{m}\end{array}$.

Conclusion:

Substitute $10\text{m}$ for $r$, $988\text{kg}$ for $m$ and $19.8\text{m}/\text{s}$ for $v$, $9.80\text{m}/{\text{s}}^2$ for $g$ in equation (I) to find $F$.

    $\begin{array}{c}F=\frac{\left(988\text{kg}\right){\left(19.8\text{m}/\text{s}\right)}^2}{10\text{m}}-\left(988\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\\ =29051.152\text{N}\end{array}$

Therefore, The force exerted at the top of the loop is $\underset{\_}{29051.152\text{N}}$.

(c)

To determine

The minimum height above the bottom of the loop.

Answer to Problem 92P

The minimum height above the bottom of the loop is $\underset{\_}{25\text{m}}$.

Explanation of Solution

The force exerted on the car by the track at the top should be equal to the weight of the car so that,

    $\begin{array}{l}\frac{m{v}^2}{r}=mg\\ v^2=rg\end{array}$        (V)

Using equation (I), the energy of the car at the top will be,

    $\begin{array}{c}E=mgh+\frac{1}{2}m{v}^2\\ E=mgh+\frac{1}{2}m\left(gr\right)\\ =mgh+\frac{1}{2}mg\left(\frac{h}{2}\right)\\ =\frac{5}{4}mgh\end{array}$        (VI)

The minimum height from which the car starts is $H$,

According to the conservation of energy,

    $mgH=\frac{5}{4}mgh$        (VII)

Solve equation (VII) for $H$,

    $H=\frac{5}{4}h$        (VIII)

Conclusion:

Substitute $20\text{m}$ for $h$ in equation (VIII) to find $H$.

    $\begin{array}{c}H=\frac{5}{4}\left(20\text{m}\right)\\ =25\text{m}\end{array}$

Therefore, The minimum height above the bottom of the loop is $\underset{\_}{25\text{m}}$.