Chapter 5, Problem 54P

To determine

The reason for the given situation in the problem is not possible.

Answer to Problem 54P

The speed of the child is too large, so static friction does not have the strength to keep the child in place on the incline. Hence the situation is impossible.

Explanation of Solution

Assume that, the friction points up the incline, the net force is directed left towards the centre of the circular path in which the child travels, and a centripetal force needed for the circular motion is directed towards the centre of the circular path. The radius of the path of child is $R=d\cos \theta$.

Free body diagram of the child is shown in Figure.

Write the expression for net force in the horizontal direction, from the free body diagram.

    $f_{\text{s}}\cos \theta -n\sin \theta =\frac{m{v}^2}{R}$        (I)

Here, $f_{\text{s}}$ is the force due to static friction, $n$ is the normal force, $\theta$ is the angle of inclination, $m$ is the mass, $v$ is the speed, and $R$ is the radius of the circular path.

Write the expression for net force in the horizontal direction, from the free body diagram.

    $f_{\text{s}}\sin \theta +n\cos \theta =mg$        (II)

Here, $g$ is the acceleration due to gravity.

Multiply equation (I) by $\cos \theta$.

    $f_{\text{s}}{\cos }^2\theta -n\sin \theta \cos \theta =\frac{m{v}^2}{R}\cos \theta$        (III)

Multiply equation (II) by $\sin \theta$.

    $f_{\text{s}}{\sin }^2\theta +n\sin \theta \cos \theta =mg\sin \theta$        (IV)

Add equation (III) and (IV), and solve for $f_{\text{s}}$.

    $\begin{array}{c}{f}_{\text{s}}{\cos }^2\theta -n\sin \theta \cos \theta +f_{\text{s}}{\sin }^2\theta +n\sin \theta \cos \theta =mg\sin \theta +\frac{m{v}^2}{R}\cos \theta \\ f_{\text{s}}{\cos }^2\theta +f_{\text{s}}{\sin }^2\theta =mg\sin \theta +\frac{m{v}^2}{R}\cos \theta \\ f_{\text{s}}\left({\cos }^2\theta +{\sin }^2\theta \right)=mg\sin \theta +\frac{m{v}^2}{R}\cos \theta \\ f_{\text{s}}=mg\sin \theta +\frac{m{v}^2}{R}\cos \theta \end{array}$        (V)

Multiply equation (I) by $-\sin \theta$.

    $-f_{\text{s}}\sin \theta \cos \theta +n{\sin }^2\theta =-\frac{m{v}^2}{R}\sin \theta$        (VI)

Multiply equation (II) by $\cos \theta$.

    $f_{\text{s}}\sin \theta \cos \theta +n{\cos }^2\theta =mg\cos \theta$        (VII)

Add equation (VI) and (VII), and solve for $n$.

    $\begin{array}{c}-f_{\text{s}}\sin \theta \cos \theta +n{\sin }^2\theta +f_{\text{s}}\sin \theta \cos \theta +n{\cos }^2\theta =mg\cos \theta -\frac{m{v}^2}{R}\sin \theta \\ n{\sin }^2\theta +n{\cos }^2\theta =mg\cos \theta -\frac{m{v}^2}{R}\sin \theta \\ n\left({\sin }^2\theta +{\cos }^2\theta \right)=mg\cos \theta -\frac{m{v}^2}{R}\sin \theta \\ n=mg\cos \theta -\frac{m{v}^2}{R}\sin \theta \end{array}$        (VIII)

Equation (V) and (VIII) are consistent only when $f_{\text{s}}\le {\mu }_{\text{s}}n$.

Use equation (V) and (VIII) in $f_{\text{s}}\le {\mu }_{\text{s}}n$, and solve for $v$.

    $\begin{array}{c}mg\sin \theta +\frac{m{v}^2}{R}\cos \theta \le {\mu }_{\text{s}}\left(mg\cos \theta -\frac{m{v}^2}{R}\sin \theta \right)\\ \frac{v^2}{R}\left(\cos \theta +{\mu }_{\text{s}}\sin \theta \right)\le g\left({\mu }_{\text{s}}\cos \theta -\sin \theta \right)\\ v^2\left(\cos \theta +{\mu }_{\text{s}}\sin \theta \right)\le gR\left({\mu }_{\text{s}}\cos \theta -\sin \theta \right)\end{array}$        (IX)

Here, ${\mu }_{\text{s}}$ is the coefficient of static friction.

Use $d\cos \theta$ for $R$ in the equation (IX).

    $\begin{array}{c}{v}^2\left(\cos \theta +{\mu }_{\text{s}}\sin \theta \right)\le gd\cos \theta \left({\mu }_{\text{s}}\cos \theta -\sin \theta \right)\\ v^2\le \frac{gd\cos \theta \left({\mu }_{\text{s}}\cos \theta -\sin \theta \right)}{\left(\cos \theta +{\mu }_{\text{s}}\sin \theta \right)}\\ v\le \sqrt{\frac{gd\cos \theta \left({\mu }_{\text{s}}\cos \theta -\sin \theta \right)}{\left(\cos \theta +{\mu }_{\text{s}}\sin \theta \right)}}\end{array}$        (X)

Here, $d$ is the distance to the child from the centre of the cone.

If $v$ is too large, this situation is cannot exist.

Conclusion:

Substitute, $9.80\text{m}/{\text{s}}^2$ for $g$, $5.32\text{m}$ for $d$, $20.0°$ for $\theta$, and $0.700$ for ${\mu }_{\text{s}}$ in the equation (X).

    $\begin{array}{c}v\le \sqrt{\frac{\left(9.80\text{m}/{\text{s}}^2\right)\left(5.32\text{m}\right)\cos 20.0°\left[\left(0.700\right)\cos 20.0°-\sin 20.0°\right]}{\left(\cos 20.0°+{\mu }_{\text{s}}\sin 20.0°\right)}}\\ \le 3.62\text{m}/\text{s}\end{array}$

Since, the speed of the child is too large, so static friction does not have the strength to keep the child in place of the incline. Hence the situation is impossible.