Chapter 6, Problem 141P

(a)

To determine

The minimum possible speed of the bob if the bob is at the top of the circle.

Answer to Problem 141P

The minimum possible speed of the bob if the bob is at the top of the circle is $\underset{\_}{\sqrt{gr}}$.

Explanation of Solution

The bob as a particle in uniform circular motion, it exerts a centripetal acceleration towards the peg.

The free body diagram of the bob at the top of the circle is shown in the Figure 1.

From the free body diagram, write the expression for net force in the vertical direction.

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

Here, $T$ is the tension, $m$ is the mass of the bob, $g$ is the acceleration due to gravity, $v$ is he speed, and $r$ is the radius.

The minimum speed of the bob occurs corresponds to a tension of zero in the cord.

Substitute, $0$ for $T$ in the equation (I), and solve for $v$.

    $\begin{array}{c}0+mg=\frac{m{v}^2}{r}\\ v=\sqrt{rg}\end{array}$        (II)

Conclusion:

Therefore, the minimum possible speed of the bob if the bob is at the top of the circle is $\underset{\_}{\sqrt{gr}}$.

(b)

To determine

The distance from a point B moved by the bob if the string breaks at point A.

Answer to Problem 141P

The distance from a point B moved by the bob if the string breaks at point A is $\underset{\_}{2r}$.

Explanation of Solution

Let C be the point at the same height as B reached by the bob when the string breaks. Assume the bob is in the x direction and moving with constant acceleration.

Consider the Figure 2.

Write the expression for distance in the x-direction

    $\Delta x=v_{xi}t+\frac{1}{2}{a}_x{t}^2$        (III)

Here, $v_{xi}$ is the initial velocity in the x-direction, $a_x$ is the acceleration in the x direction, and $t$ is the time.

Substitute, $\sqrt{gr}$ for $v_{xi}$, and $0$ for $a_x$ in the equation (III).

    $\begin{array}{c}\Delta x=\sqrt{gr}t+\frac{1}{2}\left(0\right)t^2\\ \Delta x=\sqrt{gr}t\end{array}$        (IV)

Write the expression for distance in the y-direction

    $\Delta y=v_{yi}t+\frac{1}{2}{a}_y{t}^2$        (V)

Here, $v_{yi}$ is the initial velocity in the y-direction, and $a_y$ is the acceleration in the y direction.

Substitute, $0$ for $v_{yi}$, $-g$ for $a_y$ , and $-2r$ for $\Delta y$ in the equation (V), and solve for $t$.

    $\begin{array}{c}-2r=\left(0\right)t+\frac{1}{2}\left(-g\right)t^2\\ 4r=g{t}^2\\ t=2\sqrt{\frac{r}{g}}\end{array}$        (VI)

Use equation (VI) in (IV).

    $\begin{array}{c}\Delta x=\sqrt{gr}\left(2\sqrt{\frac{r}{g}}\right)\\ =2r\end{array}$        (VII)

Conclusion:

Therefore, the distance from a point B moved by the bob if the string breaks at point A is $\underset{\_}{2r}$.

(c)

To determine

The minimum value of $h$ for which the bob will make it to point A without going slack.

Answer to Problem 141P

The minimum value of $h$ for which the bob will make it to point A without going slack is $\underset{\_}{\frac{5r}{2}}$.

Explanation of Solution

The system of the bob and Earth as an isolated system for energy with no non conservative forces acting, then the total energy is zero.

    $\Delta K+\Delta U=0$        (VIII)

Here, $\Delta K$ is the changing kinetic energy, $\Delta U$ is the change in potential energy.

Use $\frac{1}{2}m{v}^2$ for $K$, and $mgh$ for $U$ in the equation (VIII).

    $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+mg\left(2r-h\right)=0$        (IX)

Here, $h$ is the initial height of the bob.

Substitute, $\sqrt{gr}$ for $v_f$, and $0$ for $v_i$ in the equation (IX), and solve for $h$.

    $\begin{array}{c}\frac{1}{2}m{\left(\sqrt{gr}\right)}^2-\frac{1}{2}m\left(0\right)+mg\left(2r-h\right)=0\\ mgh=\frac{1}{2}m{\left(\sqrt{gr}\right)}^2+mg\left(2r\right)\\ h=\frac{1}{2}r+2r\\ =\frac{5}{2}r\end{array}$

Since $v_f=v_{\text{min}}$.

Conclusion:

Therefore, the minimum value of $h$ for which the bob will make it to point A without going slack is $\underset{\_}{\frac{5r}{2}}$.