Chapter 6, Problem 138P

(a)

To determine

The minimum speed of bob that can be achieved at the top of circle.

Answer to Problem 138P

The minimum speed is $\sqrt{rg}$.

Explanation of Solution

Length of string is $L$, height from the bottom to the starting point of motion is $h$, and the height of horizontal peg from the bottom of pendulum’s swing is $r$.

At the top of circular path, the centripetal force due to the circular motion will balances the weight of pendulum.

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

Here, $m$ is the mass of pendulum, $g$ is the gravitational acceleration, $v$ is the linear velocity of motion of pendulum, and $r$ is the radius of circular path of motion after the string hits the peg.

Conclusion:

Rewrite the above equation in terms of $v$.

$v=\sqrt{rg}$

The minimum speed is $\sqrt{rg}$.

(b)

To determine

The distance from point B to a point at same height of point B if the sting breaks when the bob is at point A.

Answer to Problem 138P

The distance is $2r$.

Explanation of Solution

Length of string is $L$, height from the bottom to the starting point of motion is $h$, and the height of horizontal peg from the bottom of pendulum’s swing is $r$.

The vertical distance between point $A$ and $B$ is $2r$.Thus, the vertical distance covered by bob during its motion from point $A$ to $B$ be $2r$ itself.

Write the second equation for Newton’s law of motion in vertical direction during its motion from point $A$ to level of point $B$.

$2r=u_{y}t+\frac{1}{2}g{t}^2$

Here, $2r$ is the vertical distance travelled by the bob, $u_y$ is the vertical component of velocity of bob at point $A$, and $t$ is time taken to reach the level of point $B$ from point $A$.

Since the string breaks at point A, bob executes free fall motion and the velocity of bob at point $A$ is zero. Thus, the vertical component of velocity of bob at point $A$ is zero.

Rewrite the above equation in terms of $t$ by substituting $0\text{m}/\text{s}$ for $u_y$.

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

Write the second equation for Newton’s law of motion in horizontal direction during its motion from point $A$ to level of point $B$.

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

Here, $x$ is the horizontal range is $x$, horizontal component of velocity at point $A$ is $u_x$, and $a_x$ is the horizontal component of acceleration.

Conclusion:

Substitute $\sqrt{rg}$ for $u_x$, $0\text{m}/{\text{s}}^{\text{2}}$ for $a_x$, and $\sqrt{\frac{4r}{g}}$ for $t$ in the above equation.

$\begin{array}{c}x=\left(\sqrt{rg}\right)\left(\sqrt{\frac{4r}{g}}\right)+\frac{1}{2}\left(0\text{m}/{\text{s}}^{\text{2}}\right)t^2\\ =2r\end{array}$

Rewrite the above equation in terms of $v$.

$v=\sqrt{rg}$

Therefore, the distance is $2r$.

(c)

To determine

The minimum value of $h$ required for bob to point $A$ without going slack.

Answer to Problem 138P

Minimum value of $h$ is $\frac{5}{2}r$.

Explanation of Solution

Length of string is $L$, height from the bottom to the starting point of motion is $h$, and the height of horizontal peg from the bottom of pendulum’s swing is $r$.

Write the law of conservation of energy from the starting point to point $A$.

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

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

Write the equation for $U_i$.

$U_i=mgh$ (II)

Write the equation for $U_f$.

$U_f=mg\left(2r\right)$ (III)

Write the equation for $K_f$.

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

Conclusion:

Rewrite equation (I) by substituting $0\text{J}$ for $K_i$, equations (II), (III), and (IV).

$\begin{array}{c}mgh+0\text{J}=mg\left(2r\right)+\frac{1}{2}m{v}^2\\ gh=2gr+\frac{1}{2}{v}^2\end{array}$

Rewrite the above equation in terms of $h$ by substituting $\sqrt{gr}$ for $v$.

$\begin{array}{c}gh=2gr+\frac{1}{2}{\left(\sqrt{gr}\right)}^2\\ h=\frac{5}{2}r\end{array}$

Therefore, the minimum value of $h$ is $\frac{5}{2}r$.