Chapter 11, Problem 97P

(a)

To determine

The stretch in the string when the ball moves in a horizontal circle as compared to when the ball is at rest.

Answer to Problem 97P

The stretch in the string when the ball moves in a horizontal circle as compared to when the ball is at rest is $6.17\times {10}^{-4}\text{m}$.

Explanation of Solution

The mass of the ball of the tetherball set is $0.411\text{kg}$. The diameter of the nylon string is $2.50\text{mm}$. The Young’s modulus of nylon is $4.00\text{GPa}$. The density of the nylon is $1150{\text{kg/m}}^{\text{3}}$. The length of the string is $2.200\text{m}$.

Write the equation of force in y-direction when the ball is hanging straight vertical.

$\begin{array}{c}{T}_1-mg=0\\ T_1=mg\end{array}$ (I)

Here, $T_1$ is the tension when the ball is hanging vertical, $m$ is the mass, $g$ is the acceleration due to gravity.

Write the equation of the force in y-direction when the ball is swinging.

$\begin{array}{c}{T}_2\cos \theta -mg=0\\ T_2=\frac{mg}{\cos \theta }\end{array}$ (II)

Here, $T_2$ is the tension when the ball is swinging, $\theta$ is the angle made by the string.

Write the formula for the young’s modulus.

$Y=\frac{\left(T_2-T_1\right)/A}{\Delta L/L}$

Here, $Y$ is the young’s modulus, $A$ is the area, $\Delta L$ is the stretch, $L$ is the length.

Write the equation for the area.

$A=\frac{\pi d^2}{4}$ (III)

Here, $d$ is the diameter.

Substitute equation (I), (II), and (II) in the above equation and re-write to get an expression for $\Delta L$.

$\begin{array}{c}\Delta L=\frac{4L\left(T_2-T_1\right)}{\pi d^{2}Y}\\ =\frac{4L}{\pi d^{2}Y}\left(\frac{mg}{\cos \theta }-mg\right)\\ =\frac{4Lmg}{\pi d^{2}Y}\left(\frac{1}{\cos \theta }-1\right)\end{array}$ (IV)

Conclusion:

Substitute $0.411\text{kg}$ for $m$ and $2.50\text{mm}$ for $d$, $4.00\text{GPa}$ for $Y$, $65°$ for $\theta$, $2.200\text{m}$ for $L$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find the stretch in the string.

$\begin{array}{c}\Delta L=\frac{4\left(2.200\text{m}\right)\left(0.411\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{\pi {\left(2.50\text{mm}\right)}^2\left(4.00\text{GPa}\right)}\left(\frac{1}{\cos 65°}-1\right)\\ =\frac{4\left(2.200\text{m}\right)\left(0.411\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{\pi {\left(2.50\times {10}^{-3}\text{m}\right)}^2\left(4.00\text{GPa}\right)}\left(\frac{1}{\cos 65°}-1\right)\\ =6.17\times {10}^{-4}\text{m}\end{array}$

The stretch in the string when the ball moves in a horizontal circle as compared to when the ball is at rest is $6.17\times {10}^{-4}\text{m}$.

(b)

To determine

The kinetic energy of the ball.

Answer to Problem 97P

The kinetic energy of the ball is $8.61\text{J}$.

Explanation of Solution

The mass of the ball of the tetherball set is $0.411\text{kg}$. The diameter of the nylon string is $2.50\text{mm}$. The Young’s modulus of nylon is $4.00\text{GPa}$. The density of the nylon is $1150{\text{kg/m}}^{\text{3}}$. The length of the string is $2.200\text{m}$.

Write the formula for the force in x-direction.

$T_2\sin \theta =\frac{m{v}^2}{r}$ (V)

Here, $m$ is the mass, $T_2$ is the tension when the ball is swinging, $\theta$ is the angle made by the string, $v$ is the velocity, $r$

Re-write the above equation in terms of kinetic energy.

$\begin{array}{c}{T}_2\sin \theta =\frac{2}{r}\left(\frac{m{v}^2}{2}\right)\\ =\frac{2K}{r}\end{array}$

Here, $K$ is the kinetic energy.

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

$K=\frac{T_{2}r\sin \theta }{2}$ (VI)

Write the equation for $r$.

$r=L\sin \theta$ (VII)

Here, $L$ is the length of the string.

Substitute equation (VII) and (II) in equation (VI).

$\begin{array}{c}K=\frac{mg\left(L\sin \theta \right)\sin \theta }{2\cos \theta }\\ =\frac{mgL\tan \theta \sin \theta }{2}\end{array}$ (VIII)

Conclusion:

Substitute $0.411\text{kg}$ for $m$, $65°$ for $\theta$, $2.200\text{m}$ for $L$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find the kinetic energy.

$\begin{array}{c}K=\frac{\left(0.411\text{kg}\right)\left(2.200\text{m}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\tan 65°\sin 65°}{2}\\ =8.61\text{J}\end{array}$

The kinetic energy of the ball is $8.61\text{J}$.

(c)

To determine

The time taken by a transverse wave pulse to travel the length of the string from the ball to the top of the poll.

Answer to Problem 97P

The time taken by a transverse wave pulse to travel the length of the string from the ball to the top of the poll is $0.0536\text{s}$.

Explanation of Solution

The mass of the ball of the tetherball set is $0.411\text{kg}$. The diameter of the nylon string is $2.50\text{mm}$. The Young’s modulus of nylon is $4.00\text{GPa}$. The density of the nylon is $1150{\text{kg/m}}^{\text{3}}$. The length of the string is $2.200\text{m}$.

Write the formula for the velocity of a wave through a string.

$v=\sqrt{\frac{T_2}{\mu }}$ (IX)

Here, $v$ is the speed of wave through a string, $T_2$ is the tension when the ball is swinging, $\mu$ is the linear mass density of the string.

Write the formula for the linear mass density of the string.

$\begin{array}{c}\mu =\frac{\rho AL}{L}\\ =\rho A\end{array}$ (X)

Here, $\rho$ is the density of the string, $A$ is the cross-sectional area, $L$ is the length of the string.

Substitute equation (X) in equation (IX).

$v=\sqrt{\frac{T_2}{\rho A}}$ (XI)

Write the formula for the time taken for a transverse wave to travel the string.

$t=\frac{L+\Delta L}{v}$ (XII)

Here, $t$ is the time taken, $\Delta L$ is the stretch.

Substitute equation (XI), (II) and (III) in equation (XII).

$\begin{array}{c}t=\left(L+\Delta L\right)\sqrt{\frac{\rho A}{T_2}}\\ =\left(L+\Delta L\right)\sqrt{\frac{\rho \pi d^2\cos \theta }{4mg}}\end{array}$ (XIII)

Conclusion:

Substitute $0.411\text{kg}$ for $m$ and $2.50\text{mm}$ for $d$, $65°$ for $\theta$, $2.200\text{m}$ for $L$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $6.17\times {10}^{-4}\text{m}$ for $\Delta L$, $1150{\text{kg/m}}^{\text{3}}$ for $\rho$ to find time taken.

$\begin{array}{c}t=\left(2.200\text{m}+6.17\times {10}^{-4}\text{m}\right)\sqrt{\frac{\left(1150{\text{kg/m}}^{\text{3}}\right)\pi {\left(2.50\text{mm}\right)}^2\cos 65°}{4\left(0.411\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}}\\ =\left(2.200\text{m}+6.17\times {10}^{-4}\text{m}\right)\sqrt{\frac{\left(1150{\text{kg/m}}^{\text{3}}\right)\pi {\left(2.50\times {10}^{-3}\text{m}\right)}^2\cos 65°}{4\left(0.411\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}}\\ =0.0536\text{s}\end{array}$

The time taken by a transverse wave pulse to travel the length of the string from the ball to the top of the poll is $0.0536\text{s}$.