Question

A ball of mass $\mathit{m}\mathbf{=}\mathbf{300}\mathit{g}$is connected by a strong string of length $\mathit{L}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$to a pivot and held in place with the string vertical. A wind exerts constant force $\mathit{F}$to the right on the ball as shown in Figure P8.82. The ball is released from rest. The wind makes it swing up to attain maximum height above its starting point before it swings down again. (a) Find $\mathit{H}$as a function of $\mathit{F}$. Evaluate $\mathit{H}$for (b) $\mathit{F}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}$and (c) $\mathit{F}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}$. How does $\mathit{H}$behave (d) as $\mathit{F}$approaches zero and (e) as approaches infinity? (f) Now consider the equilibrium height of the ball with the wind blowing. Determine it as a function of $\mathit{F}$. Evaluate the equilibrium height for (g) $\mathit{F}\mathbf{=}\mathbf{10}\mathit{N}$and (h) $\mathit{F}$going to infinity.

Short answer

(a)The value $H$of as a function of$F$ is $H=\frac{2L}{\left(1+{\left(\frac{mg}{F}\right)}^2\right)}$.

(b) The value of$H$ for $F=1.0\text{\hspace{0.33em}N}$is $0.166m$ .

(c) The value of$H$ for$F=10.0\text{\hspace{0.33em}N}$ is $1.47m$.

(d) The value of$H$ when $F$approaches zero is$\infty$ (undefined).

(e) The value of$H$ when $F$approaches infinity is $1.6m$.

(f) The equilibrium height of the ball with the wind blowing as a function of $F$is $h=\frac{F^{2}L}{mg\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}$.

(g) The equilibrium height of the ball for $F=10.0N$is $2.61m$.

(h) The equilibrium height of the ball for $F$going to infinity is $\infty$(undefined).

Step-by-step solution

Given information

The mass of the ballconnected by a strong stringis,$m=300\text{\hspace{0.33em}g}=0.3\text{\hspace{0.33em}kg}$.

The length of the string is,$L=80.0\text{\hspace{0.33em}cm}=0.8\text{\hspace{0.33em}m}$.

The force exerted by the windto the right on the ball is, $F$.

The maximum height attained by the ball is, $H$.

Work-Energy Principle

When an object is displaced in any direction due to applied force, the value of work done on the object is balanced by the total change in the energy of the object.

The change in total energy of an moving object is a combination of the kinetic energy change and the potential energy change.

Step 3(a): H as a function of F

Assuming, the maximum horizontal displacementattained by the ball is x.

Then, using Pythagoras theorem, the horizontal displacement of the ball is given by,

$\begin{array}{c}x=\sqrt{L^2-{\left(L-H\right)}^2}\\ x=\sqrt{L^2-L^2-H^2+2LH}\\ x=\sqrt{2LH-H^2}\end{array}$

The work done by the force exerted by the wind on the ball is given by,

$\begin{array}{l}{W}_F=F\times x\\ W_F=F\sqrt{2LH-H^2}\end{array}$

Applying work-energy theorem,

$\begin{array}{c}\text{Work\hspace{0.33em}done\hspace{0.33em}by\hspace{0.33em}force}=\text{Change\hspace{0.33em}in\hspace{0.33em}total\hspace{0.33em}energy}\\ W_F=\Delta KE+\Delta PE\\ F\sqrt{2LH-H^2}=0+mgH\end{array}$

Squaring both sides,

$\begin{array}{c}{F}^2\left(2LH-H^2\right)={\left(mgH\right)}^2\\ 2LH{F}^2=m^2{g}^2{H}^2+F^2{H}^2\\ 2LH{F}^2=\left(F^2+m^2{g}^2\right)H^2\\ H=\frac{2F^{2}L}{F^2+m^2{g}^2}\end{array}$

Rearranging terms,

$\begin{array}{c}H=\frac{2F^{2}L}{F^2\left(1+{\left(\frac{mg}{F}\right)}^2\right)}\\ H=\frac{2L}{\left(1+{\left(\frac{mg}{F}\right)}^2\right)}\mathrm{...}\left(1\right)\end{array}$

Hence, the value of $H$as a function of $F$is$H=\frac{2L}{\left(1+{\left(\frac{mg}{F}\right)}^2\right)}$

Step 4(b): Evaluate H for F = 1.0 N

Putting the values of $L,m,g$and $F=1.0N$in the equation (1),

$\begin{array}{c}H=\frac{2\times 0.8\text{\hspace{0.33em}m}}{\left(1+{\left(\frac{0.3\text{\hspace{0.33em}kg}\times 9.8{\text{\hspace{0.33em}m/s}}^2\times \frac{1\text{\hspace{0.33em}N}}{1\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2}}{1\text{\hspace{0.33em}N}}\right)}^2\right)}\\ H=\frac{1.6\text{\hspace{0.33em}m}}{\left(1+{\left(2.94\right)}^2\right)}\\ H=\frac{1.6\text{\hspace{0.33em}m}}{9.64}\\ H=0.166\text{\hspace{0.33em}m}\end{array}$

Hence, the value of $H$for$F=1.0N$ is $0.166m$.

Step 5(c): Evaluate H for F= 10.0 N

Putting the values of $L,m,g$and $F=10.0N$in the equation (1),

$\begin{array}{c}H=\frac{2\times 0.8\text{\hspace{0.33em}m}}{\left(1+{\left(\frac{0.3\text{\hspace{0.33em}kg}\times 9.8{\text{\hspace{0.33em}m/s}}^2\times \frac{1\text{\hspace{0.33em}N}}{1\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2}}{10\text{\hspace{0.33em}N}}\right)}^2\right)}\\ H=\frac{1.6\text{\hspace{0.33em}m}}{\left(1+{\left(0.294\right)}^2\right)}\\ H=\frac{1.6\text{\hspace{0.33em}m}}{1.086}\\ H=1.47\text{\hspace{0.33em}m}\end{array}$

Hence, the value of $H$for$F=10.0N$ is 1.47m .

Step 6(d): Evaluate  Hwhen F approaches zero 

From equation (1), the value of$H$when$F\to 0$is given by,

$\begin{array}{l}H=\frac{2L}{\left(1+{\left(\frac{mg}{0}\right)}^2\right)}\\ H=\frac{2L}{0}\\ H=\infty \end{array}$

Hence, the value of $H$when $F$approaches zero is $\infty$(undefined).

Step 7(e): Evaluate H when F approaches infinity

From equation (1), the value of $H$when $F\to \infty$is given by,

$\begin{array}{l}H=\frac{2L}{\left(1+{\left(\frac{mg}{\infty }\right)}^2\right)}\\ H=\frac{2L}{\left(1+{\left(0\right)}^2\right)}\\ H=2L\end{array}$

Putting the value of L,

$\begin{array}{l}H=2\times 0.8\text{\hspace{0.33em}m}\\ H=1.6\text{\hspace{0.33em}m}\end{array}$

Hence, the value of $H$when $F$approaches infinity is 1.6m.

Step 8(f): Equilibrium height of the ball with the wind blowing as a function of F

Assume, at theequilibrium, the height of the ball is $h$,the tension on the string is $T$and the angle made by string from vertical is$\theta$.

The free-body diagram of the ball at the equilibrium is given by,

Then, the force components of the tension on the string can be given as,

$T\sin \theta =F$

And

$T\cos \theta =mg$

$Then,$

$\begin{array}{c}\frac{T\sin \theta }{T\cos \theta }=\frac{F}{mg}\\ \tan \theta =\frac{F}{mg}\end{array}$

From right-angled triangle formula, the value of is given by,

$\sin \theta =\frac{F}{\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}$

At the equilibrium conditionof the ball with the wind blowing, applying work-energy principle,

$\begin{array}{c}\text{Work\hspace{0.33em}done\hspace{0.33em}by\hspace{0.33em}forces}=\text{Change\hspace{0.33em}in\hspace{0.33em}kinetic\hspace{0.33em}energy}\\ F\left(L\sin \theta \right)+mg\left(-h\right)+T\left(0\right)=0\\ FL\sin \theta =mgh\\ h=\frac{FL\sin \theta }{mg}\end{array}$

Putting the value of $\sin \theta$

$\begin{array}{c}h=\frac{FL}{mg}\times \frac{F}{\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}\\ h=\frac{F^{2}L}{mg\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}\mathrm{...}\left(2\right)\end{array}$

Hence, the equilibrium height of the ball with the wind blowing as a function of is

$h=\frac{F^{2}L}{mg\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}$

Step 9(g): The equilibrium height for F = 10.0 N

Putting the values of $L,m,g$and $F=10.0N$in the equation (2),

$\begin{array}{l}h=\frac{{\left(10.0\text{\hspace{0.33em}N}\right)}^2\times 0.8\text{\hspace{0.33em}m}}{0.3\text{\hspace{0.33em}kg}\times 9.8{\text{\hspace{0.33em}m/s}}^2\times \frac{1\text{\hspace{0.33em}N}}{1\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2}\sqrt{{\left(10.0\text{\hspace{0.33em}N}\right)}^2+{\left(0.3\text{\hspace{0.33em}kg}\times 9.8{\text{\hspace{0.33em}m/s}}^2\times \frac{1\text{\hspace{0.33em}N}}{1\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2}\right)}^2}}\\ h=\frac{80{\text{\hspace{0.33em}N}}^2\cdot \text{m}}{2.94\text{\hspace{0.33em}N}\sqrt{100{\text{\hspace{0.33em}N}}^2+{\text{8.6436\hspace{0.33em}N}}^2}}\\ h=\frac{80{\text{\hspace{0.33em}N}}^2\cdot \text{m}}{2.94\text{\hspace{0.33em}N}\times \text{10.42\hspace{0.33em}N}}\\ h=2.61\text{\hspace{0.33em}m}\end{array}$

Hence, the equilibrium height of the ball for$F=10.0N$is$2.61m$

Step 10(h): The equilibrium height for F going to infinity.

Fromequation (2),

$\begin{array}{c}h=\frac{F^{2}L}{mg\sqrt{{\left(F\right)}^2+{\left(mg\right)}^2}}\\ h=\frac{F^{2}L}{Fmg\sqrt{1+{\left(\frac{mg}{F}\right)}^2}}\\ h=\frac{FL}{mg\sqrt{1+{\left(\frac{mg}{F}\right)}^2}}\end{array}$

Then the value of $h$when $F\to \infty$is given by,

$\begin{array}{l}h=\frac{\infty L}{mg\sqrt{1+{\left(\frac{mg}{\infty }\right)}^2}}\\ h=\frac{1}{0}\\ h=\infty \end{array}$

Hence, the equilibrium height of the ball for $F$going to infinity is $\infty$(undefined).