The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of v0. If l = 2 ft, xB = 0.3 ft and yB = 0.4 ft determine the initial velocity v so that the ball will enter in the basket. Hint: use a computer to solve the resulting set of equations.

Question

The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of [latex]{ v }_{ 0 }[/latex]. If l= 2 ft, [latex]{ x }_{ B }= 0.3 \mathrm{\ ft}[/latex] and [latex]{ y }_{ B }= 0.4 \mathrm{\ ft}[/latex] determine the initial velocity v so that the ball will enter in the basket. Hint: use a computer to solve the resulting set of equations.

Accepted Answer

Let position 1 be at A. [latex]\quad\quad u_{1}= u_{0}[/latex]

Let position 2 be the point described by the angle [latex] heta[/latex] where the path of the ball changes from circular to parabolic. At position 2 the tension Q in the cord is zero.

Relationship between [latex] u_{2}[/latex] and [latex] heta[/latex] based on Q = 0. Draw the free body diagram.

[latex]+\swarrow \Sigma F=0:\quad Q+m g\sin heta=m a_{n}={\frac{m u_{2}^{2}}{\ell}}[/latex]

With Q=0, [latex]\quad u_{2}^{2}=g \ell \sin heta \quad ext { or } \quad u_{2}=\sqrt{g \ell \sin heta}[/latex] (1)

Relationship among [latex] u_{0}, u_{2}[/latex], and θ based on conservation of energy.

[latex]\begin{aligned}T_{1}+V_{1} & =T_{2}+V_{2} \\frac{1}{2} m u_{0}^{2}-m g \ell & =\frac{1}{2} m u_{2}^{2}+m g \ell \sin heta \ u_{0}^{2} & = u_{2}^{2}+2 g \ell(1+\sin heta) & ext{(2)}\end{aligned}[/latex]

x and y coordinates at position 2:

[latex]\begin{aligned}& x_{2}=\ell \cos heta & ext{(3)}\& y_{2}=\ell \sin heta & ext{(4)}\end{aligned}[/latex]

Let [latex]t_{2}[/latex] be the time when the ball is in position 2 .

Motion on the parabolic path. The horizontal motion is

[latex]\begin{aligned}\dot{x} & =- u_{2} \sin heta \x & =x_{2}-\left( u_{2} \sin heta ight)\left(t-t_{2} ight) & ext{(5)}\\end{aligned}[/latex]

At Point B,

[latex]\begin{aligned}& \quad\quad x =x_{B} \quad ext { and } \quad t=t_{B} . \quad ext { From Eq. (5), } \&\left(t_{B}-t_{2} ight) =\frac{\ell \cos heta-x_{B}}{ u_{ heta} \sin heta} & ext{(6)}\end{aligned}[/latex]

Vertical motion:

[latex]\begin{aligned}\dot{y} & = u_{2} \cos heta-g\left(t-t_{2} ight) \y & =y_{2}+\left( u_{2} \cos heta ight)\left(t-t_{2} ight)-\frac{1}{2} g\left(t-t_{2} ight)^{2}\end{aligned}[/latex]

At Point B,

[latex]y_{B}=\ell \sin heta+\left( u_{2} \cos heta ight)\left(t_{B}-t_{2} ight)-\frac{1}{2} g\left(t_{B}-t_{2} ight)^{2}[/latex] (7)

Data: [latex]\quad\ell=2 \mathrm{\ ft}, \quad x_{B}=0.3 \mathrm{\ ft}, \quad y_{B}=0.4 \mathrm{\ ft}, \quad g=32.2 \mathrm{\ ft} / \mathrm{s}^{2}[/latex]

With the numerical data,

Eq. (1) becomes [latex]\quad u_{2} =\sqrt{64.4 \sin heta}[/latex] (1)′

Eq. (6) becomes [latex]\quad t_{B}-t_{2} =\frac{2 \cos heta-0.3}{ u_{2} \sin heta}[/latex] (6)′

Eq. (7) becomes [latex]\quad y_{B}=2 \sin heta+\left( u_{2} \cos heta ight)\left(t_{B}-t_{2} ight)-16.1\left(t_{B}-t_{2} ight)^{2}[/latex] (7)′

Method of solution. From a trial value of [latex] heta[/latex], calculate [latex] u_{2}[/latex] from Eq. (1)′, [latex]t_{B}-t_{2}[/latex] from Eq. (6)', and [latex]y_{B}[/latex] from Eq. (7)'. Repeat until [latex]y_{B}=0.4 \mathrm{\ ft}[/latex] as required.

Try [latex] heta=30^{\circ}[/latex]

[latex]\begin{aligned} u_{2} & =\sqrt{64.4 \sin 30^{\circ}}=5.6745 \mathrm{\ ft} / \mathrm{s} \t_{B}-t_{2} & =\frac{2 \cos 30^{\circ}-0.3}{5.6745 \sin 30^{\circ}}=0.50473 \mathrm{~s} \y_{B} & =2 \sin 30^{\circ}+\left(5.6745 \cos 30^{\circ} ight)(0.50473)-(16.1)(0.50473)^{2} \& =-0.62116 \mathrm{\ ft}\end{aligned}[/latex]

Try [latex] heta=45^{\circ}[/latex].

[latex]\begin{aligned} u_{2} & =\sqrt{64.4 \sin 45^{\circ}}=6.7482 \t_{B}-t_{2} & =\frac{2 \cos 45^{\circ}-0.3}{6.7482 \sin 45^{\circ}}=0.23351 \mathrm{~s} \y_{B} & =2 \sin 45^{\circ}+\left(6.7482 \cos 45^{\circ} ight)(0.23351)-(16.1)(0.23351)^{2} \& =1.65060 \mathrm{\ ft}\end{aligned}[/latex]

Try [latex] heta=37.5^{\circ}[/latex].

[latex]\begin{aligned} u_{2} & =\sqrt{64.4 \sin 37.5^{\circ}}=6.2613 \mathrm{\ ft} / \mathrm{s} \t_{B}-t_{2} & =\frac{2 \cos 37.5^{\circ}-0.3}{6.2613 \sin 37.5^{\circ}}=0.33757 \mathrm{~s} \y_{B} & =2 \sin 37.5^{\circ}+\left(6.2613 \cos 37.5^{\circ} ight)(0.33757)-(16.1)(0.33757)^{2} \& =1.05972 \mathrm{\ ft}\end{aligned}[/latex]

Let [latex]u= heta-30^{\circ}[/latex]. The following sets of data points have be determined:

[latex]\left(u, y_{B} ight)=\left(0^{\circ},-0.62114 \mathrm{\ ft} ight),\left(7.5^{\circ}, 1.05972 \mathrm{\ ft} ight),\left(15^{\circ}, 1.65060 \mathrm{\ ft} ight)[/latex]

The quadratic curve fit of this data gives

[latex]y_{B}=-0.62114+0.29678 u-0.009688711 u^{2}[/latex]

Setting [latex]\quad y_{B}=0.4 \mathrm{\ ft}[/latex] gives the quadratic equation

[latex]-0.009688711 u^{2}+0.29678 u-1.02114=0[/latex]

Solving for u, [latex]\quad u=3.95^{\circ} ext { and } 26.68^{\circ}[/latex]

Rejecting the second value gives [latex]\quad heta=30^{\circ}+u=33.95^{\circ}.[/latex]

Try [latex] heta=33.95^{\circ}. [/latex]

[latex]\begin{aligned}& u_{2}=\sqrt{64.4 \sin 33.95^{\circ}}=5.997 \mathrm{\ ft} / \mathrm{s} \& t_{B}-t_{2}=\frac{2 \cos 33.95^{\circ}-0.3}{5.9971 \sin 33.95^{\circ}}=0.40578 \mathrm{~s} \& y_{B}=2 \sin 33.95^{\circ}+\left(5.997 \cos 33.95^{\circ} ight)(0.40578)-(16.1)(0.40578)^{2} \&=0.48462 \mathrm{\ ft}\end{aligned}[/latex]

The new quadratic curve-fit is based on the data points

[latex]\left(u, y_{B} ight)=\left(0^{\circ},-0.62114 \mathrm{\ ft} ight),\left(3.95^{\circ}, 0.48462 \mathrm{\ ft} ight),\left(7.5^{\circ}, 1.05972 \mathrm{\ ft} ight) .[/latex]

The quadratic curve fit of this data is

[latex]y_{B}=-0.62114+0.342053907 u-0.015725232 u^{2}[/latex]

Setting [latex]y_{B}=0.4 \mathrm{\ ft}[/latex] gives

[latex]-0.015725232 u^{2}+0.342053907 u-1.02114=0[/latex]

Solving for u,

[latex]u =3.572^{\circ} \quad heta=30^{\circ}+3.572^{\circ}=33.572^{\circ}[/latex]

Try [latex] heta=33.572^{\circ}. [/latex]

[latex]\begin{aligned} u_{2} & =\sqrt{64.4 \sin 33.572^{\circ}}=5.9676 \mathrm{\ ft} / \mathrm{s} \t_{B}-t_{2} & =\frac{2 \cos 33.572^{\circ}-0.3}{5.9676 \sin 33.572^{\circ}}=0.41406 \mathrm{~s} \y_{B} & =2 \sin 33.572^{\circ}+\left(5.9676 \cos 33.572^{\circ} ight)(0.41406)-(16.1)(0.41406)^{2} \& =0.40445 \mathrm{\ ft}\end{aligned}[/latex]

which is close enough to 0.4 ft.

Substituting [latex] heta=33.572^{\circ}[/latex] and [latex] u_{2}=5.9676 \mathrm{\ ft} / \mathrm{s}[/latex] into Eq. (2) along with other data gives

[latex]\begin{array}{r} u_{0}^{2}=(5.9676)^{2}+(2)(32.2)(2)\left(1+\sin 33.572^{\circ} ight)=235.64 \mathrm{\ ft}^{2} / \mathrm{s}^{2} \\mathbf{v}_{0}=15.35 \mathrm{\ ft} / \mathrm{s} \longrightarrow\blacktriangleleft\end{array}[/latex]

Question 13.201: The 2-lb ball at A is suspended by an inextensible cord and ...

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