Chapter 13.1, Problem 13.45P

A small block slides at a speed $v=8$ ft/s on a horizontal surface at a height $h=3$ ft above the ground. Determine (a) the angle $\theta$ at which it will leave the cylindrical surface BCD, (b) the distance x at which it will hit the ground. Neglect friction and air resistance.

To determine

(a)

The angle $\theta$ at which it will leave the cylindrical surface BCD by using given data.

Answer to Problem 13.45P

The value of angle,? = 27.4⁰.

Explanation of Solution

Given information:

v=8ft/s

h=3ft

Trajectory of the block diagram is,

Consider the point of the block leaves the surface C.

Then the normal reaction of the surface to the block is zero.

Force balance in the direction normal to the curvature at the point C is,

$\begin{array}{l}\sum F_n=m{a}_n\\ mg\cos \theta =m{a}_n\\ a_n=g\cos \theta \end{array}$

Expression for the force acting normal acceleration is,

$a_n=\frac{v_C^2}{r}$

Substitute this value we get,

$\begin{array}{l}\frac{v_C^2}{r}=g\cos \theta \\ v_C^2=g\left(r\cos \theta \right)\end{array}$

Trigonometry we observe that ,

$\begin{array}{l}r\cos \theta =h^{\text{'}}\\ v_C^2=g{h}^{\text{'}}-----(1)\end{array}$

Write the equation of kinetic energy at the points B and C we get,

$T_B=\frac{1}{2}m{v}_B^2------(2)$

$T_C=\frac{1}{2}m{v}_C^2------(3)$

Then the equation for loss in the potential energy is,

$U_{\left(B-C\right)}=\left(mg\right)\times \left(\Delta h\right)$

Substituting $\Delta h=h-h^{\text{'}}$ in above equation we get,

$U_{\left(B-C\right)}=\left(mg\right)\times \left(h-h^{\text{'}}\right)----(4)$

Motion of the block from B to C and take the energy balance equation.

$T_B+U_{\left(B-C\right)}=T_C$

Substituting equation (2), (3), and (4) we get,

$\begin{array}{l}\frac{1}{2}m{v}_B^2+mg\left(h-h^{\text{'}}\right)=\frac{1}{2}m{v}_C^2\\ v_C^2=v_B^2+2g\left(h-h^{\text{'}}\right)\end{array}$

Substitute equation (1),

$\begin{array}{l}g{h}^{\text{'}}=v_B^2+2g\left(h-h^{\text{'}}\right)\\ h^{\text{'}}=\frac{v_B^2+2gh}{3g}\\ h^{\text{'}}=\frac{{\left(8ft/s\right)}^2+2\left(32.2ft/s^2\right)\left(3ft\right)}{3\left(32.2ft/s^2\right)}\\ =2.663ft\end{array}$

Find the angle $\theta$ using the geometry,

$\begin{array}{l}\cos \theta =\frac{h^{\text{'}}}{h}\\ \cos \theta =\frac{2.663}{3}\\ \theta ={\cos }^{-1}\left(0.889\right)\\ ={27.4}^0\end{array}$

To determine

(b)

The distance “x” by using given information.

Answer to Problem 13.45P

The value of distance ,x=3.81ft.

Explanation of Solution

Consider the equation (1)

$\begin{array}{l}{v}_C^2=g{h}^{\text{'}}\\ v_C=\sqrt{g{h}^{\text{'}}}\\ v_C=\sqrt{\left(32.2ft/s^2\right)\left(2.663ft\right)}\\ =9.26ft/s\end{array}$

Determine the horizontal component of the velocity at the point C.

$\begin{array}{l}{v}_{Cx}=v_c\cos \theta \\ v_{Cx}=\left(9.26ft/s\right)\cos {27.4}^0\\ =8.22ft/s\end{array}$

Find the vertical component of the velocity at point C.

$\begin{array}{l}{v}_{Cy}=v_c\sin \theta \\ v_{Cx}=\left(9.26ft/s\right)\sin {27.4}^0\\ =4.26ft/s\end{array}$

Then the equation of motion in the vertical direction for the block.

Consider upward as positive.

$\begin{array}{l}-h^{\text{'}}=-v_{Cy}t+\frac{1}{2}\left(-g\right)t^2\\ h^{\text{'}}=v_{Cy}t+\frac{1}{2}g{t}^2\end{array}$

Substituting the above values we get,

$\begin{array}{l}2.663=\left(4.26\right)t+\frac{1}{2}\left(32.2\right)t^2\\ 16.1t^2+4.26t-2.663=0\\ t=0.295\sec ,t=-0.56\sec \end{array}$

Find the horizontal distance covered by the block.

$\begin{array}{l}{x}_h=v_{Cx}t\\ =\left(8.22ft/s\right)\left(0.295\sec \right)\\ =2.43ft\end{array}$

Find the total horizontal distance covered from the point B.

$\begin{array}{l}x=h\sin \theta +x_h\\ =\left(3ft\right)\sin {27.4}^0+\left(2.43ft\right)\\ =3.81ft\end{array}$