Chapter 16, Problem 16.161RP

To determine

(a)

The force in the patellar tendon.

Explanation of Solution

Given information:

Cylinder weight with hole = 16 lb

Cylinder weight without hole = 15 lb

Angular velocity = 5 rad/s

Schematic of cylinder 1

Figure A

Unit vectors along X and Y directions be $i$ and $j$ and along angular direction $k$

Angular acceleration of cylinder be $\alpha$ In vector form it is $\alpha =\alpha k$

Let mass centre of the cylinder be G at a distance b from centre A.

Let point P coordinates with point C be $\left(x,y\right)$. So, its position vector will become $r_{P/C}=xi+yj$

Let point G coordinates with point C be $\left(0,\left(r-b\right)\right)$. So, its position vector will become $r_{G/C}=\left(r-b\right)j$

Let point A coordinates with point C be $\left(0,r\right)$. So, its position vector will become $r_{A/C}=rj$

Point C acceleration,

$a_C={\left(a_C\right)}_x\to +{\left(a_C\right)}_y\uparrow$

Here, ${\left(a_C\right)}_x$ = Point C acceleration in horizontal direction

${\left(a_C\right)}_y$ = Point C acceleration in vertical direction

Here, ${\left(a_C\right)}_x$ = 0 since cylinder rolls on surface that is curved.

Hence,

$a_C=0+{\left(a_C\right)}_y\uparrow$

$a_C={\left(a_C\right)}_y\uparrow$

At any location on the cylinder, acceleration is

$a_P=a_C+\left(\alpha \times r_{P/C}\right)-{\omega }^2{r}_{P/C}$

$\therefore a_P={\left(a_C\right)}_y\uparrow +\left(\alpha k\times \left(xi+yj\right)\right)-{\omega }^2\left(xi+yj\right)$

$a_P={\left(a_C\right)}_y\uparrow +\alpha x\left(k\times i\right)+\alpha y\left(k\times j\right)-{\omega }^{2}xi-{\omega }^{2}yj$

$a_P={\left(a_C\right)}_y\uparrow +\alpha x\left(-j\right)+\alpha y\left(i\right)+{\omega }^{2}x\leftarrow +{\omega }^{2}y\downarrow$

$a_P={\left(a_C\right)}_y\uparrow +\alpha x\downarrow +\alpha y\to +{\omega }^{2}x\leftarrow +{\omega }^{2}y\downarrow$

Point G acceleration,

$a_G=a_C+\left(r_{G/C}\times \alpha \right)-{\omega }^2{r}_{G/C}$

$a_G={\left(a_C\right)}_y\uparrow +\left(\left(r-b\right)j\times \alpha k\right)-{\omega }^2\left(r-b\right)j$

$={\left(a_C\right)}_y\uparrow +\left(\left(r-b\right)\alpha \right)\left(j\times k\right)-{\omega }^2\left(r-b\right)j$

$={\left(a_C\right)}_y\uparrow +\left(\left(r-b\right)\alpha \right)\left(i\right)-{\omega }^2\left(r-b\right)j$

$a_G={\left(a_C\right)}_y\uparrow +\left(r-b\right)\alpha \to +\left(r-b\right){\omega }^2\downarrow$ .............(Equation A)

Point A acceleration,

$a_A=a_C+\left(r_{A/C}\times \alpha \right)-{\omega }^2{r}_{A/C}$

$a_A={\left(a_C\right)}_y\uparrow +\left(\left(r\right)j\times \alpha k\right)-{\omega }^2\left(r\right)j$

$={\left(a_C\right)}_y\uparrow +\left(\left(r\right)\alpha \right)\left(j\times k\right)-{\omega }^2\left(r\right)j$

$={\left(a_C\right)}_y\uparrow +\left(r\alpha \right)\left(i\right)-{\omega }^2\left(r\right)j$

$a_A={\left(a_C\right)}_y\uparrow +r\alpha \to +r{\omega }^2\downarrow$

Substitute ${\left(a_C\right)}_y\uparrow =a_A-r\alpha \to -r{\omega }^2\downarrow$ in equation A

$a_G=a_A-r\alpha \to -r{\omega }^2\downarrow +\left(r-b\right)\alpha \to +\left(r-b\right){\omega }^2\downarrow$

$=a_A-r\alpha \to -r{\omega }^2\downarrow +r\alpha \to -b\alpha \to +r{\omega }^2\downarrow -b{\omega }^2\downarrow$ ....(Equation B)

$=a_A+b\alpha \leftarrow +b{\omega }^2\uparrow$

Point A acceleration,

$a_A={\left(a_A\right)}_x+{\left(a_A\right)}_y$

$a_A=r\alpha \to +{\left(a_A\right)}_y\uparrow$

Substitute the above in equation B,

$a_G=r\alpha \to +{\left(a_A\right)}_y\uparrow +b\alpha \leftarrow +b{\omega }^2\uparrow$

$=\left(r-b\right)\alpha \to +{\left(a_A\right)}_y\uparrow +b{\omega }^2\uparrow$ ......(Equation C)

Point A velocity,

$v_A=\left(r\omega \right)\to$

Point A acceleration vertical component,

${\left(a_A\right)}_y=\left(\frac{v_A^2}{R+r}\right)\downarrow$

${\left(a_A\right)}_y=\left(\frac{{\left(r\omega \right)}^2}{R+r}\right)\downarrow$

${\left(a_A\right)}_y=\left(\frac{r^2{\omega }^2}{R+r}\right)\downarrow$

Substitute the above in equation C,

$a_G=\left(r-b\right)\alpha +{\left(a_A\right)}_y\uparrow +b{\omega }^2\uparrow$

$=\left(r-b\right)\alpha \to +\left(\frac{r^2{\omega }^2}{R+r}\right)\downarrow +b{\omega }^2\uparrow$

$=\left(r-b\right)\alpha \to +{\omega }^2\left(\frac{r^2}{R+r}-b\right)\downarrow$

Effective force,

$m{a}_G=m\left[\left(r-b\right)\alpha \to +{\omega }^2\left(\frac{r^2}{R+r}-b\right)\downarrow \right]$

$=m\left(r-b\right)\alpha \to +m{\omega }^2\left(\frac{r^2}{R+r}-b\right)\downarrow$

Cylinder free body diagram

Figure B

Moment at point C from above figure,

$0=\left[m{\omega }^2\left(\frac{r^2}{R+r}-b\right)\right]\left[0\right]+I\alpha +\left[\left(r-b\right)\right]\left[m\left(r-b\right)\alpha \right]$

$I\alpha +\left[m{\left(r-b\right)}^2\alpha \right]=0$ .... (Equation D)

$\left(I+m{\left(r-b\right)}^2\right)\alpha =0$

From equation D, angular acceleration is zero.

Conclusion:

The cylinder angular acceleration is zero.

To determine

(b)

Components of the reaction force between the cylinder and the ground.

Explanation of Solution

Given information:

Cylinder weight with hole = 16 lb

Cylinder weight without hole = 15 lb

Angular velocity = 5 rad/s

Forces horizontal component from figure B,

$C_x=m(r-b)\alpha$

Here, $\alpha =0$

Hence, $C_x=m(r-b)0$

$C_x=0$

Forces vertical component from figure B,

$C_y=W-\frac{W}{g}\left(\frac{r^2}{R+r}-b\right){\omega }^2$ ...........(Equation E)

Cylinder as combination of hole and solid is shown below

Solid cylinder area,

$A_1=\Pi r^2$

Solid centre of gravity in vertical direction,

$\overline{y_1}=0$

Hole area,

$A_2=-\frac{\Pi r^2}{16}$

Hole centre of gravity in vertical direction,

$\overline{y_2}=\frac{2}{3}r$

Equilibrium equation,

$b\sum A=\sum A\overline{y_1}$

$b\left(A_1+A_2\right)=A_1\overline{y_1}+A_2\overline{y_2}$

$b\left(\Pi r^2-\frac{\Pi r^2}{16}\right)=\left(\Pi r^2\right)0+\left(-\frac{\Pi r^2}{16}\right)\left(\frac{2}{3}r\right)$

$b\left(\frac{15\Pi r^2}{16}\right)=-\frac{\Pi r^3}{24}$

$b=\frac{2}{45}r$

Substituting in equation E we get,

$C_y=\left\{15lb-\frac{15lb}{32.2ft/s^2}\left(\frac{{\left(12in.\right)}^2}{\left(36in.\right)+\left(12in.\right)}-\frac{2}{45}\left(12in.\right)\right)\left(\frac{1ft}{12in.}\right){\left(5rad/s\right)}^2\right\}$

$C_y=12.61lb$

Conclusion:

The horizontal component reaction is zero and vertical reaction is $C_y=12.61lb$.