Chapter 16, Problem 16.161RP

A cylinder with a circular hole is rolling without slipping on a fixed curved surface as shown. The cylinder would have a weight of 16 lb without the hole, but with the hole it has a weight of 15 lb. Knowing that at the instant shown the disk has an angular velocity of 5 rad/s clockwise, determine (a) the angular acceleration of the disk, (b) the components of the reaction force between the cylinder and the ground at this instant.

Fig. P16.161

To determine

The angular acceleration of the disk.

Answer to Problem 16.161RP

The angular acceleration of the disk is $\underset{\_}{\alpha =0}$.

Explanation of Solution

Given information:

The weight of the cylinder without hole is $W_C=16\text{lb}$.

The weight of the cylinder with hole is $W=15\text{lb}$.

The angular velocity of the disk is $\omega =5\text{rad}/\text{s}$.

Calculation:

Consider the acceleration due to gravity as $g=32.2\text{ft}/{\text{s}}^{\text{2}}$.

Consider that the mass center of the cylinder is G and it lies at a distance b from center A, and C is the contact point between cylinder and the curved surface is the origin of the coordinate system.

Consider that the radius of cylinder is r and the radius of the curved surface is R.

Sketch the geometry of the cylinder as shown in Figure 1.

Refer to Figure 1.

Calculate the position vector $\left(\text{r}\right)$ as shown below.

The position vector of P with respect to C.

${\text{r}}_{P/C}=x\text{i}+y\text{j}$

The position vector of G with respect to C.

${\text{r}}_{G/C}=\left(r-b\right)\text{j}$

The position vector of A with respect to C.

${\text{r}}_{A/C}=r\text{j}$

The cylinder rolls without slipping on a fixed curved surface hence, the horizontal component of acceleration of point C is ${\left(a_C\right)}_x=0$.

Calculate the acceleration of point C $\left(a_C\right)$ as shown below.

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

Substitute 0 for ${\left(a_C\right)}_x$.

$\begin{array}{c}{a}_C=0+{\left(a_C\right)}_y\uparrow \\ ={\left(a_C\right)}_y\uparrow \end{array}$

The acceleration of the disk at A is ${\left(a_A\right)}_x=r\alpha$.

Calculate the acceleration of point A $\left(a_A\right)$ as shown below.

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

Substitute $r\alpha$ for ${\left(a_A\right)}_x$.

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

Calculate the acceleration of point P $\left(a_P\right)$ as shown below.

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

Here, $\alpha$ is the angular acceleration of cylinder.

Substitute ${\left(a_C\right)}_y\uparrow$ for $a_C$, $\alpha \text{k}$ for $\alpha$, and $x\text{i}+y\text{j}$ for $r_{P/C}$.

$\begin{array}{c}{a}_P={\left(a_C\right)}_y\uparrow +\left(\left(x\text{i}+y\text{j}\right)\times \alpha \text{k}\right)-{\omega }^2\left(x\text{i}+y\text{j}\right)\\ ={\left(a_C\right)}_y\uparrow -\alpha x\text{j}+\alpha y\text{i}-{\omega }^{2}x\text{i}-{\omega }^{2}y\text{j}\\ ={\left(a_C\right)}_y\uparrow +\left(\alpha x\right)\downarrow +\left(\alpha y\right)\to +\left({\omega }^{2}x\right)\leftarrow +\left({\omega }^{2}y\right)\downarrow \end{array}$

Calculate the acceleration of point G $\left(a_G\right)$ as shown below.

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

Substitute ${\left(a_C\right)}_y\uparrow$ for $a_C$, $\alpha \text{k}$ for $\alpha$, and $\left(r-b\right)\text{j}$ for $r_{G/C}$.

$\begin{array}{c}{a}_G={\left(a_C\right)}_y\uparrow +\left(\left(r-b\right)\text{j}\times \alpha \text{k}\right)-{\omega }^2\left(r-b\right)\text{j}\\ ={\left(a_C\right)}_y\uparrow +\left(r-b\right)\alpha \text{i}-{\omega }^2\left(r-b\right)\text{j}\\ ={\left(a_C\right)}_y\uparrow +\left(r-b\right)\alpha \to +\left(r-b\right){\omega }^2\downarrow \end{array}$ (1)

Calculate the acceleration of point A $\left(a_A\right)$ as shown below.

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

Substitute ${\left(a_C\right)}_y\uparrow$ for $a_C$, $\alpha \text{k}$ for $\alpha$, and $r\text{j}$ for $r_{A/C}$.

$\begin{array}{c}{a}_A={\left(a_C\right)}_y\uparrow +\left(\left(r\right)\text{j}\times \alpha \text{k}\right)-{\omega }^2\left(r\right)\text{j}\\ ={\left(a_C\right)}_y\uparrow +r\alpha \text{i}-{\omega }^{2}r\text{j}\\ ={\left(a_C\right)}_y\uparrow +r\alpha \to +r{\omega }^2\downarrow \end{array}$ (2)

Subtract Equation (2) from Equation (1) as shown below.

$\begin{array}{c}{a}_G-a_A={\left(a_C\right)}_y\uparrow +r\alpha \to +r{\omega }^2\downarrow -\left({\left(a_C\right)}_y\uparrow +\left(r-b\right)\alpha \to +\left(r-b\right){\omega }^2\downarrow \right)\\ =r\alpha \to +r{\omega }^2\downarrow -r\alpha \to -b\alpha \to -r{\omega }^2\downarrow -b{\omega }^2\downarrow \\ =-b\alpha \to -b{\omega }^2\downarrow \\ =b\alpha \leftarrow +b{\omega }^2\uparrow \end{array}$

Substitute $r\alpha \to +{\left(a_A\right)}_y\uparrow$ for $a_A$.

$\begin{array}{l}{a}_G-\left(r\alpha \to +{\left(a_A\right)}_y\uparrow \right)=b\alpha \leftarrow +b{\omega }^2\uparrow \\ a_G=r\alpha \to +{\left(a_A\right)}_y\uparrow +b\alpha \leftarrow +b{\omega }^2\uparrow \\ =\left(r-b\right)\alpha \to +{\left(a_A\right)}_y\uparrow +b{\omega }^2\uparrow \end{array}$ (3)

Calculate the velocity of point A $\left(v_A\right)$ as shown below.

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

Calculate the vertical component of acceleration of point A ${\left(a_A\right)}_y$ as shown below.

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

Substitute $r\omega$ for $v_A$.

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

Calculate the acceleration of point G $\left(a_G\right)$ as shown below.

Substitute $\frac{r^2{\omega }^2}{R+r}\downarrow$ for ${\left(a_A\right)}_y$ in Equation (3).

$\begin{array}{c}{a}_G=\left(r-b\right)\alpha \to +{\left(a_A\right)}_y\uparrow +b{\omega }^2\uparrow \\ =\left(r-b\right)\alpha \to +\left(\frac{r^2{\omega }^2}{R+r}\right)\downarrow +b{\omega }^2\uparrow \\ =\left(r-b\right)\alpha \to +{\omega }^2\left(\frac{r^2}{R+r}-b\right)\downarrow \end{array}$

Calculate the effective force at the mass center $\left(m{a}_G\right)$ as shown below.

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

Sketch the Free Body Diagram of the cylinder as shown in Figure 2.

Refer to Figure 2.

Apply the Equilibrium of moment about C as shown below.

$\begin{array}{l}{\displaystyle \sum M_C}=I_G\alpha +mad\\ 0=\left[m{\omega }^2\left(\frac{r^2}{R+r}-b\right)\right]\left[0\right]+I\alpha +\left[\left(r-b\right)\right]\left[m\left(r-b\right)\alpha \right]\\ I\alpha +\left[m{\left(r-b\right)}^2\alpha \right]=0\\ \left(I+m{\left(r-b\right)}^2\right)\alpha =0\end{array}$

$\alpha =0$

Hence, the angular acceleration of the disk is $\underset{\_}{\alpha =0}$.

To determine

The components of the reaction force between the cylinder and the ground.

Answer to Problem 16.161RP

The component of the reaction along x direction is $\underset{\_}{C_x=0}$.

The component of the reaction along y direction is $\underset{\_}{C_y=12.61}$.

Explanation of Solution

Given information:

The weight of the cylinder without hole is $W_C=16\text{lb}$.

The weight of the cylinder with hole is $W=15\text{lb}$.

The angular velocity of the disk is $\omega =5\text{rad}/\text{s}$.

Calculation:

Refer to part (a).

The angular acceleration of the disk is $\alpha =0$.

Refer to Figure 2.

Apply the Equilibrium of force along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=m{a}_x\\ C_x=m\left(r-b\right)\alpha \end{array}$

Substitute 0 for $\alpha$.

$\begin{array}{c}{C}_x=m\left(r-b\right)0\\ =0\end{array}$

Hence, the component of the reaction along x direction is $\underset{\_}{C_x=0}$.

Apply the Equilibrium of force along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=m{a}_y\\ C_y-W=-\frac{W}{g}\left(\frac{r^2}{R+r}-b\right){\omega }^2\\ C_y=W-\frac{W}{g}\left(\frac{r^2}{R+r}-b\right){\omega }^2\end{array}$ (4)

Sketch the cylinder with hole as shown in Figure 3.

Refer to Figure 3.

Calculate the area of solid cylinder $\left(A_1\right)$ as shown below.

$A_1=\pi r^2$

Calculate the center of gravity of solid cylinder in vertical direction $\left({\overline{y}}_1\right)$ as shown below.

${\overline{y}}_1=0$

Calculate the area of hole $\left(A_2\right)$ as shown below.

$A_2=\frac{\pi r^2}{16}$

Calculate the center of gravity of hole in vertical direction $\left({\overline{y}}_2\right)$ as shown below.

${\overline{y}}_2=\frac{2}{3}r$

Calculate the distance $\left(b\right)$ from the mass distribution of the cylinder as shown below.

$b=\frac{A_1{\overline{y}}_1-A_2{\overline{y}}_2}{\left(A_1-A_2\right)}$

Substitute $\pi r^2$ for $A_1$, $\frac{\pi r^2}{16}$ for $A_2$, 0 for ${\overline{y}}_1$, and $\frac{2}{3}r$ for ${\overline{y}}_2$.

$\begin{array}{c}b=\frac{\left(\pi r^2\right)0-\left(\frac{\pi r^2}{16}\right)\left(\frac{2}{3}r\right)}{\left(\pi r^2-\frac{\pi r^2}{16}\right)}\\ =\frac{-\frac{\pi r^3}{24}}{\frac{15\pi r^2}{16}}\\ =-\frac{2}{45}r\\ =\frac{2}{45}r\end{array}$

Calculate the component of the reaction along y direction as shown below.

Substitute $15\text{lb}$ for W, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g, $12\text{in}\text{.}$ for r, $36\text{ in}\text{.}$ for R, $\frac{2}{45}r$ for b, and $5\text{rad}/\text{s}$ for $\omega$ in Equation (4).

$\begin{array}{c}{C}_y=\left[\begin{array}{l}15\text{lb}-\frac{15\text{lb}}{32.2\text{ft}/{\text{s}}^{\text{2}}}\left(\frac{{\left(12\text{in}\text{.}\right)}^2}{\left(36\text{in}\text{.}\right)+\left(12\text{in}\text{.}\right)}-\frac{2}{45}\left(12\text{in}\text{.}\right)\right)\times \\ \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right){\left(5\text{rad}/\text{s}\right)}^2\end{array}\right]\\ =12.61\text{lb}\end{array}$

Therefore, the component of the reaction along y direction is $\underset{\_}{C_y=12.61}$.