Chapter 15.4, Problem 15.114P

To determine

The accelerations of points A, B, and C of the drums.

Answer to Problem 15.114P

The acceleration at point A $\left({\text{a}}_A\right)$ is $\underset{\_}{48\text{in}{\text{./s}}^{\text{2}}(\uparrow )}$.

The acceleration at point B $\left({\text{a}}_B\right)$ is $\underset{\_}{85.4\text{in}{\text{./s}}^{\text{2}}}$ with the angle of $\underset{\_}{69.4°\left(\text{II-Quad}\right)}$.

The acceleration at point C $\left({\text{a}}_C\right)$ is $\underset{\_}{82.8\text{in}{\text{./s}}^{\text{2}}}$ with the angle of $\underset{\_}{65°\left(\text{III-Quad}\right)}$.

Explanation of Solution

Given information:

The radius of inner drum (r) is 3 in..

The radius of outer drum (R) is 5 in..

The velocity of the cord at D is $8\text{in}\text{./s}$.

The acceleration of the cord at D is $30\text{in}{\text{./s}}^{\text{2}}$.

Calculation:

The velocity at point D and B is equal $v_B=v_D$.

Determine the angular velocity $\text{ω}$ using the relation.

$\begin{array}{c}{v}_A=l_{AB}\omega \\ =\left(R-r\right)\omega \end{array}$

Here, $l_{AB}$ is the length of the point AB.

Substitute 8 in. /s for $v_A$, 5 in. for R, and 3 in. for r.

$\begin{array}{l}8=\left(5-3\right)\omega \\ \omega =\frac{8}{2}\\ \omega =4\text{rad/s}\end{array}$

The value of acceleration at point A is ${\text{a}}_A=\left[a_A\uparrow \right]$ for no slipping.

Write the expression for the acceleration at point A $\left({\text{a}}_A\right)$.

${\text{a}}_B={\text{a}}_A+{\left({\text{a}}_{A/B}\right)}_t+{\left({\text{a}}_{A/B}\right)}_n$

Here, ${\text{a}}_A$ is the acceleration at point A, ${\left({\text{a}}_{A/B}\right)}_t$ is the tangential component of acceleration at point A with respect to B, and ${\left({\text{a}}_{A/B}\right)}_n$ is the normal component of acceleration at point A with respect to B.

Substitute $\left[a_A\uparrow \right]$ for ${\text{a}}_A$, $\left[{\left({\text{a}}_B\right)}_t\leftarrow +{\left({\text{a}}_B\right)}_n\uparrow \right]$ for ${\text{a}}_A$, $\left[\left(R-r\right)\alpha \to \right]$ for ${\left({\text{a}}_{B/A}\right)}_t$, and $\left[\left(R-r\right){\omega }^2\uparrow \right]$ for ${\left({\text{a}}_{B/A}\right)}_n$.

$\left[a_A\uparrow \right]=\left[{\left({\text{a}}_B\right)}_t\leftarrow +{\left({\text{a}}_B\right)}_n\uparrow \right]+\left[\left(R-r\right)\alpha \to \right]+\left[\left(R-r\right){\omega }^2\uparrow \right]$ (1)

Substitute $30\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_B\right)}_t$, 5 in. for R, and 3 in. for r.

$\left[a_A\uparrow \right]=\left[30\leftarrow +{\left({\text{a}}_A\right)}_n\uparrow \right]+\left[\left(5-3\right)\alpha \to \right]+\left[\left(5-3\right){\omega }^2\uparrow \right]$ . (2)

Determine the angular acceleration $\alpha$ using the relation.

Equate the horizontal components in Equation (2).

$\begin{array}{l}0=-30+2\alpha \\ 30=2\alpha \\ \alpha =\frac{30}{2}\\ \alpha =15{\text{rad/s}}^{\text{2}}\end{array}$

Write the expression for the acceleration at point A $\left({\text{a}}_A\right)$ with respect to G.

${\text{a}}_A={\text{a}}_G+{\left({\text{a}}_{A/G}\right)}_t+{\left({\text{a}}_{A/G}\right)}_n$

Substitute $\left[a_A\uparrow \right]$ for ${\text{a}}_A$, $\left[a_G\to \right]$ for ${\text{a}}_G$, $\left[r\alpha \leftarrow \right]$ for ${\left({\text{a}}_{A/G}\right)}_t$, and $\left[r{\omega }^2\uparrow \right]$ for ${\left({\text{a}}_{A/G}\right)}_n$.

$\left[a_A\uparrow \right]=\left[a_G\to \right]+\left[r\alpha \leftarrow \right]+\left[r{\omega }^2\uparrow \right]$

Substitute 3 in. for r.

$\left[a_A\uparrow \right]=\left[a_G\to \right]+\left[3\alpha \leftarrow \right]+\left[3{\omega }^2\uparrow \right]$ (3)

Equate the horizontal components in Equation (3).

$0=a_G-3\alpha$

Substitute $15{\text{rad/s}}^{\text{2}}$ for $\alpha$.

$\begin{array}{l}0=a_G-4\times 15\\ a_G=45\text{in}{\text{./s}}^{\text{2}}\end{array}$

Equate the vertical components in Equation (3).

Determine the acceleration at point A $\left({\text{a}}_A\right)$.

$a_A=3{\omega }^2$

Substitute $4\text{rad/s}$ for $\omega$.

$\begin{array}{c}{\text{a}}_A=3\times 4^2\\ =48\text{in}{\text{./s}}^{\text{2}}(\uparrow )\end{array}$

Therefore, the acceleration at point A $\left({\text{a}}_A\right)$ is $\underset{\_}{48\text{in}{\text{./s}}^{\text{2}}(\uparrow )}$.

Determine the acceleration at point B $\left({\text{a}}_B\right)$.

${\text{a}}_B={\text{a}}_G+{\left({\text{a}}_{B/G}\right)}_t+{\left({\text{a}}_{B/G}\right)}_n$

Substitute $\left[45\text{in}{\text{./s}}^{\text{2}}\to \right]$ for ${\text{a}}_G$, $\left[R\alpha \to \right]$ for ${\left({\text{a}}_{A/G}\right)}_t$, and $\left[R{\omega }^2\uparrow \right]$ for ${\left({\text{a}}_{A/G}\right)}_n$.

${\text{a}}_B=\left[45\to \right]+\left[R\alpha \to \right]+\left[R{\omega }^2\uparrow \right]$

Substitute 5 in. for R, $15{\text{rad/s}}^{\text{2}}$ for $\alpha$, and $4\text{rad/s}$ for $\omega$.

$\begin{array}{c}{\text{a}}_B=\left[45\to \right]+\left[5\times 15\leftarrow \right]+\left[5\times 4^2\uparrow \right]\\ =\left[45\to \right]+\left[75\leftarrow \right]+\left[80\uparrow \right]\\ =\left[30\text{in}{\text{./s}}^{\text{2}}\leftarrow \right]+\left[80\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the magnitude of the acceleration at point B.

${\text{a}}_B=\sqrt{{\left({\text{a}}_B\right)}_x^2+{\left({\text{a}}_B\right)}_y^2}$

Substitute $\text{30}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_B\right)}_x$ and $80\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_B\right)}_y$.

$\begin{array}{c}{\text{a}}_B=\sqrt{{30}^2+{80}^2}\\ =85.4\text{in}{\text{./s}}^{\text{2}}\end{array}$

Determine the direction of the acceleration at point B.

$\tan {\theta }_B=\frac{{\left({\text{a}}_B\right)}_y}{{\left({\text{a}}_B\right)}_x}$

Substitute $80\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_B\right)}_y$ and $\text{30}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_B\right)}_x$.

$\begin{array}{l}\tan {\theta }_B=\frac{80}{30}\\ {\theta }_B={\tan }^{-1}\left(2.667\right)\\ {\theta }_B=69.4°\left(\text{II}-\text{Quad}\right)\end{array}$

Therefore, the acceleration at point B $\left({\text{a}}_B\right)$ is $\underset{\_}{85.4\text{in}{\text{./s}}^{\text{2}}}$ with the angle of $\underset{\_}{69.4°\left(\text{II-Quad}\right)}$.

Determine the acceleration at point C $\left({\text{a}}_C\right)$.

${\text{a}}_C={\text{a}}_G+{\left({\text{a}}_{C/G}\right)}_t+{\left({\text{a}}_{C/G}\right)}_n$

Substitute $\left[45\text{in}{\text{./s}}^{\text{2}}\to \right]$ for ${\text{a}}_G$, $\left[R\alpha \to \right]$ for ${\left({\text{a}}_{C/G}\right)}_t$, and $\left[R{\omega }^2\leftarrow \right]$ for ${\left({\text{a}}_{C/G}\right)}_n$.

${\text{a}}_C=\left[45\leftarrow \right]+\left[R\alpha \uparrow \right]+\left[R{\omega }^2\leftarrow \right]$

Substitute 5 in. for R, $15{\text{rad/s}}^{\text{2}}$ for $\alpha$, and $4\text{rad/s}$ for $\omega$.

$\begin{array}{c}{\text{a}}_C=\left[45\to \right]+\left[5\times 15\uparrow \right]+\left[5\times 4^2\leftarrow \right]\\ =\left[45\to \right]+\left[75\uparrow \right]+\left[80\leftarrow \right]\\ =\left[35\text{in}{\text{./s}}^{\text{2}}\leftarrow \right]+\left[75\text{in}{\text{./s}}^{\text{2}}\uparrow \right]\end{array}$

Determine the magnitude of the acceleration at point C.

${\text{a}}_C=\sqrt{{\left({\text{a}}_C\right)}_x^2+{\left({\text{a}}_C\right)}_y^2}$

Substitute $\text{35}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_C\right)}_x$ and $\text{75}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_C\right)}_y$.

$\begin{array}{c}{\text{a}}_C=\sqrt{{35}^2+{75}^2}\\ =82.8\text{in}{\text{./s}}^{\text{2}}\end{array}$

Determine the direction of the acceleration at point C.

$\tan {\theta }_C=\frac{{\left({\text{a}}_C\right)}_y}{{\left({\text{a}}_C\right)}_x}$

Substitute $\text{75}\text{in}{\text{./s}}^{\text{2}}$ for ${\left({\text{a}}_C\right)}_y$ and $\text{35}\text{in}{\text{./s}}^{\text{2}}$ in ${\left({\text{a}}_C\right)}_x$.

$\begin{array}{l}\tan {\theta }_C=\frac{75}{35}\\ {\theta }_C={\tan }^{-1}\left(2.14\right)\\ {\theta }_C=65\left(\text{III}-\text{Quad}\right)\end{array}$

Therefore, the acceleration at point C $\left({\text{a}}_C\right)$ is $\underset{\_}{82.8\text{in}{\text{./s}}^{\text{2}}}$ with the angle of $\underset{\_}{65°\left(\text{III-Quad}\right)}$.