Chapter 17.2, Problem 17.89P

A 1.8-kg collar A and a 0.7-kg collar B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about its vertical axis of symmetry. The two collars are connected by a cord running over a pulley that is attached to the frame at O. At the instant shown, the velocity v_A of collar A has a magnitude of 2.1 m/s and a stop prevents collar B from moving. The stop is suddenly removed and collar A moves toward E. As it reaches a distance of 0.12 m from O, the magnitude of its velocity is observed to be 2.5 m/s. Determine at that instant the magnitude of the angular velocity of the frame and the moment of inertia of the frame and pulley system about CD.

Fig. P17.89

To determine

Find the magnitude of the angular velocity of the frame and the moment of inertia of the frame and pulley system about CD.

Answer to Problem 17.89P

The magnitude of the angular velocity of the frame is $\underset{\_}{18.83\text{rad/s}}$.

The magnitude of the moment of inertia of the frame and pulley system about CD is $\underset{\_}{0.0508\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

Given information:

The mass $\left(m_A\right)$ of the collar A is 1.8 kg.

The mass $\left(m_B\right)$ of the collar B is 0.7 kg.

The velocity $\left(v_A\right)$ of collar A is 2.1 m/s.

The distance $\left[{\left(r_A\right)}_2\right]$ of the collar A from frame O at final position is 0.12 m.

The velocity $\left[{\left(v_A\right)}_2\right]$ of the collar A at final position is 2.5 m/s.

Calculation:

Write the equation of the velocity component $\left(v_A\right)$ of the collar A using kinematics.

$v_A^2={\left(v_A\right)}_r^2+{\left(v_A\right)}_{\theta }^2$ (1)

Here, ${\left(v_A\right)}_r$ is the velocity of the collar A in radial direction and ${\left(v_A\right)}_{\theta }$ is the velocity of the collar A in angular direction.

Write the equation of the velocity ${\left(v_A\right)}_{\theta }$ of the collar A in angular direction.

${\left(v_A\right)}_{\theta }=r_A\omega$

Here, $r_A$ is the distance of the collar A from frame O and $\omega$ is the angular velocity of the frame.

Let $\Delta r_A$ be the constraint in collar A and $\Delta y_B$ be the constraint in collar B. Therefore, the constraint of cable AB.

$\Delta r_A=\Delta y_B$                                                                                                                        (2)

Let ${\left(v_A\right)}_1$ be the velocity of the collar A at initial position, ${\left[{\left(v_A\right)}_r\right]}_1$ be the velocity of the collar A in radial direction at initial position and ${\left[{\left(v_A\right)}_{\theta }\right]}_1$ be the velocity of the collar A in angular direction at initial position. Therefore, ${\left(v_A\right)}_1=2.1\text{m}/\text{s}$, ${\left[{\left(v_A\right)}_r\right]}_1=0$, and ${\left[{\left(v_A\right)}_{\theta }\right]}_1=\left(\Delta r_A\right){\omega }_1$.

Write the equation of the velocity component $\left[{\left(v_A\right)}_1\right]$ of the collar A at initial position.

${\left(v_A\right)}_1^2={\left({\left(v_A\right)}_r\right)}_1^2+{\left({\left(v_A\right)}_{\theta }\right)}_1^2$

Substitute $2.1\text{m}/\text{s}$ for ${\left(v_A\right)}_1$, and 0 for ${\left[{\left(v_A\right)}_r\right]}_1$.

$\begin{array}{l}{\left(2.1\right)}^2={\left(0\right)}^2+{\left({\left(v_A\right)}_{\theta }\right)}_1^2\\ {\left[{\left(v_A\right)}_{\theta }\right]}_1^2={\left(2.1\text{m}/\text{s}\right)}^2\\ {\left[{\left(v_A\right)}_{\theta }\right]}_1=2.1\text{m}/\text{s}\end{array}$

Find the angular velocity $\left({\omega }_1\right)$ at initial position using the equation.

${\left[{\left(v_A\right)}_{\theta }\right]}_1=\left(\Delta r_A\right){\omega }_1$

Substitute $2.1\text{m}/\text{s}$ for ${\left({\left(v_A\right)}_{\theta }\right)}_1$, and 0.1 m for $\Delta r_A$.

$\begin{array}{l}2.1=\left(0.1\right){\omega }_1\\ {\omega }_1=\frac{2.1\text{m}/\text{s}}{0.1\text{m}}\\ {\omega }_1=21\text{rad}/\text{s}\end{array}$

The velocity of the collar A in radial direction at initial position ${\left({\left(v_A\right)}_r\right)}_1$ will be equal to the velocity of the collar B $\left(v_B\right)$. Therefore, $v_B=0$.

Find the equation of the kinetic energy $\left(T_1\right)$ at initial position.

$T_1=\frac{1}{2}I{\omega }_1^2+\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2$

Here, I is the moment of inertia of the frame,

Substitute $21\text{rad}/\text{s}$ for ${\omega }_1$, 1.8 kg for $m_A$, $2.1\text{m}/\text{s}$ for $v_A$, 0.7 kg for $m_B$, and 0 for $v_B$.

$\begin{array}{c}{T}_1=\frac{1}{2}I{\left(21\text{rad}/\text{s}\right)}^2+\frac{1}{2}\left(1.8\text{kg}\right){\left(2.1\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(0.7\text{kg}\right){\left(0\right)}^2\\ =220.5I+3.969+0\\ =220.5I+3.969\end{array}$

At initial position, potential energy will be zero. Therefore, $V_1=0$.

Write the equation of the angular momentum ${\left(H_O\right)}_1$ at initial position.

${\left(H_O\right)}_1=I{\omega }_1+m_A{\left({\left(v_A\right)}_r\right)}_1{r}_A$

Substitute $21\text{rad}/\text{s}$ for ${\omega }_1$, 1.8 kg for $m_A$, $2.1\text{m}/\text{s}$ for ${\left({\left(v_A\right)}_r\right)}_1$, and 0.1 m for $r_A$.

$\begin{array}{c}{\left(H_O\right)}_1=I\left(21\text{rad}/\text{s}\right)+\left(1.8\text{kg}\right)\left(2.1\text{m}/\text{s}\right)\left(0.1m\right)\\ =21I+0.378\end{array}$

Let ${\left[{\left(v_A\right)}_r\right]}_2$ be the velocity of the collar A in radial direction at final position and ${\left[{\left(v_A\right)}_{\theta }\right]}_2$ be the velocity of the collar A in angular direction at final position.

Write the equation of the velocity ${\left[{\left(v_A\right)}_r\right]}_2$ of the collar A in radial direction at final position.

${\left[{\left(v_A\right)}_{\theta }\right]}_2={\left(r_A\right)}_2{\omega }_2$.

Here, ${\omega }_2$ is the angular velocity of the frame.

Substitute 0.12 m for ${\left(r_A\right)}_2$

$\begin{array}{c}{\left[{\left(v_A\right)}_{\theta }\right]}_2=\left(0.12\right){\omega }_2\\ =0.12{\omega }_2\end{array}$

Find the equation of the velocity component ${\left(v_A\right)}_2$ of the collar A at final position.

$\begin{array}{l}{\left(v_A\right)}_2^2={\left({\left(v_A\right)}_r\right)}_2^2+{\left({\left(v_A\right)}_{\theta }\right)}_2^2\\ {\left({\left(v_A\right)}_r\right)}_2^2={\left(v_A\right)}_2^2-{\left({\left(v_A\right)}_{\theta }\right)}_2^2\end{array}$

Substitute $2.5\text{m}/\text{s}$ for ${\left(v_A\right)}_2$, and $0.12{\omega }_2$ for ${\left[{\left(v_A\right)}_{\theta }\right]}_2$.

$\begin{array}{c}{\left[{\left(v_A\right)}_r\right]}_2^2={\left(2.5\text{m}/\text{s}\right)}^2-{\left(0.12{\omega }_2\right)}^2\\ =6.25-0.0144{\omega }_2^2\end{array}$

Find the velocity $\left(v_B\right)$ of the collar B using the equation:

$\begin{array}{l}{v}_B^2=6.25-0.0144{\omega }_2^2\\ v_B=\sqrt{6.25-0.0144{\omega }_2^2}\end{array}$

Find the change in radial direction or constraint in collar A $\left(\Delta r_A\right)$ using the equation.

$\Delta r_A={\left(r_A\right)}_2-{\left(r_A\right)}_1$

Substitute 0.12 m for ${\left(r_A\right)}_2$ and 0.1 m for ${\left(r_A\right)}_1$.

$\begin{array}{c}\Delta r_A=0.12-0.1\\ =0.02\text{m}\end{array}$

Find the constraint in collar B using Equation (2).

$\Delta r_A=\Delta y_B$

Substitute $0.02\text{m}$ for $\Delta r_A$.

$\Delta y_B=0.02\text{m}$

Find the equation of the kinetic energy $\left(T_2\right)$ at final position.

$T_2=\frac{1}{2}I{\omega }_2^2+\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2$

Substitute 1.8 kg for $m_A$, $2.5\text{m}/\text{s}$ for $v_A$, 0.7 kg for $m_B$, and $\sqrt{6.25-0.0144{\omega }_2^2}$ for $v_B$.

$\begin{array}{c}{T}_2=\frac{1}{2}I{\omega }_2^2+\frac{1}{2}\left(1.8\text{kg}\right){\left(2.5\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(0.7\text{kg}\right){\left(\sqrt{6.25-0.0144{\omega }_2^2}\right)}^2\\ =\frac{1}{2}I{\omega }_2^2+\frac{1}{2}\left(1.8\text{kg}\right){\left(2.5\text{m}/\text{s}\right)}^2+\left(0.35\text{kg}\right)\left(6.25-0.0144{\omega }_2^2\right)\\ =0.5I{\omega }_2^2+5.625+2.1875-0.00504{\omega }_2^2\\ =\left(0.5I-0.00504\right){\omega }_2^2+7.8125\end{array}$

Find the potential energy $\left(V_2\right)$ at final position using the equation:

$V_2=m_{B}g\left(\Delta y_B\right)$

Substitute 0.7 kg for $m_B$, $9.81{\text{m}/\text{s}}^2$ for g, and 0.02 m for $\Delta y_B$.

$\begin{array}{c}{V}_2=\left(0.7\right)\left(9.81\right)\left(0.02\right)\\ =0.1373\text{J}\end{array}$

Find the angular momentum at final position ${\left(H_O\right)}_2$ using the equation:

${\left(H_O\right)}_2=I{\omega }_2+m_A{\left({\left(v_A\right)}_{\theta }\right)}_2{\left(r_A\right)}_2$

Substitute 1.8 kg for $m_A$, $0.12{\omega }_2$ for ${\left({\left(v_A\right)}_{\theta }\right)}_2$, and 0.12 m for ${\left(r_A\right)}_2$.

$\begin{array}{c}{\left(H_O\right)}_2=I{\omega }_2+\left(1.8\text{kg}\right)\left(0.12{\omega }_2\right)\left(0.12m\right)\\ =I{\omega }_2+0.02592{\omega }_2\\ =\left(I+0.02592\right){\omega }_2\end{array}$

Consider the conservation of angular momentum.

${\left(H_O\right)}_1={\left(H_O\right)}_2$

Substitute $21I+0.378$ for ${\left(H_O\right)}_1$, and $\left(I+0.02592\right){\omega }_2$ for ${\left(H_O\right)}_2$.

$\begin{array}{l}21I+0.378=\left(I+0.02592\right){\omega }_2\\ {\omega }_2=\frac{21I+0.378}{I+0.02592}\end{array}$ (3)

Take $21I+0.378$ as N and $I+0.02592$ as D.

${\omega }_2=\frac{N}{D}$

Consider the conservation of energy.

$T_1+V_1=T_2+V_2$

Substitute $220.5I+3.969$ for $T_1$, 0 for $V_1$, $\left(0.5I-0.00504\right){\omega }_2^2+7.8125$ for $T_2$, and 0.13734 J for $V_2$.

$\begin{array}{l}\left(220.5I+3.969\right)+0=\left(\left(0.5I-0.00504\right){\omega }_2^2+7.8125\right)+0.13734\text{J}\\ 220.5I+3.969=\left(0.5I-0.00504\right){\omega }_2^2+7.8125+0.13734\\ 220.5I-\left(0.5I-0.00504\right){\omega }_2^2-7.8125-0.13734+3.969=0\\ 220.5I-\left(0.5I-0.00504\right){\omega }_2^2-3.98084=0\end{array}$

Substitute $\frac{N}{D}$ for ${\omega }_2$.

$\begin{array}{l}220.5I-\left(0.5I-0.00504\right){\left(\frac{N}{D}\right)}^2-3.98084=0\\ 220.5I-\left(0.5I-0.00504\right)\frac{N^2}{D^2}-3.98084=0\\ 220.5I{D}^2-\left(0.5I-0.00504\right)N^2-3.98084D^2=0\\ 220.5I{D}^2-0.5I{N}^2+0.00504N^2-3.98084D^2=0\end{array}$

Substitute $21I+0.378$ for N and $I+0.02592$ for D in Equation (6).

$\begin{array}{l}\left\{\begin{array}{l}220.5I{\left(I+0.02592\right)}^2-0.5I{\left(21I+0.378\right)}^2\\ +0.00504{\left(21I+0.378\right)}^2-3.98084{\left(I+0.02592\right)}^2\end{array}\right\}=0\\ \left\{\begin{array}{l}220.5I\left(I^2+0.05184I+0.0006718\right)\\ -0.5I\left(441I^2+15.876I+0.14288\right)\\ +0.00504\left(441I^2+15.876I+0.14288\right)\\ -3.98084\left(I^2+0.05184I+0.0006718\right)\end{array}\right\}=0\end{array}$

$\begin{array}{l}0I^3+1.73452I^2-0.04965167I-0.001954378=0\\ 1.73452I^2-0.04965167I-0.001954378=0\end{array}$ (4)

Solve equation (4),

$I=0.0508\text{kg}\cdot {\text{m}}^{\text{2}}$

Thus, the magnitude of the moment of inertia of the frame and pulley system about CD $\underset{\_}{0.0508\text{kg}\cdot {\text{m}}^2}$.

Find the magnitude of the angular velocity of the frame $\left({\omega }_2\right)$ using Equation (3).

${\omega }_2=\frac{21I+0.378}{I+0.02592}$

Substitute $0.050804$ for I.

$\begin{array}{c}{\omega }_2=\frac{21\left(0.0508\right)+0.378}{0.0508+0.02592}\\ =\frac{1.445}{0.07672}\\ =18.83\text{rad}/\text{s}\end{array}$

Thus, the magnitude of the angular velocity of the frame is $\underset{\_}{18.83\text{rad/s}}$.