Chapter 6.9, Problem 167RP

To determine

The final angular velocity

Answer to Problem 167RP

The final angular velocity is $1.811\text{ rad}/\text{s}$.

Explanation of Solution

Draw the free body diagram of the door as shown in Figure (1).

Write the expression to determine the moment of inertia of the gate about $O\left(I_O\right)$.

  $I_O=4\left(\frac{1}{3}m{l}^2\right)$        (I)

Here, mass of the rectangular plate is $m$ and length of the plate is $l$.

Apply conservation of angular momentum using Figure (1).

  $\begin{array}{l}{\displaystyle {\int }_{t_1}^{t_2}{M}_O}dt={\left(H_O\right)}_2-{\left(H_O\right)}_1\\ M_O=F\cos 15°\left(0.8\text{m}\right)\end{array}$        (II)

Here, applied force is $F$, moment of external forces about $O$ is $M_O$, initial and final time taken are $t_1$ and $t_2$, and initial and final angular momentum are ${\left(H_O\right)}_2$ and ${\left(H_O\right)}_1$.

Write the expression to determine the moment about point $O$.

  $M_O{t}_2=I_O\omega$        (III)

Conclusion:

Substitute $60\text{kg}$ for $m$ and $1.2\text{m}$ for $l$ in Equation (I).

  $\begin{array}{c}{I}_O=4\left(\frac{1}{3}\times 60\text{kg}\times {\left(1.2\text{m}\right)}^2\right)\\ =4\times 28.8\\ =115.2\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $90\text{N}$ for $F$ in Equation (II).

  $\begin{array}{l}{M}_O=90\text{N}\times \cos 15°\left(0.8\text{m}\right)\\ M_O=69.55\text{N}\text{.m}\end{array}$

Substitute 3 s for $t_2$, $69.55\text{N}\text{.m}$ for $M_O$ and $115.2\text{kg}\cdot {\text{m}}^2$ for $I_O$ in Equation (III).

  $\begin{array}{c}\omega =\frac{69.55\text{N}\text{.m}\times 3\text{ s}}{115.2\text{kg}\cdot {\text{m}}^2}\\ =1.811\text{ rad}/\text{s}\end{array}$

Thus, the final angular velocity is $1.811\text{ rad}/\text{s}$.