Chapter 19.1, Problem 19.13P

To determine

(a)

The angle $\theta .$

Answer to Problem 19.13P

$\theta ={1.288}^{°}$

Explanation of Solution

Given information:

Angle $\theta =5^{°}$

Length of cord $l=40in.$

The equation of a simple harmonic motion in angular displacement is given as:

$\theta ={\theta }_m\sin \left({\omega }_{n}t+\varphi \right)$ _______________ (1)

The initial condition for the system is:

$\begin{array}{l}t=0;\theta =5^{°}\\ t=0;\dot{\theta }=0\end{array}$

Now, taking derivative of equation (1) and substitute 0 for t and 0 for $\dot{\theta }$ in equation (1);

$\begin{array}{l}\theta ={\theta }_m\sin \left({\omega }_{n}t+\varphi \right)\\ \dot{\theta }={\theta }_m{\omega }_n\cos \left({\omega }_{n}t+\varphi \right)\\ 0={\theta }_m{\omega }_n\cos \left(0+\varphi \right)\\ 0=\cos \left(\varphi \right)\\ \varphi ={\cos }^{-1}\left(0\right)\\ \varphi =\frac{\pi }{2}\end{array}$

Now, For a simple pendulum:

${\omega }_n=\sqrt{\frac{g}{l}}$

Change the length (l) into ft.

$\begin{array}{l}l=40\times \frac{1}{12}\\ l=3.333ft\end{array}$

Thus, $\begin{array}{l}{\omega }_n=\sqrt{\frac{32.2}{3.333}}\\ {\omega }_n=3.1087rad/s\end{array}$

Again amplitude

$\begin{array}{l}{\theta }_m=5^{°}or,\\ {\theta }_m=5^{°}\times \frac{\pi }{180}\\ {\theta }_m=0.08727rad\end{array}$

Now, put all the above obtained values in equation (1) with $t=1.6s$ (given);

$\begin{array}{l}\theta ={\theta }_m\sin \left({\omega }_{n}t+\varphi \right)\\ \theta =0.08727\sin \left((3.1087)\left(1.6\right)+\frac{\pi }{2}\right)\\ \theta =0.08727\times 0.257786933\end{array}$

$\begin{array}{l}\theta =0.002496rador,\\ \theta ={1.288}^{°}\end{array}$

Conclusion:

The value of the angle $\theta ={1.288}^{°}$.

To determine

(b)

The magnitudes of the velocity and acceleration of the bob.

Answer to Problem 19.13P

$\theta ={1.288}^{°}$

Explanation of Solution

Given information:

Angle $\theta =5^{°}$

Length of cord $l=40in.$

The velocity of the bob is compute by taking derivative of equation (1);

$\begin{array}{l}\theta ={\theta }_m\sin \left({\omega }_{n}t+\varphi \right)\\ \dot{\theta }={\theta }_m{\omega }_n\cos \left({\omega }_{n}t+\varphi \right)\_\_\_\_\_\_\_(2)\end{array}$

By substituting the values of ${\theta }_m,{\omega }_n,tand\varphi$ in the above equation we get the required velocity as,

$\begin{array}{l}\dot{\theta }=0.08727\left(3.1087\right)\cos \left(\left(3.1087\right)\left(1.6\right)+\frac{\pi }{2}\right)\\ \dot{\theta }=0.08727\times \left(3.1087\right)\times \left(0.9662017891\right)\\ \dot{\theta }=0.26213rad/s\end{array}$

The angular velocity of pendulum is:

$\begin{array}{l}v=l\dot{\theta }\_\_\_\_\_\_\_\_\_\_\_\_\_(3)\\ v=3.3333\times 0.26213\\ v=0.874ft/s\end{array}$

The acceleration of the motion is calculated by taking the derivative of equation (2),

$\begin{array}{l}\dot{\theta }={\theta }_m{\omega }_n\cos \left({\omega }_{n}t+\varphi \right)\\ \ddot{\theta }=-{\theta }_m{\omega }_n^2\sin \left({\omega }_{n}t+\varphi \right)\end{array}$

By substituting the values of ${\theta }_m,{\omega }_n,tand\varphi$ in the above equation we get;

$\begin{array}{l}\ddot{\theta }=-\left(0.08727\right){\left(3.1087\right)}^2\sin \left(\left(3.1087\right)\left(1.6\right)+\frac{\pi }{2}\right)\\ \ddot{\theta }=-\left(0.08727\right)\times {\left(3.1087\right)}^2\times \left(0.257786933\right)\\ \ddot{\theta }=-0.21733rad/s^2\\ \end{array}$

Now, the magnitude of acceleration at both angular and tangential direction is calculated as:

$a=\sqrt{{\left(a_n\right)}^2+{\left(a_t\right)}^2}$ _________(4)

Where, a_n = acceleration in angular direction and a_t = acceleration in tangential direction.

Acceleration in angular direction $a_n=\frac{v^2}{l}$

From equation (3);

$\begin{array}{l}{a}_n=\frac{l^2{\dot{\theta }}^2}{l}\\ a_n=l{\dot{\theta }}^2\\ a_n=3.333\times {\left(0.26207\right)}^2\\ a_n=0.22894ft/s^2\end{array}$

And, acceleration in tangential direction

$\begin{array}{l}{a}_t=l\ddot{\theta }\\ a_t=3.3333\times \left(-0.21733\right)\\ a_t=-0.72443ft/s^2\end{array}$

Put the values of $a_n$ and $a_t$ in equation (4) we get,

$\begin{array}{l}a=\sqrt{{\left(a_n\right)}^2+{\left(a_t\right)}^2}\\ a=\sqrt{{\left(0.22894\right)}^2+{\left(-0.72443\right)}^2}\\ a=0.759ft/s^2\end{array}$

Conclusion:

The velocity of the bob is $v=0.874ft/s$ and acceleration of the bob is $a=0.759ft/s^2.$