Solution - Hanging Rod

[9:59 PM | 0 comments ]

See the "top view" picture above. From the picture, we can get the x and z-component of length of the string.

$l_z = \frac{l}{2}sin\theta$

$l_x = \frac{l}{2}(1-cos\theta)$

From these two component, now we can use Pythagorean theorem to get the y-component of the string.

$l^2 = {l_x^2 + l_y^2 + l_z^2}$

$l^2 = \frac{l^2}{4}(1-cos\theta)^2 + l_y^2 + \frac{l^2}{4}sin^2\theta$

Solving it, resulting:

$l_y = - l\ cos{\frac{\theta}{2}}$

(minus sign, because the string is toward downward)
If it's rotated, it will also raise. Differentiate the equation above to get the relation of its angular velocity and raising velocity. It'll result:

$\dot{l_y} = \frac{1}{2}\ l\ sin\left(\frac{\theta}{2}\right)\dot{\theta}$

And differentiate it once again to get the relation of its angular acceleration and its raising acceleration.

$\ddot{l_y} = \frac{1}{2}l \left( sin\left(\frac{\theta}{2} \right)\ddot{\theta} + \frac{1}{2}cos\left(\frac{\theta}{2} \right) \dot{\theta}^2 \right)$

And if $\dot{\theta} = 0$ , it will be:

$\ddot{l_y} = \frac{1}{2}l\ sin\left(\frac{\theta}{2}\right)\ddot{\theta}$

Now, we can write the equation of its energy.

$E = mgl_y +\frac{1}{2}m\dot{l_y}^2+\frac{1}{2}I\dot{\theta}^2$

Because energy in this system is constant, so always $\dot{E} = 0$ :

$\dot{E} = mg\dot{l_y} + m\dot{l_y}\ddot{l_y} + I\dot{\theta}\ddot{\theta} = 0$

Substitute all $\dot{l_y}$ and $\ddot{l_y}$ to $\dot{\theta}$ and $\ddot{\theta}$ , become

$\dot{\theta} \left(\frac{1}{2}mgl sin\frac{\theta}{2} + \frac{1}{4}ml^2 sin^2\left( \frac{\theta}{2}\right)\ddot{\theta} + I\ddot{\theta} \right) = 0$

Substitute $I = \frac{1}{12} ml^2$ and solve for $\ddot{\theta}$ , resulting:

$\ddot{\theta}l = -2g\frac{sin\frac{\theta}{2}}{sin^2\left( \frac{\theta}{2}\right) + \frac{1}{3}}$

and

$\ddot{\l_y} = -g\frac{sin^2\left( \frac{\theta}{2}\right)}{sin^2\left( \frac{\theta}{2}\right) + \frac{1}{3}}$

The acceleration at point A contains of angular acceleration and raising acceleration.

$a_A = \sqrt{\left( \ddot{\theta} \frac{l}{2} \right)^2 + \ddot{l_y}^2}$

Substitute all variables above, and it will result:

$a_A = g \frac{sin\left(\frac{\theta}{2}\right) \sqrt{1+sin^2\left(\frac{\theta}{2}\right)}}{sin^2\left(\frac{\theta}{2}\right) + \frac{1}{3}}$

That's my solution. If you have any correction, question, or comment, please leave a comment below.
Thank you. :-)

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