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Explain Newton's Method.
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Q. Determine the Natural Frequency of system as shown in figure:

A] Newton's Method.

ANSLOM,

IT = $\sum$ of all moments.

$I_o$ $\theta$'' = -mg sin $\theta \times L$

$I_o$ $\theta$'' + mgl sin $\theta$ = 0

$I_o$ $\theta$'' + mgl $\theta$ = 0 . . . E O M

$\theta \neq 0$

$\therefore$ sin $\theta \neq 0$

sin $\theta$ = $\theta$

Comparing with SHM

$\theta '' + {w_o}^2 \theta = 0$

${w_n}^2 = \frac{mgl}{I_o}$

$\therefore$ $w_n = \sqrt{ \frac{mgL}{I_o}}$ rad/sec

$w_n = \sqrt{ \frac{g}{L}}$

NOTE : For pendulum system

$I_o = ml^2 = kg m^2$

$w_n = \sqrt{ \frac{g}{L}}$ rad/sec.

B] Newton's Method.

ANSLOM.

IT = $\sum$ of all moments.

$I_o \theta$'' = -mgLsin $\theta - k a^2 \theta$

$I_o \theta '' + mgl \theta + ka^2 \theta = 0$

$I_o \theta '' + (mgl + ka^2) \theta = 0 . . . EOM$

$w_n \sqrt{ \frac{mgL+ ka^2}{I_o}}$ rad/sec

{ where $I_o = ml^2$ }

C] Newton's Method.

ANSLOM.

IT = sum of all moments.

$I_o \theta$ = -mgL sin IT = $\sum$ of all moments. - k. $a^2 \theta - k b^2 \theta$

$I_o \theta + mgL sin \theta + ka^2 \theta + k. b^2 \theta = 0$

$I_o \theta + mgL \theta + ka^2 \theta + k.b^2 \theta = 0$

$I_o \theta + (mgL + ka^2 + kb^2) \theta = 0$ . . . Equation of motion.

$w_n = \sqrt{ \frac{(mgL + ka^2 + kb^2}{I_o} }$ rad/sec

where $I_o = mL^2$

D] $KE = (KE)_m + (KE)_M$

$= \frac{1}{2} mx^2 + \frac{1}{2} I_G \theta^2$

$frac{1}{2} m (ro)^2 + \frac{1}{2} I_G \theta^2$

$= \frac{1}{2} mr^2 \theta^2 + \frac{1}{2} I_G \theta^2$

$KE = \frac{1}{2} [mr^2 + I_G] \theta^2$'

We know that,

$KE = \frac{1}{2} I_eq \theta^2$ {Equivalent parameter method}

$I_eq = mr^2 + IG$

$PF = (PE) spring = \frac{1}{2} kx^2 = \frac{1}{2} k (r \theta)^2 = \frac{1}{2} kr^2 \theta^2$

We know that, $PE = \frac{1}{2} Keq \theta^2$

$Keq = kr^2$

$\therefore$ $w_n = \sqrt{ \frac{k_{eq}}{I_{eq}} } = \sqrt{ \frac{kr^2}{mr^2 + I_G} } I_G = \frac{mass \times (radius)^2}{2}$

$= \frac{mr^2}{2}$

E] $KE = (KE)_m + (KE)pulley$

$\frac{1}{2} mx^2 + \frac{1}{2} I_G \theta^2$

$\frac{1}{2} m (2r \theta)^2 + \frac{1}{2} I_G \theta^2$

$= \frac{1}{2} [4mr^2 + I_G] \theta^2$

$PE = (PE)_k - \frac{1}{2} k x^2 = \frac{1}{2} k (r \theta)^2 = \frac{1}{2} kr^2 \theta^2$

set $\frac{du}{dt} = 0$

U = KE + PE

$= \frac{1}{2} 4mr^2 \theta^2 + \frac{1}{2} I_G \theta^2 + \frac{1}{2} kr^2 \theta^2$

diff w.r.t.t

$0 = \frac{1}{2} 4mr^2.2 \theta \theta + \frac{1}{2} I_G 2 \theta \theta + \frac{1}{2} kr^2 . 2 \theta \theta$

$0 = 4mr^2 \theta + IG \theta + kr^2 \theta$

$\theta (4mr^2 + IG) + kr^2 \theta = 0$

$w_n = \sqrt{ \frac{kr^2}{4mr^2 + I_G} }$ rad/s where $I_G = \frac{m4r^2}{2} 2 Mr^2$

F] $KE = (KE)_M + (KE)$ compond pulley

$\frac{1}{2} mx^2 + \frac{1}{2} Jo \theta^2$

$= \frac{1}{2} mr^2 \theta^2 + \frac{1}{2} Jo \theta^2$

$= \frac{1}{2} [mr^2 + Jo] \theta^2$

I_eq

$PE = (PE) spring$ = $\frac{1}{2} (2k) (4r \theta)^2 = \frac{1}{2} 32 kr^2 \theta^2$

$w_n = \sqrt{ \frac{keq}{Ieq} } = \sqrt{ \frac{32kr^2}{mr^2 + J_o} }$ rad/s.