**1 Answer**

written 4.3 years ago by |

A 2.5 kg slender bar of of length 40cm is pinned to one end, a 3kg is to be attached to the bar how far from the pin support should the particle be placed such that the period of the bar's oscillation is 1 sec?

$\rightarrow$ T = 1 sec

m = 2.5kg

mp = 3kg

$w_n = 2 \pi fo$

$[w_n = 2\pi \frac{1}{T} ]$

$\therefore$ $T = \frac{2 \pi }{w_n}$

$\therefore$ $w_n = 2 \pi $ rad/sec

$KE = (KE)_{rod} + (KE)_{rob}$

$\frac{1}{2} I_o \theta^2 + \frac{1}{2} I_o \theta^2$

$= \frac{1}{2} [ \frac{mL^2}{12} + \frac{mL^2}{4} ] \theta^2 + \frac{1}{2} [ mp L^2] \theta^2$

$= \frac{1}{2} [ \frac{mL^2}{3} + mpL^2] \theta^2$

$= \frac{1}{2} [ \frac{2.5 \times 0.4^2}{3} + dxl^2] \theta^2$

$= \frac{1}{2} [ 0.1333 + 3L^2] \theta^2$

$PE = (PE)_{rod} + (PE)_{Bob}$

$= mg \frac{L}{2} (1 - cos \theta) + m_p gL (1- cos \theta)$

as, $cos \theta = 1 - \frac{\theta^2}{2}$

$PE = mg \frac{L}{2} ( 1 - 1 + \frac{\theta^2}{2}) + m_p g L [ 1 - 1 + \frac{\theta^2}{2} ]$

$= mg \frac{L}{2} \frac{\theta^2}{2} + mp.g.L \frac{\theta^2}{2}$

$= \frac{1}{2} [ \frac{m.g.L}{2} + mp.g.L] \theta^2$

$= \frac{1}{2} [ \frac{2.5 \times 9.81 \times 0.4}{2} + 3 \times 9.81 \times L] \theta^2$

$= \frac{1}{2} [4.905 + 29.43 L] \theta^2$

$w_n \sqrt{ \frac{K_{eq}}{I_{eq}}}$

$2 \pi = \sqrt{ \frac{4.905 + 29.43 L}{0.1333 + 3l^2}}$

39.478 [0.1333 + 3E] = 4.905 + 29.43 L

5.262 + 118.434 $L^2$ = 4.905 + 29.43L

118.434 $L^2$ - 29.43 + 0.357 = 0

L = 0.2357m

= 23.57 cm