(a) Uniform rod about perpendicular bisector ,consider a uniform rod of mass m,and length l and suppose the moment of inertia is to be calculated about the bisector AB,Take origin at the middle point o of the of the rod .Consider the element of the between the distance $x \ \text{and} \ x+dx$ from the origin.

Mass per unit length of the rod is $\frac{m}{l}$ ,So the mass of the element is $(\frac{m}{l})\times dx$

$I=\int{dI}$ $=\int_{\frac{-l}{2}}^{\frac{l}{2}}\frac{m}{l}dx\times x^2$ $=\frac{m}{l}[\frac{x^3}{3}]_\frac{-l}{2}^\frac{l}{2}$ $=\frac{ml^2}{12}$

(b) Moment of inertia about a rectangular phase about a line parallel to the edge and passing through the center

mass per unit arc=$\frac{m}{bl}$; mass of strip $=\frac{m}{bl}\times bdx$ $=\frac{m}{l}dx$

$I=\int_{\frac{-l}{2}}^{\frac{l}{2}}\frac{m}{l}dx\times x^2$ $=\frac{ml^2}{12}$

(C) Moment of inertia of a uniform circular plate about the axis

Let mass of the plate M; and radius R,The center is at 'o',and the axis 'xox' is parallel to the plane of the plate.

Peripheral of the ring is $2\pi x$ and the width of this is 'dx',so area of this elementary ring is $'2\pi x dx'$

Mass perv unit area =$\frac{M}{\pi R^2}$, Mass of the ring is = $\frac{M}{\pi R^2}\times 2\pi x dx$ = $\frac{2Mxdx}{R^2}$

$I=\int{dI}$ =$\int_{0}^{R}\frac{2Mx}{R^2}dx\times x^2$ =$\int_{0}^{R}\frac{2M}{R^2}dx\times x^3$ =$\frac{mR^2}{2}$

Q.1 A 3 kg uniform rotates in a vertical plane about B. The spring of constant k is 300N/m and stretched length of 120 mm. is attached to the rod. In position shown ,angular speed of the rod is 4 rad/sec clockwise.Determine angular velocity of the rod after it has rotated through 90 degree.

Solution:

$I_B=I_G+mh^2$

$=\frac{ml^2}{12}+mh^2$ $=\frac{3\times (0.75)^2}{12}+3\times (0.225)^2$ $=0.2925 kgm^2$

Unstretched length=0.12m. Initial position,length of spring $L_i=\sqrt{0.36^2+0.15^2}=0.39 m.$

Final position, length of the spring $2f=0.21 m$

Workdone: Workdone by gravity is =$-mgh=-3\times 3.91\times 0.225=-6.62 j$

Workdone by spring =$\frac{1}{2}k(x_1^2-x_2^2)$

$=\frac{1}{2}k[{(L_1-L_2)^2-(L_2-L_0)^2}]$

$=\frac{1}{2}\times 300\times [{(0.39-0.12)^2-(0.21-0.12)^2}]$ $=9.27 j$

Net workdone$=9.27+(-6.62)=3.1 j$

Change in K.E =$K.E_f-K.E_i$

$=\frac{1}{2}I_Bw_f^2-\frac{1}{2}I_Bw_i^2$ $=\frac{1}{2}I_b(w_f^2-w_i^2)$ $=\frac{1}{2}0.2925(w_f^2-16)$

Work Energy Principle:Net workdone=change of $K.E$

$3.1=0.14625(w_f^2-16)$ $=\gt21.196=w_f^2-16$ $=\gt w_f^2=37.196$ $\quad w_f=6.099$

2 Find D.O.F ;h=6; no of lower pair binary joints are j=7

D.O.F=$3(6-1)-2(7)=3(5)-14=15-14=1$

$n=6$; no of lower pair , $j=6$;

no of higher pair, $h=1$

D.O.F = $3(6-1)-2(6)-1$ $=15-12-1$ $=2$