Moments of inertia, rods mechanism

Question

I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling "moments of inertia" to try and understand the concept better, but I am having issues knowing when to apply which formulas.

I have attached my specific question and the "answer" to the question, hopefully someone can help me understand how to get the answer.

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Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.

i) Determine the position of the center of gravity of the link

ii) Find the moment of inertia of the link about the point B

iii) Find the moment of inertia of the link about the point D

iv) Find the radius of gyration about the point G

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ANSWERS:

i) Center of gravity is at D

ii) I_B=(5/12)ml² kg m²

iii) I_D=(7/24)ml² kg m²

iv) k= sqrt(17/24) l m

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My attempts

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i)

I know inherently that the centre of gravity is at D. How do i prove it mathematically?

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ii)

I get the moment of inertial about B like so:

I_b = (ml²)/3 + ((ml²/3) - (ml²)/4)

But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?

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iii)

For I_d I get the answer by doing:

I_d = ((ml²)/3 - (ml²)/4) + ((ml²)/(3*16)) + (9ml²)/(16*3)

Why do I subtract (ml²)/4) if that rod is above D?

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iv)

I'm trying to use I_G to find the radius of gyration:

So for I_G

I_g = (7/24)ml² - ml²

Here I am using the parallel axis theorem, but why do I subtract the (md²) of the AC rod from the Moment of inertia about D?

Answer 1

The point where center of mass lies so the center of gravity. So, for center of mass For rod AC. B is the mid point of AC.So,let us take A as origin So coordinates of B($\frac{l}{2}$,0).Similarly for midpoint of BG We say there are four divisions which are equally space hence the length of BD,DE,EF,FG are$\frac{l}{4}$ So the mid point is E the mid point of BG. So coordinates of E($\frac{l}{2}$,$\frac{l}{2}$) are (according to the origin taken). $$\Rightarrow X_{com}=\frac{m_{1}x_{1}+m_{2}x_{2}+.............+m_{n}x_{n}}{m_{1}+m_{2}+...........m_{n}}$$ So here mass of the mass of both the rods are same and we have worked out their individual center of masses. $$\Rightarrow \frac{m\frac{l}{2}+m\frac{l}{2}}{m+m}$$ $$\Rightarrow \frac{2m\frac{l}{2}}{2m}$$ $$X_{com}=\frac{l}{2}$$ Now same for $Y_{com}$ $$Y_{com}=\frac{m(0)+m\frac{l}{2}}{2m}$$ $$Y_{com}=\frac{l}{4}$$ So, coordinates of center of mass of the system is ($\frac{l}{2}$,$\frac{l}{4}$) Now coordinates of point G($\frac{l}{2}$,$l$),Point F($\frac{l}{2}$,$\frac{3l}{4}$),Point E($\frac{l}{2}$,$\frac{l}{2}$)And D($\frac{l}{2}$,$\frac{l}{4}$)which satisfies the $X_{com}$ and $Y_{com}$.Hence its is the center of gravity of the system.

ii) moment of inertia about the point B.

1)Due to AC rod for which B is the mid point And for its the moment of intertia is as follows: Assume the frame from B is perpendicular to the rod AC passing through the point B. So only considering AC rod The mass about the B point is $\frac{m}{l}$ mass of rod is divided among the lenght l Now imagining a point x distance from B now the moment of inertia about the imagined point is $$\Rightarrow \frac{m}{l}x^2$$ now for the whole we integrate so,$$\Rightarrow \int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{m}{l}x^2\,dx$$ $$\Rightarrow \frac{m}{l}\int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{x^3}{3}$$ $$\Rightarrow \frac{m}{3l}\left(\frac{L^3}{8}-\left(-\frac{L^3}{8}\right)\right)$$ now finally we get $$\Rightarrow \frac{m L^2}{12}$$ 2)Due to BG rod As B is at B corner so just change the limits and put it 0 to l and we get $$\frac{m L^2}{3}$$. So total moment of inertia about the point is $$\Rightarrow \frac{m L^2}{12}+\frac{m L^2}{3}$$ which is $$\Rightarrow \frac{5m L^2}{12}$$

iii)just put the limits in above integral $\frac{-L}{4}$ to $\frac{3L}{4}$ We get moment of inertia for BG to D is $$\frac{7m L^2}{48}$$ Hence we use the parallel axis theorem for AC so $$\Rightarrow \frac{mL^2}{12}+m\left(\frac{L}{4}\right)^2$$ $$\Rightarrow \frac{7mL^2}{48}$$ Again adding the both of the moment of interia that is about BG and AC $$\Rightarrow \frac{7mL^2}{48}+\frac{7mL^2}{48}$$ which is $$\frac{7mL^2}{24}$$.

Answer 2

To compute the c.g location, assume that is is at a distance Y below point B. Use the fact that the mass moment about that point should be zero to determine Y.

$$ \int_{Y-L}^Y \frac{m}{l}y \, dy = 0$$

The above equation has one unknown Y. Solving, we get $ Y = l/4 $. This is point D.

Finding $I_D$ is straightforward integration

$$ I_D=\int_{-\frac{l}{2}}^{\frac{l}{2}} \frac{m }{l} \left(\frac{l}{4}+x\right)^2\, dx+\int_{\frac{l}{4}-l}^{\frac{l}{4}} \frac{m}{l} y^2 \, dy$$

Solving (using Mathematica), I get $$ I_D= \frac{7 l^2 m}{24}$$

$I_B$ can then get obtained using the parallel axis theorem

$$ I_B = I_D + 2 m \left(\frac{l}{4}\right)^2 = \frac{5 l^2 m}{12}$$

You can similarly find $I_G$

$$I_G= I_D + 2 m \left(\frac{3 l}{4}\right)^2 = \frac{17 l^2 m}{12}$$

The radius of gyration follows from this value.