A straight rod of length L is made of a mater | vedclass

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(A) The mass per unit length is given by $dm = \lambda|x|dx$.
First,calculate the moment of inertia about the axis $AA$ passing through the center of

Vedclass · Accepted Answer

$1.5 \ kg \cdot m^2$

Vedclass · Answer

(A) The mass per unit length is given by $dm = \lambda|x|dx$.
First,calculate the moment of inertia about the axis $AA$ passing through the center of the rod:
$I_{AA} = \int x^2 dm = \int_{-L/2}^{L/2} x^2 (\lambda|x|) dx = 2\lambda \int_{0}^{L/2} x^3 dx = 2\lambda \left[ \frac{x^4}{4} ight]_0^{L/2} = \frac{\lambda}{2} \left( \frac{L}{2} ight)^4 = \frac{\lambda L^4}{32}$.
Given $\lambda = 16 \ kg/m^2$ and $L = 1 \ m$,$I_{AA} = \frac{16 	imes 1^4}{32} = 0.5 \ kg \cdot m^2$.
Now,calculate the total mass $M$ of the rod:
$M = \int_{-L/2}^{L/2} \lambda|x| dx = 2\lambda \int_0^{L/2} x dx = 2\lambda \left[ \frac{x^2}{2} ight]_0^{L/2} = \lambda \left( \frac{L}{2} ight)^2 = \frac{\lambda L^2}{4}$.
With $\lambda = 16$ and $L = 1$,$M = \frac{16 	imes 1^2}{4} = 4 \ kg$.
Using the parallel axis theorem to find the moment of inertia about axis $BB$ passing through one end:
$I_{BB} = I_{CM} + M h^2$,where $h = L/2 = 0.5 \ m$.
$I_{BB} = 0.5 + 4 	imes (0.5)^2 = 0.5 + 4 	imes 0.25 = 0.5 + 1 = 1.5 \ kg \cdot m^2$.

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