Question

10. The object shown in figure is constructed of uniform
rods of same material. The centre of mass of the
whole arrangement will lie on the straight line
2L
----->X
2L-
L
(1) y = -2x +
2
(2) y = -x + L
(4) y = -X-L
)
(3) y = -2x + L

Answer 1

☆Centre of mass:

》Point where whole mass of object is assumed to be concentrated.

☆Solution:

》First convert the continous mass system into particle mass system.

》See the above figure!

• As both uniform rods AB, BC are identical material($\lambda$)
• Mass =$\lambda (2L)$= m
• There CM will at L in y and x axis respectively,at their centre.

》$\ R_{cm} = \dfrac{m(l)\hat{i} + m(l)\hat{j}}{m+m}$

》$\ R_{cm} = \dfrac{(l)\hat{i} + (l)\hat{j}}{2}$

》$\ R_{cm} = \dfrac{L}{2}\hat{i} + \dfrac{L}{2}\hat{j}$

------

• $\ tan\theta = \dfrac{dy}{dx}$

•$\ tan\theta = \dfrac{L}{L}=1$

》But the line drawn cutting centre of masses of all systems will have negative slope= -1.

According to y=mx+c

》y = (-1)x + L

》$\boxed{\red{\bf{ y = –1x + L}}}$

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