AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

The figure below represents the three uniform thin rods in the form of triangle.The centre of mass of the system,
rCOM=

m1r1+m2r2+m3r3

(m1+m2+m3)

(rCOM)i=

1

2

(

√3

2

×1×2)

^

j

(rCOM)i=

1

√3

^

j

The new length of the rod due to expansion is,
lf=li(1+αΔT)
lf=li(1+

γ

3

ΔT)
Substitute the values in the above equation.
lf=2(1+

12√3×10−6

3

×50)
lf=2(1+4×50×√3×10−6)
lf=2(1+2√3×10−4)
The final centre of mass can be calculated as,
(rCOM)f =

1

3

(

√3

2

×(1+2√3×10−4)×2)

^

j

=(

1

√3

+2×10−4)

^

j

The increase in y coordinate of the centre of mass of the system of the rods is,
δy=(rCOM)f−(rCOM)i
=(

1

√3

+2×10−4−

1

√3

)

^

j

=2×10−4m