Question

shou
centr
3. The density of a linear rod of length L varies as
p= A + Bx where x is the distance from the left end.
Locate the centre of mass.
Solution
mi
f
C is
dista
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dx
Figure 9-W3
at C​

Answer 1

Answer:

ANSWER

Consider an element dx at a distance x from one end of the rod of length L.

The center of mass of the rod is X

cm

=

∫

0

L

λdx

∫

0

L

xλdx

or X

cm

=

∫

0

L

(A+Bx)dx

∫

0

L

x(A+Bx)dx

=

[Ax+Bx

2

/2]

0

L

[Ax

2

/2+Bx

3

/3]

0

L

=

AL+BL

2

/2

AL

2

/2+BL

3

/3

=

6

3AL

2

+2BL

3

×

2AL+BL

2

2

=

3L(2A+BL)

L

2

(3A+2BL)

=

3(2A+BL)

L(3A+2BL)

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Answer 2

Bravo! Its an IIT Question!

Answer:

Centre of Mass, Xcm ⇒ $\frac{(3A + 2BL)L}{3(2A + BL)}$

Explanation:

Given;-

  Density, λ = A + Bx

   Length, = L

Let the very small distance after x be dx, mass be dm. We also know that, dm = λ × dx.

Now, we know by formula that;-

$Xcm = \frac{\int\limits^L_0 {} \, x\, dm }{\int\limits^L_0 {x} \, dm }$

        $= \frac{\int\limits^L_0 {x}(A+ Bx)\, dx }{\int\limits^L_0 {(A + Bx)} \, dx }$

        $= \frac{\int\limits^L_0 {(Ax + Bx^{2} )} \, dx }{(A + Bx)\,dx}$

        $= \frac{\int\limits^L_0 {(Adx^{2} + Bdx^{3} }) }{\int\limits^L_0 {(Adx + Bdx^{2}) } }$

        $= \frac{A\int\limits^L_0 {x} \, dx^{2} \,+ \,B\int\limits^L_0 {dx^{3} } }{A\int\limits^L_0 {dx} \,+ \, B\int\limits^L_0 {dx^{2} } }$

        $=\frac{A[\frac{x^{2} }{2}]^{L\, to \, 0} \, + \, B[\frac{x^{3} }{3}]^{L \, to \, 0} }{A[x]^{L \, to \, 0}\, + \, B[\frac{x^{2} }{2}]^{L \, to \, 0} }$

        $= \frac{\frac{AL^{2} }{2} \, + \, \frac{BL^{3} }{3} }{AL + \frac{BL^{2} }{2} }$

        $= \frac{\frac{(3AL^{2})\, + \, (2BL^{3} ) }{6} }{\frac{(2AL)\, + \, (BL^{2}) }{2} }$

        $= [\frac{3AL^{2} \, + \, 2BL^{3} }{6} ] [\frac{2}{2AL \, + \, BL^{2} } ]$

        $= \frac{3AL^{2}\, +\, 2BL^{3} }{3(2AL\, + \, BL^{2} )}$

        $= \frac{L ( 3AL \,+\, 2BL^{2}) }{L( 3(2a\, + \, BL) ) }$

        $= \frac{(3AL \,+\, 2BL^{2}) }{3 (2A\, + \, BL)}$

        $= \frac{(3A\, +\, 2BL)L}{3(2A + BL)}$

Hope it helps! ;-))