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If the linear density of rod of length 3m varies as (lambda) = 2+x then the distance of centre of gravity of the rod is: (a) 7/3m (b)12/7m (c)10/7m (d)9/7m Pls provide with the complete workout of solution and also with proper explanation..

If the linear density of rod of length 3m varies as (lambda) = 2+x then the distance of centre of gravity of the rod is: (a) 7/3m  (b)12/7m  (c)10/7m  (d)9/7m
 
Pls provide with the complete workout of solution and also with proper explanation..

Grade:11

3 Answers

Vikas TU
14149 Points
7 years ago
total mass = density * length
that is
dm = lamda * dx
dm = (x+2)dx
integrating both sides,
we get,
m = x^2/2 + 2x
   = l^2/2 + 2l
put l = 3 meter.
m = 9/2 + 6 = 21/2 m that is approx. 10.5 m
Raman Mishra
67 Points
7 years ago
Here it is given that \lambda =x+2
Considering an element dx of the rod, its mass dm will be: dm=\lambda dx=(x+2)dx
If xcg denotes the centre of gravity of the rod, then:
             x_{cg}=\frac{\int_{0}^{L}xdm}{\int_{0}^{L}dm}     =>     x_{cg}=\frac{\int_{0}^{L}x\lambda dx}{\int_{0}^{L}\lambda dx}
Substituting the value of \lambda, we get:
           x_{cg}=\frac{\int_{0}^{L}x(x+2)dx}{\int_{0}^{L}(x+2)dx}
Solving the above integral and further putting the value of L=3m, we get
          {\color{DarkRed}x_{cg}= \frac{12}{7}m }  Ans.
            
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to your problem below.
 
Given, λ = dm/dx = x + 2
hence, dm = (x + 2)dx
Let, xcg denote the center of gravity of rod.
Now, xcg = ∫xdm / ∫dm = ∫x(x + 2)dx / ∫(x+2)dx              [ both integral varying from 0 to L, where L = 3m]
Hence, xcg = (x3/3 + 2x2/2) / (x2/2 + 2x)
or, xcg = (33/3 + (3)2) / (32/2 + 2(3))
           = 36/21 = 12/7 m
 
Hope it helps.
Thanks and regards,
Kushagra

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