Question

# If the linear density of the rod of length L starting from one end (x=0) varies as $$\lambda=A+bx$$, then its centre of mass of the rod will be at

A
Xcm=L(2A+BL)3(3A+2BL)
B
Xcm=L(3A+2BL)3(2A+BL)
C
Xcm=L(3A+2BL)3
D
Xcm=L(2A+3BL)3

Solution

## The correct option is A $$\displaystyle{X_{cm}=\frac{L(3A+2BL)}{3(2A+BL)}}$$Total mass of rod is $$M=\int_0^L{A+Bx}dx=(Ax+Bx^2/2)|{}_0^L=(2A+BL)L/2$$Now position of COM is given as $$X_cm=\dfrac{\int_0^L(A+Bx)xdx}{M}=\dfrac{(Ax^2/2+Bx^3/3)|{}_0^L}{M}=\dfrac{(3A+2BL)L^2/6}{(2A+BL)L/2}=\dfrac{L(3A+2BL)}{3(2A+BL)}$$Physics

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