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Home » The density of a non-uniform rod of length 1 m is given by where a and b are constant and 0 ≤ x ≤ 1. The centre of mass of the rod will be ata) b) c) d)
The density of a non-uniform rod of length 1 m is given by \rho(x)=a\left(1+b x^{2}\right) where a and b are constant and 0 ≤ x ≤ 1. The centre of mass of the rod will be at
a) \frac{3(2+b)}{4(3+b)}
b) \frac{4(2+b)}{3(3+b)}
c) \frac{3(3+b)}{4(2+b)}
d) \frac{4(3+b)}{3(2+b)}
CBSE | Class 11 | NCERT Exemplar | Physics | System of particles and rotational motion
The density of a non-uniform rod of length 1 m is given by \rho(x)=a\left(1+b x^{2}\right) where a and b are constant and 0 ≤ x ≤ 1. The centre of mass of the rod will be at
a) \frac{3(2+b)}{4(3+b)}
b) \frac{4(2+b)}{3(3+b)}
c) \frac{3(3+b)}{4(2+b)}
d) \frac{4(3+b)}{3(2+b)}

Correct answer is a) \frac{3(2+b)}{4(3+b)}

Considering a differential part of the rod at a distance of x, we have,

\begin{array}{l} l=\frac{d m}{d x}=a\left(1+b x^{2}\right) \Rightarrow d m=a\left(1+b x^{2}\right) d x \end{array}

We know,

\mathrm{X}_{\mathrm{COM}}=\frac{\int_{0}^{1}(\mathrm{dm}) \mathrm{x}}{\int_{0}^{1}(\mathrm{dm})}=\frac{\int_{0}^{1} \mathrm{a}\left(1+\mathrm{bx}^{2}\right) \mathrm{x} \cdot \mathrm{dx}}{\int_{0}^{1} \mathrm{a}\left(1+\mathrm{bx}^{2}\right) \mathrm{dx}}

X_{C O M}=\frac{a \int_{0}^{1}\left(x+b x^{3}\right) d x}{a \int_{0}^{1}\left(1+b x^{2}\right) \cdot d x}=\frac{\left[\frac{x^{2}}{2}+\frac{b x}{4}\right]_{0}^{1}}{\left[x+\frac{b x^{3}}{3}\right]_{0}^{1}}

\mathrm{X}_{\mathrm{COM}}=\frac{\frac{1}{2}+\frac{\mathrm{b}}{4}}{1+\frac{\mathrm{b}}{3}}=\frac{3(2+\mathrm{b})}{4(3+\mathrm{b})}

i

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