A uniform thin rod AB of length L has linear mass density , where is measured from A. If the CM of the rod lies at a distance of L from A, then and are related as . Find the value of .

Question

A uniform thin rod AB of length L has linear mass density $μ (x) = a + \frac{b x}{L}$ , where $' x'$ is measured from A. If the CM of the rod lies at a distance of $(\frac{7}{12})$ L from A, then $a$ and $b$ are related as $b = x a$ . Find the value of $' x'$ .

Smart Achievers · Accepted Answer

(2)

Centre of mass of the rod is given by :

$x_{cm} = \frac{\int_{0}^{L} (a x + \frac{b x^{2}}{L}) d x}{\int_{0}^{L} (a + \frac{b x}{L}) d x} = \frac{\frac{a L^{2}}{2} + \frac{b L^{2}}{3}}{a L + \frac{b L}{2}} = \frac{L (\frac{a}{2} + \frac{b}{3})}{a + \frac{b}{2}}$

Now $\frac{7 L}{12}$ $= \frac{\frac{a}{2} + \frac{b}{3}}{a + \frac{b}{2}}$ ,

On solving we get, $b = 2 a$

Solution

Q.
A uniform thin rod AB of length L has linear mass density $μ (x) = a + \frac{b x}{L}$ , where $' x'$ is measured from A. If the CM of the rod lies at a distance of $(\frac{7}{12})$ L from A, then $a$ and $b$ are related as $b = x a$ . Find the value of $' x'$ .

Want Us to Email you About Special Offers & Updates?

Solution

Q. A uniform thin rod AB of length L has linear mass density μ(x)=a+bxL, where 'x' is measured from A. If the CM of the rod lies at a distance of (712)L from A, then a and b are related as b=xa. Find the value of 'x'.

Ans. (2) Centre of mass of the rod is given by : xcm=∫0L(ax+bx2L)dx∫0L(a+bxL)dx=aL22+bL23aL+bL2=L(a2+b3)a+b2 Now 7L12=a2+b3a+b2, On solving we get, b=2a

Want Us to Email you About Special Offers & Updates?

Q.
A uniform thin rod AB of length L has linear mass density $μ (x) = a + \frac{b x}{L}$ , where $' x'$ is measured from A. If the CM of the rod lies at a distance of $(\frac{7}{12})$ L from A, then $a$ and $b$ are related as $b = x a$ . Find the value of $' x'$ .