First slide

Methods to locate centre of mass

Question

A rod of length 3 m and its mass acting per unit length is directly proportional to distance x from one of its end, then its centre of gravity from that end will be at

Moderate

A

1.5m
B

2m
C

2.5 m
D

3.0m

Solution

Given that mass per unit length is proportional to distance x, i.e., p =k x, where k = constant
Now mass of dx length at a distance x
=p dx =k x dx
Let the centre of mass is at a distance x₀ from one end. Then mass on either side of the rod must be equal. Thus
$\int_{0}^{x_{0}} kxdx = \int_{x_{0}}^{3} kxdx$
or ${[\frac{x^{2}}{2}]}_{0}^{x_{0}} = {(\frac{x^{2}}{2})}_{x_{0}}^{3}$
or $\frac{x_{0}^{2}}{2} = \frac{9}{2} - \frac{x_{0}^{2}}{2}$
$∴ x_{0}^{2} = \frac{9}{2} or x_{0} = \frac{3}{\sqrt{2}}$
So, $x_{0} \approx 2 m$

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