First 10 cm piece of a uniform meter stick is cut and kept on the other end as shown in figure. Find the shift in the x

Engineering

Physics

Center of Mass

Summation of Vectors

Forces and Types of Forces

Question

First 10 cm piece of a uniform meter stick is cut and kept on the other end as shown in figure. Find the shift in the x - coordinate of the centre of mass.

10 cm

9 cm

45 cm

5 cm

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Let the mass of the stick is M initial centre of mass will be at (50 cm, 0) if we cut 10 cm part. Its mass will be (M/10) and put it at the end.
Then centre of mass of the piec of mass 9M/10 will be at (55cm, 0) and
and centre of mass of the piece of M/10 will be at (95cm, 0)

New centre of mass $= \frac{\frac{9}{10} M \times 55 + \frac{M}{10} \times 95}{\frac{9}{10} M + \frac{M}{10}} = \frac{(9 \times 55 + 95) M}{10 M} = \frac{9 \times 55 + 95}{10} = 59 cm$

So shift will be 9 cm