First slide

Motion of centre of mass

Question

Two masses $m_{1} =$ 1 kg and $m_{2}$ = 2 kg and connected by a light inextensible string and suspended by means of a weightless pulley as shown in the figure. Assuming that both the masses start from rest, the distance travelled by the centre of mass in two seconds is (Take g = 10 ${ms}^{- 2}$ )

Moderate

A

$\frac{20}{9} m$
B

$\frac{40}{9}$
C

$\frac{2}{3} m$
D

$\frac{1}{3} m$

Solution

Here $m_{1} = 1 kg, m_{2} = 2 kg$
The acceleration of the system is
$a = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2}} = \frac{(2 - 1) g}{(1 + 2)} = \frac{g}{3} = \frac{10}{3}$
Acceleration of the centre of mass is
$a_{cm} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}} = \frac{1 (- a) + 2 (a)}{1 + 2}$
= $\frac{(\frac{- g}{3}) + 2 (\frac{g}{3})}{3} = \frac{g}{9} = \frac{10}{9} {ms}^{- 2}$
The distance travelled by the centre of mass in two seconds is
$S = \frac{1}{2} a_{cm} t^{2} = \frac{1}{2} \times \frac{10}{9} \times {(2)}^{2} = \frac{20}{9} m$

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