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  • Arrangement of two masses  are as shown in th

Arrangement of two masses m_1 \: \: and \: \: m_2 are as shown in the fig 

Assume strings and pulley are massless and frictionless . Find acceleration of centre of mass of system let m_1 = 3 Kg \: \: and \: \: m_2 = 1 Kg

 

Option: 1

g/2


Option: 2

g/3 


Option: 3

g/4


Option: 4

g


Answers (1)

best_answer

As we have learned

Acceleration of centre of mass -

{\vec{a}_{CM}}=\frac{m_{1}\vec{a_{1}}+{m_{2}}\vec{a}_{2}........}{m_{1}+m_{2}........}

- wherein

m1, m2 are mass of all the particles \vec{a_{1}}, \; \vec{a_{2}}.... are their respective acceleration.

 

 acceleration of system a = \frac{(m_1-m_2)}{(m_1+m_2)}g

\vec a_{cm} = \frac{(m_1\vec a _{1}+m_2\vec a_{2})}{(m_1+m_2)}

\vec a_{cm} = \frac{(m_1 (-a) +m_2a)}{(m_1+m_2)}\Rightarrow - \left ( \frac{m_1-m_2 }{m_1+m_2} \right )a

The magnitude of acceleration of the centre of mass is a_{cm} = \left ( \frac{m_1-m_2 }{m_1+m_2} \right )a

a = \frac{(3-1)^2}{(3+1)^2}g \\\\ a = g/4

 

 

 

 

 

 

Posted by

Pankaj Sanodiya

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