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Three rods each of length L and Mass M are arranged as shown . The distance of the centre of mass of system from point A is  

 

Option: 1

L /4 


Option: 2

L/4 


Option: 3

(11/12) L 


Option: 4

(5/6)L


Answers (1)

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As we have learned

Centre of Mass of a uniform rod -

At a distance \frac{l}{2}  from one of the ends of rod. l is the length of the rod.

 

For a uniform rod centre of mass lies at midpoint of the rod 

From figure 

x_1 = \frac{L}{2} \: \: \: x_2 = \frac{L}{2}+\frac{L}{2} = L

 

x _3 = \frac{L}{2}+ \frac{L}{4} + \frac{L}{2} = \frac{5L}{4}\\\\ X _{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}\\\\ x_{cm} = \frac{\left ( M\times \frac{L}{2})+\left (( M\times L \right ) \right )+\left ( M \times \frac{5L}{4} \right )}{M+M+M} = \frac{\frac{11}{4}ML}{3M}\\\\ x_{cm} =\frac{11L}{12}

 

 

 

 

Posted by

avinash.dongre

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