JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    The centre of mass of three particles of masses \[1\,\,kg,\,\,2\,\,kg\] and \[3\,\,kg\] is at \[(3,\,\,3,\,\,3)\] with reference to a fixed coordinate system. Where should a fourth particle of mass \[4\,\,kg\] be placed, so that the centre of mass of the system of all particles shifts to a point\[(1,\,\,1,\,\,1)?\]

    A) \[(-1,\,\,-1,\,\,-1)\]                        

    B) \[(-2,\,\,-2,\,\,-2)\]

    C)  \[(2,\,\,2,\,\,2)\]                            

    D)  \[(1,\,\,1,\,\,1)\]

    Correct Answer: B

    Solution :

    Centre of mass of a solid body is given by                 \[{{x}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{x}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}},\,\,{{y}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{y}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}}\] \[{{z}_{CM}}=\frac{\sum\limits_{i=1}^{n}{\Delta {{m}_{i}}{{z}_{i}}}}{\sum\limits_{j=1}^{n}{\Delta {{M}_{j}}}}\] \[1\times {{x}_{1}}+2\times {{x}_{2}}+3\times {{x}_{3}}=(1+2+3)3...\]     ... (i) and        \[{{x}_{1}}={{x}_{2}}={{x}_{3}}=3\]                 \[{{x}_{CM}}={{y}_{CM}}={{z}_{CM}}=1\](given) \[1(1+2+3+4)=1{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}+4{{x}_{4}}\]        ? (ii) Solving Eqs. (i) and (ii), we get                 \[4{{x}_{4}}=10-18\Rightarrow {{x}_{4}}=-2\] Similarly,\[{{y}_{4}}=-2,\,\,{{z}_{4}}=-2\] The fourth particle must be placed at the point\[(-2,\,\,-2,\,\,-2)\].


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