CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    2 kg mass is rotating on a circular path of radius 0.8 m with an angular velocity of 44 rad/sec. If the radius of the path becomes 1 m. Then value of angular velocity is:

    A)  \[35.32\,\text{rad/sec}\]            

    B)  \[28.16\,\text{rad/sec}\]

    C)  \[14-08\,\text{rad/sec}\]           

    D)  \[7\,\text{rad/sec}\]

    Correct Answer: B

    Solution :

    Mass\[m=2\,kg\] Initial radius of the path \[{{r}_{1}}=0.8\,m\]Initial angular velocity \[{{\omega }_{1}}=44\,\text{rad/sec}\] Final radius of the path \[{{r}_{2}}=1\,m\] The moment of inertia of rotating body is given by \[{{I}_{1}}={{m}_{1}}r_{1}^{2}=2\times {{(0.8)}^{2}}=1.28\,kg-{{m}^{2}}\] Similarly,\[{{I}_{2}}=mr_{2}^{2}=2\times {{(1)}^{2}}=2kg-{{m}^{2}}\] Now from the law of conservation of angular momentum is                 \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\]                 \[{{\omega }_{2}}=\frac{{{I}_{1}}}{{{I}_{2}}}\times {{\omega }_{1}}\] \[=\frac{1.28}{2}\times 44=28.16\,\text{rad/sec}\]


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