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JEE Main & Advanced Physics Rotational Motion Question Bank Self Evaluation Test - System of Particles Rotational Motion

  • question_answer
    The instantaneous angular position of a point on a rotating wheel is given by the equation\[\theta (t)=2{{t}^{3}}-6{{t}^{2}}\]. The torque on the wheel becomes zero at

    A) \[t=1s\]

    B) \[t=0.5s\]

    C) \[t=0.25s\]

    D) \[t=2s\]

    Correct Answer: A

    Solution :

    [a] When angular acceleration (\[\alpha \]) is zero then torque on the wheel becomes zero. \[\theta (t)=2{{t}^{3}}-6{{t}^{2}}\Rightarrow \frac{d\theta }{dt}=6{{t}^{2}}-12t\] \[\Rightarrow \alpha =\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=12t-12=0\,\,\,\therefore t=1\sec .\]


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