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Units And Measurement

Question
CBSEENPH11020608

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When a string is cut, the initial angular acceleration of the rod is,

  • 3g / 2L

  • g/L

  • 2g/L

  • 2g/3L

Solution

A.

3g / 2L

Torque on the rod = moment of weight of the rod about P

straight tau space equals space mg space straight L over 2              ... (i)
Moment of inertia of rod about,
straight P space equals space ML squared over 3              ... (ii)
As straight tau = I straight alpha
From equations (i) and (ii), we get
Mg space straight L over 2 space equals space ML squared over 3 straight alpha
therefore space space space space straight alpha space equals space fraction numerator 3 straight g over denominator 2 straight L end fraction