$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical,its lower half separates without exerting any reaction at the breaking point. Then the maximum angle '$\theta$' made by the hinged upper half with the vertical is ......... $^o$.

  • A
    $30$
  • B
    $45$
  • C
    $60$
  • D
    $90$

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$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

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$A$ uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of $20^o$ to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above,is there a torque acting on it,and if so,in what direction?

$A$ uniform rod $AB$ of length $1 \ m$ and mass $4 \ kg$ is sliding along two mutually perpendicular frictionless walls $OX$ and $OY$. The velocity of the two ends of the rod $A$ and $B$ are $3 \ m/s$ and $4 \ m/s$ respectively, as shown in the figure. Which of the following statement$(s)$ is/are correct?

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