JEE Main 2017 · 2 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

A slender uniform rod of mass M and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is :

Axes z (up) and x; a rod pivoted at the origin makes angle θ with the z-axis and is rotating down.

\dfrac{3g}{2l}\sin\theta

\dfrac{2g}{3l}\sin\theta

\dfrac{3g}{2l}\cos\theta

\dfrac{2g}{3l}\cos\theta

Answer and worked solution

Official answer: A

Torque about the pivot: $Mg\dfrac l2\sin\theta$ . With $I=\dfrac{Ml^2}3$ :
$\alpha=\frac{3g}{2l}\sin\theta.$

Topics:Torque and rotational dynamics

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 6