JEE Advanced 2019 · Paper 2, Physics question 1

The rod pivots about its foot, with $I=\frac{ML^2}3$. Let $\theta$ be the angle from the vertical.

(A) Angular speed. Energy conservation:
$$\tfrac12\cdot\tfrac{ML^2}3\omega^2=Mg\tfrac L2(1-\cos\theta)\implies\omega^2=\frac{3g}L(1-\cos\theta).$$
At $60^\circ$, $\omega^2=\dfrac{3g}{2L}$. True.

(B) Angular acceleration. From the torque of the weight:
$$\alpha=\frac{Mg\frac L2\sin\theta}{ML^2/3}=\frac{3g\sin\theta}{2L}=\frac{3\sqrt3\,g}{4L},$$
not $\frac{2g}L$. False.

(C) Radial acceleration of the centre of mass.
$$\omega^2\cdot\frac L2=\frac{3g}4.$$
True.

(D) Normal reaction. The centre of mass is at height $y=\frac L2\cos\theta$, so
$$\ddot y=-\frac L2\left(\omega^2\cos\theta+\alpha\sin\theta\right)=-\frac L2\left(\frac{3g}{4L}+\frac{9g}{8L}\right)=-\frac{15g}{16}.$$
Then $N-Mg=M\ddot y$, giving $N=\dfrac{Mg}{16}$. True.

(A), (C), (D)