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(D) To find the angle $	heta$ of the pendulum string,we analyze the forces acting on the bob in the non-inertial frame of the trolley.$1$. The forces

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$	heta=	an ^{-1}\left(\frac{a_0+g \sin \alpha}{g \cos \alpha}ight)$

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(D) To find the angle $	heta$ of the pendulum string,we analyze the forces acting on the bob in the non-inertial frame of the trolley.$1$. The forces acting on the bob are tension $T$,gravity $mg$,and the pseudo force $ma_0$ acting in the direction opposite to the acceleration of the trolley.$2$. Resolve the forces into components parallel and perpendicular to the string.$3$. The effective acceleration vector $\vec{g}_{eff}$ is the vector sum of gravity $\vec{g}$ and the pseudo acceleration $-\vec{a}_0$. The string aligns itself with the direction of $\vec{g}_{eff}$.$4$. The components of gravity are $mg \cos \alpha$ (perpendicular to the incline) and $mg \sin \alpha$ (parallel to the incline).$5$. The pseudo force $ma_0$ acts parallel to the incline,directed downwards along the slope.$6$. Balancing forces perpendicular to the string:$T = mg \cos \alpha \cos 	heta + (mg \sin \alpha + ma_0) \sin 	heta$$7$. Balancing forces parallel to the string:$mg \cos \alpha \sin 	heta = (mg \sin \alpha + ma_0) \cos 	heta$$8$. Rearranging the equation:$	an 	heta = \frac{ma_0 + mg \sin \alpha}{mg \cos \alpha} = \frac{a_0 + g \sin \alpha}{g \cos \alpha}$Therefore,$	heta = 	an^{-1} \left( \frac{a_0 + g \sin \alpha}{g \cos \alpha} ight)$.

$A$ pendulum of mass $m$ hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up an inclined plane of inclination $\alpha$ with acceleration $a_0$ is (the string and bob remain fixed with respect to the trolley):

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