A pendulum of mass m hangs from a support fix | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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):

Similar Questions

What is weightlessness? Explain with examples.

The ratio of the weight of a man in a stationary lift to his weight in a lift moving downwards with an acceleration '$a$' is $3:2$. Then '$a$' = . . . . . . .

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration $a$. Then:

$A$ body of mass $1.0 \, kg$ is falling with an acceleration of $10 \, m/s^2$. Its apparent weight will be ......... $kg \, wt$ $(g = 10 \, m/s^2)$.

$A$ body of mass $2 \, kg$ is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity $g$,the reading on the spring balance will be ....... $kg$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module