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

$A$ man of mass $60\, kg$ records his weight on a weighing machine placed inside a lift. The ratio of weights of the man recorded when the lift is ascending with a uniform speed of $2\, m/s$ to when it is descending with a uniform speed of $4\, m/s$ will be:

$A$ lift is moving down with an acceleration $a$. $A$ man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively:

The weight of a man in a stationary lift is $w_1$ and when it is moving downwards with uniform acceleration $a$ is $w_2$. If the ratio $w_1 : w_2 = 4 : 3$,then the value of $a$ is ($g =$ acceleration due to gravity).

$A$ $60\, kg$ man stands on a spring scale in a lift. At some instant,he finds the scale reading has changed from $60\, kg$ to $50\, kg$ for a while and then comes back to the original mark. What should we conclude?

$A$ vessel containing water is moving with a constant speed towards the right along a straight horizontal path. If the vessel is given a constant retardation towards the right along a straight line,which of the following diagrams represents the surface of the liquid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module