A smooth rod of length l is kept inside a tro | vedclass

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

Question

(D) To keep the rod in equilibrium with respect to the trolley,we analyze the forces in the non-inertial frame of the trolley.Let the rod have mass $m

Vedclass · Accepted Answer

$g \cot 	heta$

Vedclass · Answer

(D) To keep the rod in equilibrium with respect to the trolley,we analyze the forces in the non-inertial frame of the trolley.Let the rod have mass $m$ and length $l$. The forces acting on the rod in the trolley's frame are:$1$. Gravitational force $mg$ acting downwards at the center of mass.$2$. Pseudo force $ma$ acting horizontally backwards at the center of mass.$3$. Normal reactions $N_1$ and $N_2$ at the ends of the rod,perpendicular to the rod's surface.For rotational equilibrium about the center of mass,the torques due to the forces must balance. The center of mass is at a distance $l/2$ from both ends.The torque due to the pseudo force $ma$ about the center of mass is $	au_{pseudo} = (ma) \cdot (l/2) \sin 	heta$.The torque due to the gravitational force $mg$ about the center of mass is $	au_{gravity} = (mg) \cdot (l/2) \cos 	heta$.For the rod to be in equilibrium,the net torque about the center of mass must be zero:$(ma) \cdot (l/2) \sin 	heta = (mg) \cdot (l/2) \cos 	heta$Canceling $(l/2)$ from both sides:$ma \sin 	heta = mg \cos 	heta$Solving for $a$:$a = g \frac{\cos 	heta}{\sin 	heta} = g \cot 	heta$

$A$ smooth rod of length $l$ is kept inside a trolley at an angle $\theta$ as shown in the figure. What should be the acceleration $a$ of the trolley so that the rod remains in equilibrium with respect to it?

Similar Questions

$A$ 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?

In a lift moving up with an acceleration of $5 \,m/s^2$, a ball is dropped from a height of $1.25 \,m$. The time taken by the ball to reach the floor of the lift is ... (nearly) $(g=10 \,m/s^2)$ (in $\,s$)

$A$ mass $M$ is placed on a smooth wedge of angle $\theta$ resting on a frictionless surface. To keep the mass $M$ at rest relative to the wedge,the wedge must be given a horizontal acceleration $a$ equal to:

$A$ point mass $m$ is suspended from a light thread of length $l$, fixed at $O$, and is whirled in a horizontal circle at a constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are

Vedclass Test Series

Exam Paper Generator

Online Exam Module