A bead of mass m can slide on a frictionless | vedclass

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Question

(A) Let the position of the bead be at an angle $	heta$ from the horizontal axis passing through $O$. The distance of the bead from $A$ is $l_A = \sq

Vedclass · Accepted Answer

It will move with variable speed.

Vedclass · Answer

(A) Let the position of the bead be at an angle $	heta$ from the horizontal axis passing through $O$. The distance of the bead from $A$ is $l_A = \sqrt{(r\cos	heta)^2 + (r\sin	heta - 0.5r)^2} = \sqrt{r^2 + 0.25r^2 - r^2\sin	heta} = r\sqrt{1.25 - \sin	heta}$.Similarly,the distance from $B$ is $l_B = r\sqrt{1.25 + \sin	heta}$.The potential energy of the system is $U = \frac{1}{2}k(l_A^2 + l_B^2) = \frac{1}{2}kr^2(1.25 - \sin	heta + 1.25 + \sin	heta) = 1.25kr^2$.Since the potential energy $U$ is constant regardless of the position $	heta$,the total mechanical energy $E = K + U$ implies that the kinetic energy $K$ must also be constant.Therefore,the speed of the bead remains constant throughout the motion.Since the speed is constant and the path is a circle,the bead moves with constant speed,which contradicts option $A$.Angular momentum about $O$ is not conserved because the spring forces exert a torque about $O$.Since the potential energy is constant,the bead does not experience a restoring force towards $C$,so it does not oscillate simple harmonically.Thus,none of the provided options are strictly correct in the context of standard physics problems,but checking the energy expression,$U = 1.25kr^2$,which is constant. Given the options,if we re-evaluate the potential energy,it is indeed constant,meaning the bead moves with constant speed. However,looking at the options provided,there might be a typo in the question's options. Re-checking the potential energy: $U = \frac{1}{2}k(l_A^2 + l_B^2) = 1.25kr^2$. The speed is constant. Thus,the bead moves with constant speed.

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