Physics - W.E.P. Question with Solution | TestHub
The two particles of mass m and 2 m , respectively are connected by a light rod of negligible mass and slide with negligible friction on a circular path of radius r inside a fixed vertical circular ring. If the system is released from rest at and is taken from positive x-axis in clockwise direction.
Options:(select one or more)
Answer:
Solution:
Both particles slide on the circle of radius r with fixed angular separation 90 deg (the chord length fixes this; the figure shows the 2m particle leading by 90 deg). With theta measured clockwise from the +x axis for the 2m particle, heights are y_2m = -r sin(theta) and y_m = -r cos(theta)... i.e. energy conservation from rest at theta=0 gives (3/2) m v^2 = mgr(2 sin(theta) + cos(theta) - 1), so v^2 = (2gr/3)(2 sin(theta) + cos(theta) - 1). (A) The rod is horizontal when both particles are at equal height: theta = 45 deg; then v^2 = (2gr/3)(sqrt2 + 1/sqrt2 - 1) = (2gr/3)(3/sqrt2 - 1) - matches A. TRUE. (B) and (C): v^2 is maximum when d/dtheta(2 sin(theta) + cos(theta)) = 0, i.e. tan(theta) = 2, so theta = tan^-1(2), NOT tan^-1(1/2); the maximum value is v^2 = (2gr/3)(sqrt5 - 1). So B TRUE, C FALSE. (D) The system turns back when v = 0 again: 2 sin(theta) + cos(theta) = 1 gives tan(theta/2) = 2, i.e. theta_max = 2 tan^-1(2). TRUE. Answer: A, B, D.
