Question

PLEASE DO NOT WRITE DOWN NEEDS MORE INFO, all what is required to solve the problem is provided. If it is necessary I can wait more than stipulated, as this exercise is for own practise.

Thanks

Answer 1

Since, we know that, If rod hinged like that then, (MOI) - moment of inertia I = M l^2/3 sin^2 theta Now, In our case, Since we know, MOI, I about hinged point will be, I = Ml^2/3 + M (l/4)^2 rightarrow By parallel axis - theorem I = Ml^2/3 + Ml^2/16 = 19 Ml^2/48 Therefore I = 19 M l^2/48 sin^2 theta \r\n (b) Angular acceleration, we know that, Torque = I alpha or tau = I X material implication alpha = tau/I since, tau = Mg times (3l/4/20 tau = 3 Mgl/Q thus, alpha = tau/I = 3 Mgl/8/19 Ml^2/sin^2 theta 48 = 3 Mgl/8 times 48/19 Ml^2 sin^2 theta alpha = 18/19 g/l sin^2 theta Note: - for small oscillation sin theta rightarrow theta (c) Comparing, aloha = w^2 Q Therefore W^2 = 18/19 g/l or W = squareroot 18/19 g/l T = 2 pi/W Therefore T = 2 pi squareroot 19 l/18 g Required Time period.