Question

A plank with a body of mass m placed on it starts moving straight up according to the law y = a(1 —cosωt), where y is the displacement from the initial position, ω = 11s^-1. Find:

(a) the time dependence of the force that the body exerts on the plank if a = 4.0cm; plot this dependence;

(b) the minimum amplitude of oscillation of the plank at which the body starts falling behind the plank;

(c) the amplitude of oscillation of the plank at which the body springs up to a height h = 50cm relative to the initial position (at the moment t = 0).

Answer 1

(a) For the block from Newton's second law in projection form F_y = mw_y

N - mg = my         (1)

But from   y = a(1 - cosωt)

we get  y = ω²acosωt   (2)

From eqn (1) and (2)

From Newtons' third law the force by which the body m exert on the block is directed vertically downward and equls

(b) When the body m starts, falling behind the plank or losing contact, N = 0, (because the normal reaction is the contact force). Thus from eqn. (3)

(c) We observe that the motion takes place about the mean position y = a. At the initial instant y = 0. As shown in (b) the normal reaction vanishes at a height (g/ω²) above the position of equilibrium and the body flies off as a free body. the speed of the body at a distance (g/ω²) from the equilibrium position is ω√(a² - (g/ω²)²), so that the condition of the problem gives

Hence solving the resulting quadratic equation and taking the positive roof,