A thin rod (length = 1.10 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
Sorry - I would show some work but I have no clue how to even attempt this. Please help :(
2 answers
Not sure how to apply the principle of conservation of mechanical energy.
(a) With no mass in the rod and all of the mass (M) at the top, the total kinetic energy just before it strikes the floor is
(M/2)V^2 = (M/2)(w*L)^2 = M g L
M cancels out, and the angular velocity w is
w = sqrt(2g/L) = 4.2 rad/s
(b) Just before impact, the mass atop the rod is accelerating at rate g. Angular acceleration before impact = a/L = g/L = 8.9 rad/s^2
(M/2)V^2 = (M/2)(w*L)^2 = M g L
M cancels out, and the angular velocity w is
w = sqrt(2g/L) = 4.2 rad/s
(b) Just before impact, the mass atop the rod is accelerating at rate g. Angular acceleration before impact = a/L = g/L = 8.9 rad/s^2