Question

A uniform rod of mass M = 250.0 g and length L = 50.0 cm stands vertically on a horizontal table. It is released from rest to fall.

Answer 1

Calculate the vertical acceleration of the moving end of the rod as it makes an angle θ = 45° with respect to the vertical (take upward as positive and downward as negative).

Calculate the normal force exerted by the table on the rod as it makes an angle θ = 45° with respect to the vertical.

Answer 2

my question is that is vertical acceleration constant? since the gravitational acc. is constant i thought vertical acc would be constant as well.

Answer 3

I think what the question seeks is that you find moment of inertia of the rod,
from its end, which is I=M L/12 + M (L/2)^2 and then calculate
torque due to gravity acting on the center of mass, m g L sin(45),
finally equating the torque with the product of I and the angular acceleration.

Normal force on table seems to be m g sin(45) but am not sure, due to complexity motion.

Answer 4

torque vector tau = g M (L/2) sin(45) = I alpha.
alpha vector = angular acceleration = acceleration / (L/2) = a / (L/2)

vector a has a horizontal component proportional to sin(45) and a vertical
acceleration component proportional to - cos(45)

Sorry it is hard to make these vectors and scalar clear here.

Answer 5

steps makes good sense but i dont quite get how we find I of different shapes? I hope we dont solve the integration everytime?