Posted by Jimmy on .
A long uniform rod of length 6.66 m and mass 6.74 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as in the figure.
At the instant the rod is horizontal, find its angular speed. The acceleration of gravity is 9.8 m/s2 . Answer in units of rad/s.
At the same instant find the magnitude of its angular acceleration. Answer in units of rad/s2.
Still at the same instant find the magnitude of the acceleration of its center of mass. Answer in units of m/s2.
Finally, find the magnitude of the reaction force at the pivot (still at the same moment). Answer in units of N.
I do not see the picture but assume that the rod falls with one end fixed.
The easy way to approach this is to use conservation of energy. The center of gravity of the rod has fallen 3.33 meters by the time it reaches horizontal.
It has therefore lost potential energy equal to m g (3.3) = 6.74 * 9.8 * 3.33
= 220 Joules
Therefore since the hinge is frictionless I have 220 J of kinetic energy
Call the angular velocity w, then the velocity, v, of the center of mass is 3.33 w
Ke = (1/2) m v^2 + (1/2) I w^2
I for a rod about its center is (1/12) m L^2
220 = (1/2)(6.74)(3.33w)^2 +(1/24)(6.74)(6.66)^2 w^2
220 = 49.8 w^2
w= 2.1 rad/s
(check my math!)
Now what is the angular acceleration?
The moment about the hinge is 3.33* mg
= 3.33 * 6.74 * 9.8
= 220 N m
That equals the moment of inertia about the hinge times A, the angular acceleration. I about hinge = (1/3) m L^2
220 = (1/3)(6.74)(6.66^2)A
so A = 2.21 rad/s^2
acceleration of middle = 3.33A = 7.35 m/s^2
force = mass * acceleration
force up = F at hinge
Force down = m g = 6.74*9.8 = 66 N
66 - F = 6.74*7.35
F = 16.51 N
that force * 3.33 should be (1/12) 6.74 (6.66^2) A about center
That gives A = 2.21, sure enough