Daniel B. answered • 01/16/21
A retired computer professional to teach math, physics
I am assuming that originally the rod is horizontal.
Notation:
1) Let me denote the length of the rod as L, rather than l, because the latter looks like 1.
2) All definite integrals below are between 0 and the length L.
a)
The angular acceleration is the ration between the torque and the moment of inertia.
The torque is the product of the force of gravity, Mg, and the distance, L/4,
between the pivot P and the center of gravity.
(This is because the rod is horizontal making it perpendicular to the force of gravity.)
Thus the torque is
MgL/4
Now we calculate the moment of inertia.
The moment of inertia dI of a point mass dm of length dx at distance r from the pivot P is
dI = r²dm
In terms of the x-coordinate -- distance from the left end --
r² = (x - L/4)²
dm = b dx, where b is the linear density of the rod
The whole moment of inertial, I, is then the sum over all the points of mass:
I = ∫(x - L/4)²b dx = b ∫(x² - xL/2 + L²/16)dx = b (L³/3 - L³/4 + L³/16) = 7bL³/48 = 7ML²/48
The angular acceleration, being the ration between the torque and the moment of inertia, is
(MgL/4)/(7ML²/48) = 12g/7L
b)
We compute the angular velocity, ω, of the rod when it becomes vertical from conservation of energy.
The loss of potential energy (from the rod being horizontal to being vertical)
gets converted to the rotational energy of the rod.
The height of the center of gravity got reduced by L/4.
Therefore the loss of potential energy is
MgL/4
The rotational energy is
Iω²/2 = 7ML²ω²/96
The equality between the two energies:
7ML²ω²/96 = MgL/4
ω = √(24g/7L)
The velocity of the bottom is the product of angular velocity and the distance of the bottom from the pivot P:
√(24g/7L) 3L/4 = 3√(3gL/14)
