A thin uniform rod (length = 1.3 m, mass = 4.1 kg) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is
1/3 m L2.
The rod is released when it is 55° below the horizontal. What is the angular acceleration of the rod at the instant it is released?
angular acceleration
= (Torque)/(Moment of Inertia)
= (M*g*cos55*L/2)/[(1/3)*M*L^2)
= (3/2)*cos55*g/L = 6.5 s^-2
To find the angular acceleration of the rod at the instant it is released, we can use the principle of conservation of energy.
The potential energy of the rod when it is released is converted into rotational kinetic energy. The formula for rotational kinetic energy is given by:
K_rot = (1/2) I ω^2
where K_rot is the rotational kinetic energy, I is the moment of inertia of the rod, and ω is the angular velocity of the rod.
Since the rod is released from rest, its initial angular velocity ω_0 is 0. The final angular velocity ω can be calculated using the conservation of energy equation:
mgh = (1/2) I ω^2
where m is the mass of the rod, g is the acceleration due to gravity, and h is the vertical height from the pivot point to the lower end of the rod.
In this case, the height h can be calculated by multiplying the length of the rod L by the sine of the angle θ:
h = L sin(θ)
Substituting this into the conservation of energy equation:
mgh = (1/2) I ω^2
mLg sin(θ) = (1/2) I ω^2
Now, we can substitute the values given:
m = 4.1 kg
L = 1.3 m
g = 9.8 m/s^2
θ = 55°
I = (1/3) mL^2
Substituting the values into the equation:
(4.1 kg)(1.3 m)(9.8 m/s^2) sin(55°) = (1/2) [(1/3) (4.1 kg)(1.3 m)^2] ω^2
Simplifying the equation:
21.481 N = (1/6) (8.533 kg m^2) ω^2
Now, we can solve for ω^2:
ω^2 = (21.481 N) x (2) / (8.533 kg m^2)
ω^2 = 5.393 rad^2/s^2
Finally, we can calculate the angular acceleration α using the equation:
α = ω / t
Since the rod is released, the initial angular velocity ω_0 is 0, and we assume it reaches the final angular velocity ω instantaneously. Therefore, the time t is also 0.
Thus, the angular acceleration α of the rod at the instant it is released is 0 rad/s^2.