A thin uniform rod (length = 1.3 m, mass = 3.2 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)mL2. The rod is released when it is 50° below the horizontal. What is the angular acceleration of the rod at the instant it is released?
To find the angular acceleration of the rod at the instant it is released, we can use the rotational analog of Newton's second law, which states that:
Torque = Moment of inertia × Angular acceleration
In this case, the torque acting on the rod is due to the gravitational force. The equation for torque caused by the gravitational force is given by:
Torque = mg × d × sin(θ)
Where:
m = mass of the rod
g = acceleration due to gravity (approximately 9.8 m/s^2)
d = distance from the pivot point to the center of mass of the rod
θ = angle the rod makes with the horizontal (in radians)
First, we need to calculate the distance (d) from the pivot point to the center of mass of the rod. Since the rod is uniform, the center of mass is located at the midpoint of the rod, which is 1.3 m / 2 = 0.65 m from the pivot point.
Next, we convert the given angle of 50° to radians. We know that 180° = π radians, so:
θ = 50° × (π / 180°) = π / 6 radians
Now we can substitute the given values into the torque equation:
Torque = (3.2 kg) × (9.8 m/s^2) × (0.65 m) × sin(π / 6)
The moment of inertia of the rod is given as (1/3)mL^2, which we can substitute as:
Torque = (1/3)(3.2 kg)(1.3 m)^2 × (9.8 m/s^2) × (0.65 m) × sin(π / 6)
Finally, we can rearrange the torque equation to solve for angular acceleration:
Angular acceleration = Torque / Moment of inertia
Substituting the calculated torque and moment of inertia:
Angular acceleration = [(1/3)(3.2 kg)(1.3 m)^2 × (9.8 m/s^2) × (0.65 m) × sin(π / 6)] / [(1/3)(3.2 kg)(1.3 m)^2]
Simplifying the expression gives us the angular acceleration of the rod at the instant it is released.