Question

Suppose that the rod is held at rest horizontally and then released. (throughout the remainder of this problem, your answer may include the symbol i , the moment of inertia of the assembly, whether or not you have answered the first part correctly.) what is the angular acceleration α of the rod immediately after it is released? take the counterclockwise direction to be positive. express α in terms of some or all of the variables mᵣ , m₁ , m₂, x , i , and g .

Answer 1

To find the angular acceleration of a rod immediately after it is released, we must use the equation α = (mₕ⋅ₓₑ⋅g⋅x) / I, where the torque due to gravity is divided by the moment of inertia.

Step-by-step explanation:

The question asks to calculate the angular acceleration (α) of a rod immediately after it is released from a horizontal position. To find the angular acceleration, we assume that the only force causing angular acceleration is the torque due to gravity acting at the center of mass of the system, and we use Newton's second law for rotation.

The formula for torque (τ) is the product of the force and the lever arm distance (r). Since the weight acts at the center of mass of the system, we use the distance x from the axis of rotation to the center of mass. Torque is therefore τ = mₕ⋅ₓₑ⋅g⋅x, where mₕ is the mass of the rod, m₁ and m₂ are the masses of the beads, and x is the distance from the center. In this system, the center of mass will not change since the masses are equal and placed symmetrically.

Angular acceleration can then be found by dividing the torque by the moment of inertia (I): α = τ / I. So, the angular acceleration immediately after the rod is released is α = (mₕ⋅ₓₑ⋅g⋅x) / I.