A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m and m are attached to it at distances b and 3b respectively from the pivot. The rod is held horizontal and then released. The angular acceleration of the rod at the instant it is released is
Question
A nearly massless rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m and m are attached to it at distances b and 3b respectively from the pivot. The rod is held horizontal and then released. The angular acceleration of the rod at the instant it is released is
Solution
The problem involves the concept of rotational motion, specifically the moment of inertia and torque.
Step 1: Identify the forces acting on the system.
The forces acting on the system are the gravitational forces acting on the two masses.
Step 2: Calculate the moment of inertia (I) of the system.
The moment of inertia of a system of particles is given by the sum of the products of the mass of each particle and the square of its distance from the axis of rotation.
So, I = m*(3b)^2 + 2m*b^2 = 9mb^2 + 2mb^2 = 11mb^2
Step 3: Calculate the torque (τ) acting on the system.
The torque is given by the product of the force and the perpendicular distance from the axis of rotation.
So, τ = mg3b + 2mgb = 3mbg + 2mbg = 5mbg
Step 4: Use Newton's second law for rotation to find the angular acceleration (α).
Newton's second law for rotation states that the net torque acting on a body is equal to the product of its moment of inertia and its angular acceleration.
So, α = τ/I = 5mbg / 11mb^2 = 5g / 11b
Therefore, the angular acceleration of the rod at the instant it is released is 5g / 11b.
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