A slender unform rod of mass M and length ℓ is pivoted at one end so that it can rotate in a vertical plane (see figure.) There is neglifible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it make an angle θ with the vertical is :
Question
A slender unform rod of mass M and length ℓ is pivoted at one end so that it can rotate in a vertical plane (see figure.) There is neglifible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it make an angle θ with the vertical is :
Solution
The problem involves the concept of rotational motion and torque. Here are the steps to solve it:
Step 1: Identify the forces acting on the rod. The force of gravity acts at the center of mass of the rod, which is at a distance of ℓ/2 from the pivot point.
Step 2: Calculate the torque about the pivot point. Torque (τ) is given by the equation τ = rFsinθ, where r is the distance from the pivot point to the force (ℓ/2), F is the force (mg), and θ is the angle between the force and the line of action. In this case, the angle between the force of gravity and the line of action is 90 - θ, so sin(90 - θ) = cosθ. Therefore, τ = (ℓ/2)mgcosθ.
Step 3: Use Newton's second law for rotation, which states that the net torque on an object is equal to its moment of inertia times its angular acceleration. The moment of inertia of a rod about one end is (1/3)Mℓ². Therefore, (ℓ/2)mgcosθ = (1/3)Mℓ²α, where α is the angular acceleration.
Step 4: Solve for α. α = 3gcosθ/(2ℓ).
So, the angular acceleration of the rod when it makes an angle θ with the vertical is 3gcosθ/(2ℓ).
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