A uniform thin rod of mass ‘m’ and length L is held horizontally by two vertical strings attached to the two ends. One of the string is cut. Find the angular acceleration soon after it is cut
Question
A uniform thin rod of mass ‘m’ and length L is held horizontally by two vertical strings attached to the two ends. One of the string is cut. Find the angular acceleration soon after it is cut
Solution
To solve this problem, we need to use the principles of rotational dynamics.
Step 1: Identify the forces acting on the rod After the string is cut, the only forces acting on the rod are its weight (mg) acting at the center of mass (midpoint of the rod) and the tension (T) in the remaining string.
Step 2: Apply Newton's second law in rotational form The net torque on the rod is equal to the moment of inertia times the angular acceleration (τ=Iα). The torque due to the tension is zero because the line of action passes through the axis of rotation. The torque due to the weight is (mg)(L/2) because the weight acts at a distance L/2 from the axis of rotation.
Step 3: Calculate the moment of inertia of the rod The moment of inertia of a rod about an end is given by I=(mL^2)/3.
Step 4: Solve for the angular acceleration Substituting the expressions for the torque and the moment of inertia into the equation τ=Iα gives (mg)(L/2) = ((mL^2)/3)α. Solving for α gives α = 3g/(2L).
So, the angular acceleration of the rod just after the string is cut is 3g/(2L).
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