The angular acceleration of a simple pendulum at an angle α from the v

The angular acceleration of a simple pendulum at an angle α from the vertical is proportional to

tan α

sin² α

sin α

sin 2α

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This question was previously asked in

UPSC Geoscientist – 2023

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For a simple pendulum at an angle α from the vertical, the restoring torque about the point of suspension is given by τ = -mgl sin α, where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and α is the angular displacement. According to Newton’s second law for rotation, the torque is also equal to Iβ, where I is the moment of inertia and β is the angular acceleration. For a simple pendulum, I = ml². Thus, ml²β = -mgl sin α. This gives β = -(g/l) sin α. The magnitude of the angular acceleration is |β| = (g/l) sin α. Since g and l are constants, the angular acceleration is proportional to sin α.

– Restoring torque τ = -mgl sin α.
– Torque τ = Iβ.
– Moment of inertia of a simple pendulum I = ml².
– Angular acceleration β = τ/I = -(g/l) sin α.
– Magnitude of angular acceleration is proportional to sin α.

For small angles (α << 1 radian), sin α ≈ α, and the equation of motion becomes β ≈ -(g/l)α. In this small angle approximation, the motion is Simple Harmonic Motion (SHM), and the angular acceleration is proportional to the angular displacement α. However, for any angle α, the angular acceleration is strictly proportional to sin α.

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