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Question

A pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25° from the vertical. If the initial speed of the pendulum is 1.2 m/s, what is its speed at the bottom of the swing?

Answer 1

Solve using energy conservation.

 

At top,

 

E₁ = (1/2)mv₁² + mgL(1 - coθ)

 

At bottom,

 

E₂ = (1/2)mv₂² + mg(0) = (1/2)mv₂2

 

E₁ = E₂ ⇒

 

(1/2)v₂² = (1/2)v₁² + gL(1 - cosθ)

 

v₂ = √[v₁² + 2gL(1 - cosθ)]

 

v₂ = √[(1.2)² + 2(9.8)(2.0)(1 - cos25°)] m/s ≅ 2.26 m/s