Suppose a pendulum of mass m1is suspended on the roof of a car. A light pin is attached to it and another pendulum of mass m2 is suspended from the pin. The car then starts with a constant accelaration. What are the angles @1 and @2 which the first and the second pendulum make with the vertical ?



The answer goes like this.........

From the FBD of the 2nd pendulum, by using lammi's theorem

m₂a/sin(180-@₂) = T₂/sin90 = m₂g/sin(90+@₂)

tan@₂ = a/g

T₂sin@₂ = m₂a

T₂cos@₂ = m₂g

Here T₂ = tension on the string

Now from the FBD of the 1st pendulum

for horizontal equilibrium


m₁a + T₂sin@₂ = T₁sin@₁ ......................(1)

for vertical equilibrium

m₁g + T₂cos@₂ = T₁cos@₁ .....................(2)


By (1)/(2)....

tan@₁ = (m₁a + m₂a)/(m₁g + m₂g) = a/g

Therefore

@₁ = @₂ = tan^-1(a/g)


Please tell me wherher i'm correct or not. I will be waiting for your answer. This question was posted by shank on the discussion zone.