Posted by Amar on Wednesday, October 3, 2012 at 6:54pm.
Given:R=150m,
tangential component of acceleration a(τ)=1.2 m/s²,
a=2.2 m/s²,
φ=?
a²=a(n)²+a(τ)²,
where a(n) is the normal( centripetal) component of acceleration.
a(n)= sqrt(a²-a(τ)²) = sqrt(2.2²-1.2²) = 1.84 m/s²
a(n)=ω²•R =>
ω² = a(n)/R
The angular acceleration ε= a(τ)/R.
From
φ=ε•t²/2
ω=ε•t,
we obtain
φ=ω²/2•ε= a(n)/2•a(τ)=
=1.84/2•1.2=0.77 rad.
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