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November 29, 2014

November 29, 2014

Posted by **Amar** on Wednesday, October 3, 2012 at 6:54pm.

- Rotational Motion Physics Problem -
**Elena**, Thursday, October 4, 2012 at 2:31pmGiven: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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