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November 22, 2014

November 22, 2014

Posted by **Alphonse** on Friday, October 26, 2012 at 2:10am.

[Note: If it helps you, for sake of definiteness, you may take the wheel to be rotating clockwiseÃ¢â‚¬Â¦ though this is immaterial.]

Â

a)Â Find the magnitude of the passenger's acceleration at that instant in m/s2.

b)Â What is the angle between the passenger's velocity vector and acceleration vector at that instant (in degrees)?

- Physics -
**Elena**, Friday, October 26, 2012 at 2:27amω₀=2πn₀=2π/33.2=0.19 rad/s

centripetal acceleration is a(cen) = ω₀²R=0.19²•42.2=1.52 m/s²

tangential acceleration is a(tan) =ε•R =0.0542•42.2=0.12 m/s²

acceleration is a=sqrt{ a(cen)²+a(tan)²}= sqrt{1.52²+0.12²}=1.525

α=arctan(a(tan)/a(cen)} =4.51°

the angle between the passenger's velocity vector and acceleration vector =90+4.51 =94.51°

- Physics -
**Alphonse**, Thursday, November 1, 2012 at 11:21pmThank you very much!

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