Question

Please help me! "Determine the acceleration (in m/s/s) of a rider on the Cajun Cliffhanger (a barrel ride at an amusement park) if the rider makes 8.44 revolutions around the 5.23-m diameter circle in 28.8 seconds."

Answer 1

You need to know these formulas. \[\omega=\frac{ \theta }{ t } , v= r \omega, a=\frac{ v ^{2} }{ r }\]

You also need to know that:
1) r= radius (you are given the diameter)
2) how to convert from \[rev \to \theta \] . . . . \[\theta =rev * \frac{ 2\pi} { 1 }\]

That should allow you to solve the problem. I hope this is right, I'm a little rusty.

Answer 2

Yea that'll do it.

Answer 3

The period
\[T=2\pi radians\]
\[\frac{ 8.44revs }{ 28.8\sec }*\frac{ 2\pi radians }{ 1rev }=\frac{ 211\pi radians }{ 360 \sec}=\omega\]

Answer 4

Change rotational velocity into translational velocity.
\[v=r \omega \]
\[v=\left( \frac{ 5.23m }{ 2 } \right)\left( \frac{ 211\pi rads }{ 360\sec } \right)\approx4.815\frac{ m }{ \sec }\]

Answer 5

Then the centripetal acceleration. We know it's centripetal because it's going around in a circle.
\[\alpha=\frac{ v^2 }{ r }=\frac{ 4.815 \frac{ m }{ s }}{ 5.23m*.5 }\approx1.84\frac{ m }{ s^2 }\]

Answer 6

awesome!!!! thank you guys so much!!!