What Is the Radius of the Amusement Park Ride?

AI Thread Summary
The discussion revolves around calculating the radius of an amusement park ride based on the given speed, mass, and force experienced by a rider. The participant initially attempted to use the formula for centripetal force but made an error by not squaring the velocity in their calculations. After receiving clarification, they realized the correct formula involves rearranging the variables to account for the squared velocity. The correct equation for radius is r = mv²/F, which was highlighted as the proper approach. The conversation emphasizes the importance of careful arrangement and calculation in physics problems.
rockmorg
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I have a question on what it is I might be doing wrong here... I've tried this answer in the grader and it is wrong...

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s and an 83 kg person feels a 495 N force pressing against his back. Radius of the chamber?

Here is what I have done..

v = 3.2 m/s
m = 83 kg
F = 495 N
r = ?

Fc = mv2/r
r = Fc(1/mv2)

r = 495 N (1/(83kg * 3.2 m/s2))
r = .582 m

Even just looking at it it looks wrong, heh...

Any help would be great!

Thanks,
-
Morgan
 
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Notice this:

F = m \frac{v^2}{r}

\frac{r}{F} F = \frac{r}{F} m \frac{v^2}{r}

r = m \frac{v^2}{F}
 
I was just going to say that it looks like you didn't square the velocity.
 
Ahhh thanks much - so I guess it was just a matter of arranging my equation? The mass needs to multiply out to those other variables for it to work...
 
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