Max Speed for 10 m Radius Spinning Drum Ride

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To determine the maximum speed of a spinning drum ride with a 10 m radius and a maximum acceleration of 2g, the standard formula for centripetal acceleration is applied. The formula a = v²/r is rearranged to find v, resulting in v = √(a*r). Substituting the values, the maximum speed calculated is approximately 4.47 m/s. The discussion also clarifies that the question specifically asks for linear speed, not angular speed. The importance of considering friction and the design of the ride is acknowledged in the calculations.
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Homework Statement


If a spinning drum ride cannot exceed a 10 m radius, and the riders are not to
experience an acceleration greater than 2g, what is the maximum speed at
which the drum can spin?


Homework Equations


No Idea


The Attempt at a Solution


I tried using (theta) = tan-1 of (rg/v2)
v2 = (tan(theta))/(rg)
v = (root) of (tan(theta))/(rg)
 
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Hi Alcape! :smile:

(have a theta: θ :wink:)
Alcape said:
If a spinning drum ride cannot exceed a 10 m radius, and the riders are not to
experience an acceleration greater than 2g, what is the maximum speed at
which the drum can spin?

I tried using (theta) = tan-1 of (rg/v2)
v2 = (tan(theta))/(rg)
v = (root) of (tan(theta))/(rg)

What's θ? :confused: You can assume that the drum's wall are vertical. :wink:

Now use the standard formula for centripetal acceleration :smile:
 
If the walls are vertical then it doesn't work as it's at 90 degrees and tangent of 90 degrees then it comes out to be a mathematical error
 
ah … you obviously haven't seen one of these at a fairground. :biggrin:

Yes, if the walls are smooth, it can't be done, but you can assume the walls are rough, so that friction will keep you up (or even that there's a ledge to stand on :wink:).
 
I now have the answer if you use the formula for acceleration:a=v2/r
and transpose it so that it becomes: v2 = a*r and therefore becomes:
v = √(a*r) and it comes out to be 4.47...
 
Hi Alcape! :smile:

(just got up :zzz: …)
Alcape said:
I now have the answer if you use the formula for acceleration:a=v2/r
and transpose it so that it becomes: v2 = a*r and therefore becomes:
v = √(a*r) and it comes out to be 4.47...

what about g ? :redface:

(and are you sure the question isn't asking for the angular speed?)
 
Yes I'm sure as this is question a and b asks for the angle
 

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