Centripetal force circular drum question

AI Thread Summary
A circular drum with a radius of 2.4m spins, and a person on its wall needs to avoid falling out. The speed of the drum is calculated to be 6.0 m/s, leading to a centripetal acceleration of 15 m/s². To find the coefficient of friction required to keep the person from falling, it is determined that the normal force equals the centripetal force, which is independent of the person's mass. The final calculation shows that the coefficient of friction needed is approximately 0.65. This analysis confirms that both large and small masses can be supported as long as the coefficient of friction remains constant.
Checkfate
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Okay, I need help with this question. It is killing me! :confused:

A circular drum is spinning round and round. A person is on the wall of the drum and there is no floor beneath him. If the drum has a radius of 2.4m and takes 2.5s to comlete a revoltion, what is the coefficient of friction reqired to prevent him from falling out?

I calculated the speed of the drum.

(Pi symbol did not work... PI=pi)
v=\frac{2PIr}{T}
=6.0m/s

and then used that to calculate the centripetal acceleration.

a_{c}=\frac{v^{2}}{r}
=\frac{36m^{2}/s^{2}}{2.4m}
=15m/s^{2}

But then I haven't the slightest clue how to calculate a coefficient of friction. I thought that perhaps I could treat the drum as a planet and it's gravitational constant could be the centripetal acceleration (I am guessing this is the right line of thinking :) ) but then the fact that I lack a mass to work with gets in the way. Can someone point me in the right direction? Thanks :smile:

coefficient of friction = \frac {F_{f}}{F_{N}} but I can't calculate forces without a mass!
 
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Consider the forces acting on the person: What's holding the person up? What friction force is required? How does friction relate to the normal force? What provides the normal force and how can you calculate it?
 
You don't need to know the mass. What's the normal force acting on the person?

Edit: I was late as usual. ;)
 
Thank you both.. The normal force acting on the person would be the centripetal acceleration times the person's mass, would it not? I am still stuck :( lol.

The thing holding the person up is his inertia... His body wants to continue straight ahead but the wall stops his body from doing this. The friction force required... I would guess that it depends on the person's mass! Would a whale not stay in this thing longer than a flee? Since it has more lnertia and thus would create more friction..
 
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Checkfate said:
The normal force acting on the person would be the centripetal acceleration times the person's mass, would it not?
Correct. The normal force of the wall against the person provides the centripetal force that keeps him going in a circle.

The thing holding the person up is his inertia... His body wants to continue straight ahead but the wall stops his body from doing this.
Let's be precise. What forces act on the person? Hint: I count three forces acting.

Hint 2: There's no acceleration in the vertical direction. So what does that tell you about the sum of the vertical forces?



The friction force required... I would guess that it depends on the person's mass!
Indeed it does. But neutrino is correct: You don't need to know the person's mass to solve for the coefficient of friction. (Hint: Solve the problem using symbols. Don't plug in numbers until the last step.)

Would a whale not stay in this thing longer than a flee? Since it has more lnertia and thus would create more friction.
As long as each animal has the same coefficient of friction with the wall (and assuming that the usual model of friction applies), then the answer is: Both flea and whale are equally supported! True, the whale does require more friction, but the normal force on the whale is enough to create it. (Of course, the normal force required to keep the massive whale moving in a circle might be so great that the drum just breaks apart! :smile: )
 
Okay I think I finally got it :)

Since there is no vertical acceleration, F_{f}=F_{g}=mg

The normal force = F_{N}=15m/s^{2}(m)

Coefficient of friction =\frac{F_{f}}{F_{N}}
=\frac{m(9.80m/s^{2})}{15m/s^{2}(m)}
=0.65

Right? :)

Thanks Doc Al
 
You got it. :smile:
 
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