Circular motion- speed of rider, centripetal acceleration and coefficient of friction

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Homework Statement



One of the most popular rides at carnivals is the one that resembles a large circular drum that rotates on its axis. Once the ride is spinning fast enough, the floor drops down and the riders are left pinned to the inside wall of the drum (I believe they are describing a ride called the gravitron but im not sure).

If the ride has a diameter of 5.2m and makes four complete rotations in 9.2s, determine:

a) the speed of the rider.

b) the centripetal acceleration of the rider.

c) the coefficient of friction required to keep the rider from slipping downwards along the drum wall when the floor drops down.

Homework Equations


For part a) I used the following equations:

Δd=2∏R
Δt=T time interval for one cycle is equal to a period.
v=Δd/Δt

For part b) I used the following equations:
ac=v2/r

For part c) I used the following equation:
v2=μgR

The Attempt at a Solution



a) From the diameter given d=5.2m, so from this we can divide by 2 to give R=2.6m
With this value we can plug it into the Δd=2∏R equation to give 16.34m.

We go on to find the time interval for one cycle by doing Δt=9.2s/4 rotations= 2.3s

Then to find the speed of the rider we use v=Δd/Δt=16.34m/2.3s=7.10m/s and this is our final answer.

b)For this part of the question I used v=7.10m/s which was found in the previous part a) and r=2.6m, and plugged these values into ac=v2/r= 19.4 m/s2 as our final answer.

c) For the last part I simply plugged more values into v2=μgR and solved for μ. Which turned out to be μ=1.98

I would like to just ask if anyone would be able to verify that I did this problem correctly and that all the significant figures are being respected! Thank you so much for your time :)
 

Answers and Replies

  • #2
ehild
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c) the coefficient of friction required to keep the rider from slipping downwards along the drum wall when the floor drops down.


For part c) I used the following equation:
v2=μgR
Why did you use that equation? Force of friction equals the normal force times μ.
From what does the normal force act on the man? Remember, there is no floor!

ehild
 
  • #3
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because that was the only equation that I could think of that had μ... so if theres no floor, that means theres no coefficient of friction?
 
  • #4
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There is a coefficient of friction, but the frictional force is not due to the non-existent floor, instead because of the wall. Since gravity doesn't act towards the wall, it cannot be responsible for the normal force. So the force of friction, as ehild said, would be normal force(on the wall) times μ.
 
  • #5
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ohhh ok! thank you, ill try that!
 
  • #6
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in the meantime are my answers for part a) and b) correct?
 
  • #7
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i dont have the weight of the rider to calculate the normal force... Im having a mind block at the moment
 
  • #8
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i dont have the weight of the rider to calculate the normal force... Im having a mind block at the moment
Why would you need the weight? . What causes the normal force that 'pushes' him against the wall?


Edit : The concepts for the other two parts seem right, though I haven't checked the calculations.
 
  • #9
tms
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i dont have the weight of the rider to calculate the normal force... Im having a mind block at the moment
Don't forget Newton's third law.
 
  • #10
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i still don't understand what I can do without having the value of m... help please!
 
  • #11
ehild
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You do not need the value of m. If it is easier to you, assume
it 100 kg.


ehild
 
  • #12
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how can you make an assumption like that though? :S
 
  • #13
tms
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how can you make an assumption like that though? :S
Because it ultimately does not matter. Write down the full equation for the normal force, and the full equation for the gravitational force, and you should see why.
 
  • #14
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I don't understand still :( part c) is giving me a headache
 
  • #15
ehild
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The wall exerts a normal force N on the man, and that normal force keeps him in circular motion. The normal force is equal to the centripetal force.
The force of friction is normal force times μ, and it acts along the wall, and upward, in the opposite direction as gravity. The man will not drop if the force of friction is equal to the force of gravity.

ehild
 
  • #16
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and the normal force is just mv2/R... right? after I get the normal force and times that by the μ that I got from doing v2/gR; do then go on to do μ=Ff/Fn? Cause wouldnt that just end up giving me the same value for μ that I got in the first place? Thank you very much for your time by the way!
 
  • #17
ehild
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The force of friction is Nμ, and N=mv2/r. So μmv2/r=mg. The friction acts against gravity. It is not the same equation you used.

ehild
 
Last edited:
  • #18
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and the normal force is just mv2/R... right? after I get the normal force and times that by the μ that I got from doing v2/gR; do then go on to do μ=Ff/Fn? Cause wouldnt that just end up giving me the same value for μ that I got in the first place? Thank you very much for your time by the way!

What you are doing μ=Ff/Fn=(mv2/R)/mg=v2/Rg
Which is wrong.

Ff is the value of frictional force which is perpendicular to the Fn which is equal to mg.
Fn is the pressing force = centripetal force.

The orientation is turn 90° from normal horizontal plane. Now its a vertical plane.
 
Last edited:
  • #19
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I think I got it finally.. so I did:

μ=(9.8m/s2)*(2.6m)/ (7.10m/s)2
μ=0.505

does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!
 
  • #20
881
40


I think I got it finally.. so I did:

μ=(9.8m/s2)*(2.6m)/ (7.10m/s)2
μ=0.505

does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!
Yep, that is correct!! :approve:
 
  • #21
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its about time hahah thank you so much for you help!
 

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