Nonuniform Circular Motion - Find Maximum Total Acceleration

  • Thread starter LovePhys
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Homework Statement


A turntable is rotating at a constant angular velocity of ω = 4.0 rad/s in the direction of a clockwise fashion. There is a ten-cent coin on the turntable, at a distance of 5 cm from the axis of rotation.

Consider the time interval during which the turntable is accelerated initially from rest to its final angular velocity (ωf = 4.0 rad/s) . This is achieved with a constant angular acceleration (α) for 0.5 s.

Find an expression for the magnitude of total acceleration of the coin in terms of α and ω and use this to determine the maximum total acceleration experienced by the coin.


Homework Equations


[itex]a=\sqrt{a_{r}^2+a_{t}^2}[/itex]
[itex]a_{r}=ω^2r[/itex]
[itex]a_{t}=αr[/itex]


The Attempt at a Solution


Using those equations, I find an expression for the total acceleration:
[itex]a=r\sqrt{ω^4+α^2}[/itex]
I don't understand why the problem is asking for a maximum total acceleration. I am already given the values of r, ω, and α, can I just substitute them in to find the answer? Another thing that came into my mind when the word "maximum" pops up is finding the derivative of a function of a(t) with respect to time, but apparently I do not have that function here...

Any help would be greatly appreciated.
LovePhys
 

Answers and Replies

  • #2
Doc Al
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I don't understand why the problem is asking for a maximum total acceleration. I am already given the values of r, ω, and α, can I just substitute them in to find the answer?
Well, since ω varies over the given interval, you have to determine its maximum value. But I suspect you can handle it. :smile:
 
  • #3
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@Doc Al: Thanks for the hint. I think the maximum angular velocity is also the final angular velocity ω=4rad/s, isn't it?
[itex]α=\frac{Δω}{Δt}=\frac{4}{0.5}=8 (rad/s^2) [/itex]
So finally maximum [itex] a=0.05\sqrt{4^4+8^2}≈0.894(m/s^2) [/itex]
 
  • #4
Doc Al
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@Doc Al: Thanks for the hint. I think the maximum angular velocity is also the final angular velocity ω=4rad/s, isn't it?
Exactly.

[itex]α=\frac{Δω}{Δt}=\frac{4}{0.5}=8 (rad/s^2) [/itex]
So finally maximum [itex] a=0.05\sqrt{4^4+8^2}≈0.894(m/s^2) [/itex]
Looks good to me.
 

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