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Create an expression for a rotating plate

  • Thread starter apierron
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  • #1
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A flat, circular plate (mass Mp, radius Rp, thickness Tp) rotates parallel to the surface of the Earth on a frictionless axle. As it spins, a point on the top of the disc traces a circle each time the plate completes a rotation.

in terms of the number of rotations per minute (N/min), create an expression to describe the speed of a point on the surface of the plate in units of m/s. In other words, what is v(N) if v has units of m/s? Hint: look at your units.

I'm thinking you need to use an equation that deals with centripetal force but I'm not sure

I'm hitting a brick wall here and need some major help. I don't even know where to start.
 

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  • #2
gneill
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In circular motion, what's the relationship between:

ω, r and v?
rpm and ω?
 
  • #3
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This sounds bad but I don't know what w represents
 
  • #4
gneill
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This sounds bad but I don't know what w represents
No problem! ω is the angular velocity, the rate of rotation in radians per second.
 
  • #5
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Okay so what I found is v= ωr so I came up with v(N)=2(pi)Rp/N but since he says N is rotations per minute I divided that by 60 to get v(N)=120(pi)Rp/N so that my units are in m/s Does this seem right?
 
  • #6
gneill
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Okay so what I found is v= ωr so I came up with v(N)=2(pi)Rp/N but since he says N is rotations per minute I divided that by 60 to get v(N)=120(pi)Rp/N so that my units are in m/s Does this seem right?
You're getting there. Your equation v= ωr is fine. Just what you need. There's a bit of a problem with your velocity expression though, having to do with the unit conversions.

The thing to do first is to concentrate on the conversion of rpm to ω in radians per second.

A good way to approach these sorts of conversions is to construct a series of unit conversions in the form of unity-valued ratios (fractions whose numerator and denominator express the same value but in different units).

So, for example, to convert n hours to seconds you might write:

[tex] n \; hr \times \frac{60 min}{hr} \times \frac{60 sec}{min} = n \times 3600 sec [/tex]

Note how the hour and minute units mutually cancel along the way leaving seconds as the units. Also note how each fraction represents the same value but in different units (thus 60 minutes is the same as 1 hour).

See if you can do the same thing for the conversion of N revs/min to radians per second.
 
  • #7
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so 1 rpm is 2pi radians/60 s, so now I have v(N)=60Rp/N.... better?
 
  • #8
gneill
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so 1 rpm is 2pi radians/60 s, so now I have v(N)=60Rp/N.... better?
If N starts out as the number of rpm, how does it end up in the denominator of the expression? Does it make sense that angular velocity would get smaller if rpm gets bigger?

If you've got 1 rpm → 2∏ rad/60s = 0.10472 rad/s, then N x 1rpm is?

Presumably the radial position of the point on the plate can be anywhere between the center of the plate (r = 0) and the edge of the plate (r = Rp). So you'll want to use r for the radius variable in velocity expression.
 
  • #9
HallsofIvy
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Each rotation, the point travels a distance equal to the circumference of the circle it describes- [itex]2\pi r[/itex] meters is r is measured in meters. If the disk is turning at "n" revolutions per minute, it does that n times per minute. That is, it travels a distance [itex]2n\pi r[/itex] meters per minute.
 

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