Radius of Circular Disc for Mech. Engr Problem

  • Thread starter Robust
  • Start date
In summary, the mechanical engineer requires the turning of a circular disc giving an area of 16 units. He gives the machinist a radius of 2.2567. This equation results in the area of the disc being 16*2.2567=36. The problem is that if the area is given as the cross sectional area, not the total surface area, then the thickness of the disc is necessary in order to account for the two circular surfaces and the 'strip' around the circumference.
  • #1
Robust
18
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A mechanical engineer requires the turning of a circular disc giving an area of 16 units. What radius does he give the machinist?
 
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  • #2
What is an equation for area that involves the radius?

You have the area...You have a constant [tex] \pi[/tex]...
 
  • #3
FredGarvin said:
What is an equation for area that involves the radius?

You have the area...You have a constant [tex] \pi[/tex]...
The equation I would give is: sqrt area/sqrt pi = radius; thus, if an area 16 then 4/1.77245...= 2.2567...radius. r^2*pi = 16 area

I took a bit of abuse by contempories for this formula - wanted to hear what Y'all might think of it.
 
  • #4
Looks good to me Robust.

Check that the given area is the cross sectional area, and not the total surface area of the disc. If this is so, you're spot on.

Don't be so quick to take abuse from contemporaries! Too many times I've lost face and backed down, only to find that I was right all along...
 
  • #5
brewnog said:
Looks good to me Robust.

Check that the given area is the cross sectional area, and not the total surface area of the disc. If this is so, you're spot on.

Don't be so quick to take abuse from contemporaries! Too many times I've lost face and backed down, only to find that I was right all along...
Thanks for the confirmation, but I'm not out of the woods yet on this, for the problem as originally given (project engineer) did state the total surface area, the tolerance being incidental. The PE is one of those who likes to throw these kinds of questions at us, just to keep us on our toes I suspect.

But the implications of this one does present a serious conflict as regards the pi value, showing it to be irrelevant. the radius is given consistently regardless of the pi pi value employed (recognized pi values). Here is the given formula: sqrt area/sqrt pi = radius; r^2*pi = 16 area!
 
  • #6
Robust said:
But the implications of this one does present a serious conflict as regards the pi value, showing it to be irrelevant. the radius is given consistently regardless of the pi pi value employed (recognized pi values). Here is the given formula: sqrt area/sqrt pi = radius; r^2*pi = 16 area!

I am not quite following you on this. All your PE is doing is restating the equation for the area of a circle in different ways. Nothing is different. Pi is Pi is Pi is Pi...
 
  • #7
If the PE is giving you the total surface area of the disc, this means you'll need to also know its thickness, - the total surface area comprising of two circular surfaces, and a 'strip' to go around the periphery.
 
  • #8
brewnog said:
If the PE is giving you the total surface area of the disc, this means you'll need to also know its thickness, - the total surface area comprising of two circular surfaces, and a 'strip' to go around the periphery.
It's a hypothetical question. the thickness and other parameters are immaterial. Only the radius to the circular plane is required.
 
  • #9
Robust.

Were you given the cross sectional area of the disc, in which case you're just rearranging the formula for the area of a circle, or were you given the total surface area of the disc.

If it's the latter, then the thickness is imperative, because you are considering two equal circular surfaces, and the 'strip' around the circumference. If the former is the case, your problem is pretty trivial, and Fred gave you the answer (in fact, you worked it out yourself!).
 
  • #10
Never before has so much discussion devolved from the simple formula, A = pi*r^2...;)
 
  • #11
Robust said:
the problem as originally given did state the total surface area

I'm not nitpicking here am I?
 
  • #12
brewnog said:
I'm not nitpicking here am I?
Knitpicking is fine with me - probably the more the better!
 
  • #13
pack_rat2 said:
Never before has so much discussion devolved from the simple formula, A = pi*r^2...;)
Not so simple considering the absurdity of the area to a closed continuum described by other than a whole number or ending decimal.
 
  • #14
No mentor/ just a comon sense approach

I guess with questions concerning math problems, I go back to the basic formula. A=pi*r^2. Then simply factor it, to put the term you are looking for on the correct side of the = sign. r=sqrt(A/pi). A simple and basic approach to solving problems is always the best way to solve the problem. Any time a person tries to get out of their comfort zone, problems usually arise (wrong solution.) The best way to expand your comfort zone is to study study study. And, read forums like this to keep your mind open to the possibility that there may be more than approach to solving a problem. " Critical thinking starts with you."
 

Related to Radius of Circular Disc for Mech. Engr Problem

1. What is the formula for calculating the radius of a circular disc for mechanical engineering problems?

The formula for calculating the radius of a circular disc is r = √(I/(πρt)), where r is the radius, I is the moment of inertia, ρ is the density, and t is the thickness of the disc.

2. How does the radius of a circular disc affect its strength and stability?

The radius of a circular disc directly affects its strength and stability. A larger radius means a larger moment of inertia, making the disc more resistant to bending and buckling. A smaller radius may result in a weaker and less stable disc.

3. Can the radius of a circular disc be too large or too small for a mechanical engineering problem?

Yes, the radius of a circular disc can be too large or too small for a mechanical engineering problem. It is important to consider the specific requirements and constraints of the problem to determine the appropriate radius for the disc.

4. How does material selection impact the radius of a circular disc for mechanical engineering problems?

Material selection can impact the radius of a circular disc in terms of its strength and stability. Different materials have different densities and strengths, which can affect the moment of inertia and therefore, the required radius for a disc to withstand the forces and loads in a mechanical engineering problem.

5. Are there any other factors to consider when determining the radius of a circular disc for mechanical engineering problems?

Other factors to consider when determining the radius of a circular disc include the type and placement of loads and forces, as well as any external factors such as temperature and vibration. These factors can affect the stress and strain on the disc and may impact the required radius for optimal performance.

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