Standard kilogram physics help

In summary, the problem is to show that making the height of a cylinder equal to its radius minimizes the surface area (for a given volume). This can be done by writing the formula for the surface area and volume, differentiating with respect to the radius, and setting it equal to 0. This results in two solutions, but the first solution is the correct one for a solid cylinder.
  • #1
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The standard kilogram is in the shape of a circular cylinder with its height equal to its dimager. Show that, for a circular cylinder of fixed volume, this equality gives the smallest surface area, thus minimizing the effects of surface contamination and wear.

Now some of you might have seen this problem in Fundamentals of Physics Extended Fifth Edition. Please give me some hints.
 
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  • #2
The problem is to show that making the height of a cylinder equal to its radius minimizes the surface area (for a given volume). The standard way of doing such a problem is to write the formula for the quantity, then take the derivative.

A cylinder of radius r and height h has three surfaces: the top and bottom, each of area [pi]r2, and the lateral surface which has area 2[pi]rh ("uncurl" the lateral surface and it is a rectangle with area the length of the circumferences time the height). The total area is A= 2[pi]r2+ 2[pi]rh= 2[pi](r2+ rh). The Volume is the area of the base time the height:
V= [pi]r2h so h= V/([pi]r2) and we can write
the surface area as A= 2[pi](r2+ V/([pi]r)) (V is a constant).

Differentiate with respect to r, set equal to 0 and solve for r.
You will get a formula for r that depends on V. Substitute for V with the formula above and see what happens.
 
  • #3
Yes, I obtained to results from two different surface areas.

First solution:
A minimum area is obtained when h = r for A = [pi] r^2 + 2[pi] rh.

Second solution:
A minimum surface area is obtained when h = 2r.

Can anyone please explain the result to me?
 
  • #4
Second solution: when A = 2[pi] r^2 + 2[pi] rh
 
  • #5
A= 2[pi]r2+ 2[pi]rh is the surface area of a cylinder:
the "lateral surface" area is 2[pi]rh and each end has area [pi]r2.

A= [pi]r2+ 2[pi]rh would be the area of the lateral surface and ONE end. If you were doing a problem concerning a "can" open at one end you would use that but "the standard kilogram in the shape of a circular cylinder" is a solid cylinder so there is no reason to use that.
 

1. What is a standard kilogram in physics?

A standard kilogram in physics is a unit of mass that is defined as the mass of a specific object called the International Prototype Kilogram (IPK). It is used as the basis for measuring mass in the International System of Units (SI).

2. How is the standard kilogram defined?

The standard kilogram is defined as the mass of the IPK, which is a platinum-iridium cylinder that is kept at the International Bureau of Weights and Measures (BIPM) in France. It was established in 1889 and is the only physical object that is used to define a unit of measurement.

3. Why is the standard kilogram important in physics?

The standard kilogram is important in physics because it serves as the basis for measuring mass in the SI system, which is used in scientific research and international trade. It allows for consistent and accurate measurements of mass, which is a fundamental property in many physical laws and equations.

4. How accurate is the standard kilogram?

The IPK is considered to be extremely accurate and has been used as the standard for over a century. However, there have been concerns about the stability of its mass over time. As a result, efforts are being made to redefine the kilogram in terms of fundamental constants of nature, rather than a physical object.

5. Can the standard kilogram change?

Technically, the standard kilogram can change if the mass of the IPK were to change. However, it is highly unlikely as the cylinder is stored under controlled conditions and is rarely handled. Efforts are being made to redefine the kilogram in a more stable and universal way to eliminate this possibility.

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