Maximizing the weight of a cylinder cut from a sphere

In summary: Re: max and min 4Now we need to put that respect to...The maximum volume of the cylinder occurs when $h=2R$.
  • #1
leprofece
241
0
A sphere weighs P kg what is the weight of the higher straight circular cylinder that can cut from the sphere?

Answer sqrt(3)P/3
 
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  • #2
Re: max and min 4

Assuming the mass density of the sphere is constant throughout, how are mass and volume related?

I suggest drawing a diagram of a cross section of the two objects through the axis of symmetry of the cylinder and try to find a relationship between the radius and height of the cylinder with the radius of the sphere. What do you find?
 
  • #3
Re: max and min 4

MarkFL said:
Assuming the mass density of the sphere is constant throughout, how are mass and volume related?

I suggest drawing a diagram of a cross section of the two objects through the axis of symmetry of the cylinder and try to find a relationship between the radius and height of the cylinder with the radius of the sphere. What do you find?

I can do that by pitagorean theoreme
 
  • #4
Re: max and min 4

leprofece said:
I can do that by pitagorean theoreme

Yes, that's correct. What is the relationship between mass, volume and mass density?
 
  • #5
Re: max and min 4

P or mass = d.v
 
  • #6
Re: max and min 4

leprofece said:
P or mass = d.v

Yes, and if the density is constant, then we know mass and density are proportional to one another, and so we need only compute the ratio of the volume of the cylinder to that of the sphere, and multiply this by the mass of the sphere to get the mass of the cylinder.

So, what is the relationship between the dimensions (radius and height) of the cylinder with the radius of the sphere?

What is your objective function?
 
  • #7
Re: max and min 4

MarkFL said:
Yes, and if the density is constant, then we know mass and density are proportional to one another, and so we need only compute the ratio of the volume of the cylinder to that of the sphere, and multiply this by the mass of the sphere to get the mass of the cylinder.

So, what is the relationship between the dimensions (radius and height) of the cylinder with the radius of the sphere?

What is your objective function?

m= d.v
and sphre volume is 4pir3/3
Then m = d(4pir3/3)
r2= R2+ H2
i tried to derive v respect to r
m= d (4pi(R2+ H2)3
looking at the answer i don't know how to eliminate r and h to get (sqrt of 3)m/3 that is my book answer
 
  • #8
Re: max and min 4

Look at a cross-section of the sphere and cylinder through their mutual center. You should find that your relationship (using the Pythagorean theorem) between the radius $R$ of the sphere and the radius $r$ and $h$ of the cylinder is incorrect. You have one side of the right triangle wrong.
 
  • #9
Re: max and min 4

ohhh I can't find this relationship really I give up sorry
I would appreciate your helping
 
  • #10
Re: max and min 4

Here is a diagram...do you see from where I obtained the measures of the 3 sides of the right triangle?

View attachment 1960
 

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  • #11
Re: max and min 4

MarkFL said:
Here is a diagram...do you see from where I obtained the measures of the 3 sides of the right triangle?

View attachment 1960
lets see
m= d.v
and sphre volume is 4pir3/3
Then m = d(4pir3/3)
r2= R2- (H/2)2
i tried to derive v respect to r
m= d (4pi(R2- (H/2)2])3/3
now can i derive from here??
 
Last edited:
  • #12
Re: max and min 4

I would just maximize the volume of the cylinder first, and then use the relationship between weight density and weight.

\(\displaystyle w_C\) = weight of the cylinder

\(\displaystyle w_S=P\) = weight of the sphere

Thus, we may state, given that both objects share the same weight density:

\(\displaystyle \rho=\frac{P}{V_S}=\frac{w_C}{V_C}\)

Hence:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)

You already know the volume of the sphere, so all you need now is the volume of the cylinder. So, our objective function is the volume of the cylinder:

\(\displaystyle V_C=\pi r^2h\)

Subject to the constraint:

\(\displaystyle r^2+\left(\frac{h}{2} \right)^2=R^2\)

So, using the constraint, we may write the volume of the cylinder in terms of 1 variable, and it will be simpler to substitute for $r^2$:

\(\displaystyle V_C(h)=\pi\left(R^2-\left(\frac{h}{2} \right)^2 \right)h=\pi R^2h-\frac{\pi}{4}h^3\)

Now, differentiate this with respect to $h$ and equate the result to zero to determine the critical value(s).
 
  • #13
Re: max and min 4

MarkFL said:
I would just maximize the volume of the cylinder first, and then use the relationship between weight density and weight.

\(\displaystyle w_C\) = weight of the cylinder

\(\displaystyle w_S=P\) = weight of the sphere

Thus, we may state, given that both objects share the same weight density:

\(\displaystyle \rho=\frac{P}{V_S}=\frac{w_C}{V_C}\)

Hence:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)

You already know the volume of the sphere, so all you need now is the volume of the cylinder. So, our objective function is the volume of the cylinder:

\(\displaystyle V_C=\pi r^2h\)

Subject to the constraint:

\(\displaystyle r^2+\left(\frac{h}{2} \right)^2=R^2\)

So, using the constraint, we may write the volume of the cylinder in terms of 1 variable, and it will be simpler to substitute for $r^2$:

\(\displaystyle V_C(h)=\pi\left(R^2-\left(\frac{h}{2} \right)^2 \right)h=\pi R^2h-\frac{\pi}{4}h^3\)

Now, differentiate this with respect to $h$ and equate the result to zero to determine the critical value(s).

ok I got +/- 2Rsqrt3/3 as critical point
Now we need to put that respect to P:confused:
 
  • #14
Re: max and min 4

leprofece said:
ok I got +/- 2Rsqrt3/3 as critical point
Now we need to put that respect to P:confused:

First, I would demonstrate that this critical value is at a relative maximum. My preference here would be the second derivative test.

Then, you want to evaluate $V_C$ at this critical value, and then substitute that into the formula:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)
 
  • #15
Re: max and min 4

MarkFL said:
First, I would demonstrate that this critical value is at a relative maximum. My preference here would be the second derivative test.

Then, you want to evaluate $V_C$ at this critical value, and then substitute that into the formula:

\(\displaystyle w_C=\frac{V_C}{V_S}P\)

and how does Vs remains??
must I substitute this value too there??
 
  • #16
Re: max and min 4

leprofece said:
and how does Vs remains??
must I substitute this value too there??

$V_S$ is the volume of the sphere, so you want to substitute its formula there.
 
  • #17
Re: max and min 4

MarkFL said:
$V_S$ is the volume of the sphere, so you want to substitute its formula there.

OK I GOT THE ANSWER THANKS A LOT(Angel)
 

FAQ: Maximizing the weight of a cylinder cut from a sphere

What is the formula for calculating the maximum weight of a cylinder cut from a sphere?

The formula for calculating the maximum weight of a cylinder cut from a sphere is:
W = π/3 * ρ * h * (3r² + h²)
where W is the weight, ρ is the density of the material, h is the height of the cylinder, and r is the radius of the sphere.

How do I determine the optimal height and radius for the cylinder to maximize its weight?

To determine the optimal height and radius for the cylinder, you can use the following steps:
1. Calculate the volume of the cylinder using the formula V = π * r² * h
2. Use the formula for calculating the maximum weight to create an equation with one variable (either h or r)
3. Take the derivative of the equation and set it equal to 0
4. Solve for the variable to find the optimal height and radius for the cylinder.

What materials are best for maximizing the weight of a cylinder cut from a sphere?

The density of the material will play a significant role in maximizing the weight of the cylinder. Materials with higher density, such as metals, will generally result in a higher weight when cut from a sphere. However, the strength and structural integrity of the material should also be considered when choosing the best material for this purpose.

Can the weight of the cylinder be further increased by altering its shape?

Yes, the weight of the cylinder can be further increased by altering its shape. For example, a cylinder with a tapered or conical shape will have a greater weight than a cylinder with a constant radius. However, this will also depend on the density and structural integrity of the material being used.

Are there any practical applications for maximizing the weight of a cylinder cut from a sphere?

Yes, there are practical applications for maximizing the weight of a cylinder cut from a sphere. For example, in the field of construction, this concept can be applied to design stronger and more efficient structures. It can also be useful in other industries, such as transportation or manufacturing, where maximizing weight can lead to improved performance or cost savings.

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