Calculating the density of a sphere (uncertainty included)

[AryRezvani]

Homework Statement

The radius of a solid sphere is measured to be (6.30 ± 0.26) cm, and its mass is measured to be (1.81 ± 0.08) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures. Use the following formula to calculate the uncertainty in the density:
Δρ/ρ = Δm/m + 3Δr/r

Homework Equations

Δρ/ρ = Δm/m + 3Δr/r

D = M/V

The Attempt at a Solution

I'm going to assume you use the radius and calculate the volume of a sphere (4/3pi(r^3), and then convert to m^3.

Use the above information to calculate the density, but how does uncertainty come into play here? Do I plug it into the above equations along with the numbers, or do I do it all at the end with one formula?

I'm a newbie at Physics, really trying to improve myself. Don't laugh guys :(

 

[SammyS]

Homework Statement

The radius of a solid sphere is measured to be (6.30 ± 0.26) cm, and its mass is measured to be (1.81 ± 0.08) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures. Use the following formula to calculate the uncertainty in the density:
Δρ/ρ = Δm/m + 3Δr/r

Homework Equations

Δρ/ρ = Δm/m + 3Δr/r

D = M/V

The Attempt at a Solution

I'm going to assume you use the radius and calculate the volume of a sphere (4/3pi(r^3), and then convert to m^3.

Use the above information to calculate the density, but how does uncertainty come into play here? Do I plug it into the above equations along with the numbers, or do I do it all at the end with one formula?

I'm a newbie at Physics, really trying to improve myself. Don't laugh guys :(

Well, give us some results, & we'll comment.

 

[AryRezvani]

Well, give us some results, & we'll comment.

Alrighty, so I calculated the volume of the sphere and decided to disregard uncertainty until the end of the problem since it has its own formula.

(4/3)*pi*(6.30^3) = 1047.394424 cm^3

Let's just say the results is A

Took A and used dimensional analysis to convert it to m^3

A cm^3 x 1 m^3/1 000 000 cm ^3 = .0010473944

Took the above result and divided it into the mass (1.81), and got 1728 kg/m^3

D = 1728 kg/m^3

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A little stumped on the uncertainty formula. What does the P with the curve mean?

And what exactly does it mean by change in P, ect.

 

[SammyS]

Alrighty, so I calculated the volume of the sphere and decided to disregard uncertainty until the end of the problem since it has its own formula.

(4/3)*pi*(6.30^3) = 1047.394424 cm^3

Let's just say the results is A

Took A and used dimensional analysis to convert it to m^3

A cm^3 x 1 m^3/1 000 000 cm ^3 = .0010473944

Took the above result and divided it into the mass (1.81), and got 1728 kg/m^3

D = 1728 kg/m^3

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A little stumped on the uncertainty formula. What does the P with the curve mean?

And what exactly does it mean by change in P, etc.

Sorry for the tardy reply.

That's not letter, p. That's Greek letter, ρ (rho) .

ρ in this case is used to represent the density.

So, ρ = m/V. You used the letter, D, for density in your formula, D = M/V.

 

[Nickie]

I would approach this problem by first acknowledging that there will always be some level of uncertainty in any measurement or calculation we make. This is due to the limitations of our instruments and the inherent variability in physical quantities.

In this case, the uncertainty in the radius and mass measurements will affect the accuracy of our calculated density. To account for this uncertainty, we can use the formula Δρ/ρ = Δm/m + 3Δr/r, which takes into account the relative uncertainties in both the mass and radius measurements.

To calculate the density of the sphere, we can use the formula D = M/V, where M is the mass of the sphere and V is the volume calculated from the measured radius. We can then plug in the measured values for mass and radius, along with their respective uncertainties, into this formula.

For example, using the given values, we can calculate the volume of the sphere as (4/3)(π)(6.30cm)^3 = 1.51 x 10^-3 m^3. The uncertainty in this volume would be (4/3)(π)(6.56cm)^3 - (4/3)(π)(6.04cm)^3 = 0.00072 m^3.

We can then plug in the values for mass (1.81 ± 0.08) kg and volume (1.51 x 10^-3 ± 0.00072) m^3 into the formula D = M/V to calculate the density. This gives us a density of 1197 ± 92 kg/m^3.

To determine the uncertainty in the density, we can use the formula Δρ/ρ = Δm/m + 3Δr/r, where Δm/m is the relative uncertainty in mass and Δr/r is the relative uncertainty in radius. This gives us Δρ/ρ = (0.08/1.81) + 3(0.26/6.30) = 0.045 + 0.123 = 0.168. Multiplying this by the calculated density of 1197 kg/m^3 gives us an uncertainty of approximately ± 202 kg/m^3.

In summary, to calculate the density of the sphere and its uncertainty, we need to use the formula D = M/V and take into account the relative uncertainties in both the mass and radius measurements. This will give us a more accurate and