Force of Gravity on the tip of a giant snowcone

[m00npirate]

Homework Statement

I'm studying for a physics exam and made up the following problem to practice. Since I made it up there is no way to check my answer and I would greatly appreciate any and all feedback.

Find the force of gravity exerted on a point mass m at the tip of a snowcone (portion of a ball cut out by a cone where the radius of the sphere = the length of the side of the cone) with constant density ϱ and radius R. The internal angle of the cone is pi/4 in both directions.

Homework Equations

F = -GmM(r)/r^2

The Attempt at a Solution

I used spherical coordinates
M(r)=ϱV(r)

M(r)=$$\frac{-\rho}{3}r^{3}\theta\cos\phi$$

$$dM(r)=r^{2}\sin\phi dr d\phi d\theta$$

$$F=\int_0^{\pi/4}\int_0^{\pi/4}\int_0^R \frac{-Gm\rho r^{2}}{r^{2}} \sin\phi dr d\phi d\theta$$

$$F=-\frac{\pi}{4} Gm\rho\int_0^{\pi/4} \sin\phi d\phi \int_0^R dr$$

$$F=-\frac{\pi}{4} Gm\rho R(\cos0-\cos\frac{\pi}{4})$$

$$F=-\frac{\pi}{8} (2-\sqrt{2})Gm\rho R$$

Also, I can't think of any way to find F(d) where d is the distance from the point of the cone. It might just be wishful thinking though.
Thanks in advance!

 

[LowlyPion]

I might consider solving by superposition.

The sum of the conical solid and the spherical one.

For the cone you can sum the individual disks of dm = ϱ*a(x)dx and then add the spherical section calculated from the distance of the cone height.

I guess I don't see the need to use the spherical coordinates.

 

[nlightner]

First of all, great job on attempting to solve this problem using spherical coordinates. However, there are a few things that need to be addressed in your solution.

Firstly, the equation you used for the force of gravity, F = -GmM(r)/r^2, is actually the equation for the gravitational force between two point masses. In this problem, we are only dealing with a single point mass (m) at the tip of the snowcone, so this equation is not applicable.

Secondly, the equation for the mass of a cone, M(r) = (1/3)πr^2hρ, assumes that the cone has a circular base. In this problem, the base of the cone is not circular, but rather a portion of a sphere. Therefore, this equation cannot be used.

Instead, we can approach this problem by considering the snowcone as a series of thin concentric rings, each with a different radius and height. The mass of each ring can be approximated as dm = ρ2πrdr, where r is the radius of the ring and dr is the thickness of the ring. The force of gravity on each ring can then be calculated using the equation F = -Gm(dm)/r^2.

To find the total force of gravity on the snowcone, we can integrate the force over all the rings, from the base of the cone (r = 0) to the tip of the cone (r = R). This will give us the following integral:

F = -Gm∫(0 to R) ρ2πrdr/r^2

Simplifying this integral, we get:

F = -2πGmρ∫(0 to R) dr

F = -2πGmρR

This is the force of gravity on the entire snowcone, but we are interested in the force at the tip of the cone, where r = R. Therefore, we can simply substitute R into the equation and get the final answer:

F = -2πGmρR

This is the force of gravity on the tip of the snowcone. It is important to note that this is an approximation, as we have assumed the snowcone to be a series of thin concentric rings. However, it should be a good approximation for the given problem.

To find the force at a distance d from the tip of the cone, we can use the