Question

Part B: Setting up integrals

A circular ring is made of a material with uniform density p (kg/m²).
The ring has rectangular cross section, as shown in the figure to the
right. The outer radius of the ring is R.
We would like to calculate the total mass of the ring.
First we will use the approximation that the width of the ring, a, is very much smaller than
the outer radius of the ring R(a <<R).
a) On the figure to the right, label a small "chunk"
of the ring that subtends an arc de.
b) What is the mass of this chunk (dm) in terms of
p. a, b, R, and de?
c) Express the mass of the entire ring as a single integral in the angle 0. (Don't forget limits)
d) What is the total mass of this ring?​

Answer 1

The mass of a thin circular ring with uniform density can be calculated by setting up an integral that accounts for the mass of a small 'chunk' of the ring. This is then integrated over the entire angle subtended by the ring, using the outer radius and the width of the ring as parameters.

Step-by-step explanation:

To calculate the mass of a thin circular ring with a uniform density ρ (kg/m²), where the width of the ring (a) is much smaller than its radius (R), we can follow these steps:

• Identify a small ‘chunk’ of the ring that could be approximated by a rectangular section with length equal to the arc length of an angle dθ and width a.
• The mass dm of this small chunk can be described as dm = ρ × segment area, where the segment area can be approximated by (2πR) × a × dθ.
• To find the mass of the entire ring, integrate this expression from 0 to 2π with respect to θ.
• The definite integral yields the total mass M of the ring, which is M = ρ × 2π² × a × R.

This assumes that each ring segment is of uniform density and that the radius of the ring is constant throughout the integration.

Answer 2

b) dm = ρab R dθ

c) m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) m = 2πρabR

Step-by-step explanation:

b) We want to find the mass dm of a small chunk of the ring.

Mass is density times volume:

dm = ρ dV

Since a << R, we can approximate the volume as a rolled rectangular prism. Therefore, the volume of the chunk is the area of the cross section times the arc length.

dV = ab R dθ

dm = ρab R dθ

c) The mass of the entire ring is the sum of the masses of all the chunks.

m = ∫ dm

m = ∫ ρab R dθ, from θ = 0 to θ = 2π

d) ρ, a, b, and R are constants, so:

m = ρabR ∫ dθ

Evaluating the integral:

m = ρabR (θ|0 to 2π)

m = ρabR (2π − 0)

m = 2πρabR