JPO 122 Assignment 1 – Mathematics of
Physics
Due date: 2nd August 2013
Name:
Student No:
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should I be found guilty. This work is my original effort.
Signature: __________________
Question 1
[6]
A rain gauge is canonical with dimensions as given in the figure alongside.
1
The volume of the rain collected is given by 𝑉 = 3 𝜋𝑟 2 ℎ.
a) Using similar triangles, eliminate r from this equation and show that
1
𝑅2
𝑉 = 3 𝜋 𝐻 2 ℎ3.
(2)
Triangles ADE and ABC are similar thus:
𝑟 𝑅
𝑅
= ⇒𝑟= ℎ
ℎ 𝐻
𝐻
1
𝑅2
Hence it is immediate by substitution that 𝑉 = 3 𝜋 𝐻2 ℎ3
b) Calculate the rate at which the height changes as the volume increases
(4)
1
3𝑉𝐻 2
3𝐻 2 3 1
ℎ= √
=
(
) 𝑉3
𝜋𝑅 2
𝜋𝑅 2
3
1
1
𝑑ℎ 1 3𝐻2 3 −2 1 3𝐻 2 3
= (
) 𝑉 3 = ( 2 2)
𝑑𝑉 3 𝜋𝑅 2
3 𝜋𝑉 𝑅
Question 2
[9]
A rectangular tank is open on the top side and is manufactured from sheet iron. The capacity of the
tank is 𝑉. The square base has sides of length 𝑎 and the height of the tank is ℎ
a) Give an expression for the height ℎ in terms of the volume 𝑉 of the tank
𝑉 = 𝑎2 ℎ ⇒ ℎ =
(2)
𝑉
𝑎2
b) If the volume of the tank is 𝑉 = 8 𝑚3 , show that the area of all the sheets is 𝐴 =
32
𝑎
+ 𝑎2
(3)
Each side has area 𝑎ℎ while the square base has area 𝑎2 thus total area is:
𝐴 = 𝑎2 + 4𝑎ℎ = 𝑎2 + 4𝑎
𝑉
32
= 𝑎2 +
2
𝑎
𝑎
c) Determine the length 𝑎 that would result in the smallest amount of sheet iron used.
We want to minimize the area of the sheets hence
0=
𝑑𝐴
32
3
= 2𝑎 − 2 ⇒ 𝑎3 − 16 = 0 ⇒ 𝑎 = √16 ≈ 2.52
𝑑𝑎
𝑎
(4)
Question 3
[5]
The sketch shows a thin straight uniform rod with length 𝐿 and mass 𝑀. The mass per unit length is
given by 𝜇 =
𝑀
𝐿
𝑘𝑔/𝑚. AB is an axis perpendicular to the rod PQ and 𝑑𝑥 is an infinitesimal element
of length at position 𝑥 from AB. The mass of this element is 𝑑𝑚 = 𝜇 𝑑𝑥 =
𝑀
𝑑𝑥.
𝐿
The quantity 𝑑𝐼 =
𝑥 2 𝑑𝑚 is called the moment of inertia of the element about the axis AB.
a) Calculate the moment of inertia 𝐼 of the entire rod
𝐿
𝐼 = ∫ 𝑑𝐼 = ∫ 𝑥 2 𝑑𝑚 = ∫ 𝑥 2
0
(2)
𝐿
𝑀
𝑀
1
𝑑𝑥 = [ 𝑥 3 ] = 𝑀𝐿2
𝐿
3𝐿
3
0
b) Calculate the moment of inertia of the rod about an axis parallel to AB and through the
centre of the rod PQ.
(3)
If we start with our origin in the middle of the rod then integral remains the same except our
boundary points change for [0,L] to [-L/2, L/2]
𝑀
𝑀 3 𝐿/2
1
1
1
𝐼 = ∫ 𝑑𝐼 = ∫ 𝑥 𝑑𝑚 = ∫
𝑥
𝑑𝑥 = [ 𝑥 ]
=
𝑀𝐿2 + 𝑀𝐿2 =
𝑀𝐿2
𝐿
3𝐿
24
24
12
−𝐿/2
−𝐿/2
𝐿/2
2
2