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Question

The energy W stored in a flywheel is given by

W = kr^5N^2 (^) = the power of.

Where k is a constant
R is the radius
N is the number of revs

Determine the approximate percentage change in W when r is increased by 1.3% & N is decreased by 2%.

The answer states in the book 2.5% increase.

All of them are very good answers, its difficult to pick the correct one.

Answer 1

W = kr^5N^2 ----(1)
r increase by 1.3%
and
N decrease by 2 %

Let say the rate of change in W = x %

(1+x/100)W = k(1.013r)^5(0.98N)^2 ---(2)
Divide (2) by (1)

1+x/100 = (1.013)^5(0.98)^2
1+x/100 = 1.02447
x =2.447(approximate to 2.5)
so W increase by 2.5%

Answer 2

the new W will be given by W = k(r*1.013)^5(N*0.98)^2

the new W divided by the old W is given by

[k(r*1.013)^5(N*0.98)^2]
divided by
[kr^5N^2]
equals
(1.013)^5(0.98)^2/(1)

I throw this into a calculator and get approximately 1.024470313

so this number minus 1 is 0.024470313, or 2.4470313%

which should be rounded to 2.4, so the answer is actually 2.4.

They rounded up for some reason. Books make mistakes too I guess.

Answer 3

Set this up as a ratio W1 / W or
(k * (1.013r)^5 * (.98N)^2) / ( k * r^5 * n^2) =
(k * 1.0667 * r^5 * .96 * N^2) / (k * r^5 * n^2) / =
divide by k ,N^5, r^2 = (1.0667*.96)/1 = 1.02447 or approximately a 2.5% increase

Answer 4

% change in W = [W(new) - W(old) ] / W(old)
= W(new)/W(old) - 1

W(old) = k*R^5*N^2
W(new) = k*(1.013*R)^5*(.98*N)^2
1.3% increase in R = R + .013*R = (1.013)*R
2.0% decrease in N = N - .02*N = (.98)*N

W(new)/W(old) - 1 =
= 1*(1.013)^5*(.98)^2 - 1
(since the variables cancel out)
= 1.02447 - 1 = .02447 = 2.447%