A rod extending between x=0 and x= 14.0cm has a uniform cross-
sectional area A= 9.00cm^2. It is made from a continuously changing
alloy of metals so that along it's length it's density changes
steadily from 2.70g/cm^3 to 19.3g/cm^3.
a) Identify the constants B and C required in the expression p= B +
Cx to describe the variable density.
b) The mass of the rod is given by
m= integral(pdV)= integral(pAdx)= integral 14cm/0 (14cm above integral sign and 0 below integral)(B+Cx)*(9.00cm^2)dx
(below the firs integral it says all material and below the
second integral it says all x
~this part a has gotten me confused as to what to do at all and the second part has gotten me even more confused!!
HELP!!!
Thank YOU!
Yes, p represents density in this case. The mass of the rod is
A*(Integral of) p (x) dx
where A is the cross sectional area of the rod.
For p(x), use the function you derive in part (a).
p(x) = A + bx
p(0) = A = 2.70 g/cm^3 (That already tells you what A is)
p(14) = A + B*14 = 19.3 g/cm^3
Solve for B.
first of all I don't know how to use the #s given to put into the equation or do something else with them.
I know...
x=0
x=14.0cm
A=9.00cm^2
density changes from: 2.70g/cm^3 to 19.3g/cm^3
How do I put it into p= B+Cx??
would x= 14-0??
C and B I'm not sure what they represent except that you said the mass is equal to B right?
I forgot to include what it says under the integration:
Carry out the integration to find the mass of the rod
Now about that..what in the equation for the integral is the mass???
it's not like they have a m anywhere in there..unless it's p but I started thinking it was to represent density..
and since p=m/V does that play a role in this as well?
Again I have no idea what C represents.
I forgot t use the expression B + Cx for p, and used A + Bx instead. The method of solution is the same. In your case, B = 2.70, and C = 1.186
p(x) = 2.70 + 1.186 x
Note that p = 2.70 g/cm^3 when x = 0 and 9.30 g/cm^3 when p = 14 cm, as required.
Now integrate A p(x) dx for the total mass
for this I got
m= A integral 14\0 (B+Cx)dx
(took out the A contstant A=9.00cm^2)
m= 9.00 integral 14\0 (2.70+1.186x)dx =
9.00(2.70x + 0.593x^2)|14 0 =
(the 14 is above and 0 below the | sign)
9.00[ 2.70(14) + 0.593(14)^2 ]=
1,386.252g
~I'm not sure if I did the integration right and also I'm confused as to why you said
"note that p= 2.70g/cm^3 when x=0 and 9.30g/cm^3 when p= 14cm, as required" does this apply to what I plug into the integration ?? I would think it wouldn't since I have to put in 14 and 0 in anyways and the 0 cancels out and all is left is the 14 right?
Is this alright what I got??
To identify the constants B and C required in the expression p = B + Cx to describe the variable density, we can use the given information about the density change along the rod's length.
Given:
Density at x = 0, p1 = 2.70 g/cm^3
Density at x = 14.0 cm, p2 = 19.3 g/cm^3
Since we have two data points, we can set up a system of equations using these data:
At x = 0: p1 = B + C * 0
=> p1 = B
At x = 14.0 cm: p2 = B + C * 14.0
=> p2 = B + 14C
We can solve these equations simultaneously to find the values of B and C.
Now, let's move on to part b of the question, which involves calculating the mass of the rod.
The mass of the rod can be obtained using the integral of density (p) multiplied by the differential volume (dV) or integrated with respect to x.
We are given the expression for mass: m = ∫pdV
Since the rod has a uniform cross-sectional area A throughout, we can express mass in terms of density and length:
m = ∫pAdx
The limits of the integral are stated as "all material" and "all x." This means we need to integrate with respect to x over the entire length of the rod from x = 0 to x = 14.0 cm.
So, the integral expression for the mass becomes:
m = ∫(B + Cx)(A)dx from 0 to 14.0 cm
Now, we can calculate the mass by plugging in the values of A, B, C, and integrating with respect to x. The integral of (B + Cx)dx will give us the expression for mass.