Thread: center of mass of a rod

Find the total mass of the given rod and the center of mass:

The length of a rod is 12cm and the measure of the linear density at a point is a linear function of the measure of the distance from the left end of the rod. THe linear density at the left end is 3g/cm and at the right end is 4 g/cm.

Edit: this is okay now

Answer:
mass:42 g
center of mass: 44/7 cm from the left end

Originally Posted by ^_^Engineer_Adam^_^

Find the total mass of the given rod and the center of mass:

The length of a rod is 12cm and the measure of the linear density at a point is a linear function of the measure of the distance from the left end of the rod. THe linear density at the left end is 3g/cm and at the right end is 4 g/cm.

Edit: this is okay now

Answer:
mass:42 g
center of mass: 44/7 cm from the left end

The linear density at any point, x, along the rod is: L = (4-3)/(12-0)x + 3 = 1/12x + 3

Mass = INT L dx = INT{0,12} 1/12x + 3 dx = [1/24x^2 + 3x] from {0,12} = 6 + 36 = 42

I'll have to think about the center of mass for a moment. I don't yet remember how to find this.

Originally Posted by ^_^Engineer_Adam^_^

Find the total mass of the given rod and the center of mass:

The length of a rod is 12cm and the measure of the linear density at a point is a linear function of the measure of the distance from the left end of the rod. THe linear density at the left end is 3g/cm and at the right end is 4 g/cm.

Edit: this is okay now

Answer:
mass:42 g
center of mass: 44/7 cm from the left end

Call the linear density D. Then we know that
D(x) = (4 g/cm - 3 g/cm)/(12 cm)*x + 3 g/cm = (1/12)x + 3
where x is the measure of the distance from the left end of the rod. (I've dropped the units for convenience. We'll have to remember D is in g/cm and x is in cm.)

The mass M of the rod will be:
M = Int[D(x), x, 0, 12] = [(1/12)*(1/2)*12^2 + 3*12] - [(1/12)*(1/2)*0^2 + 3*0]) = 42 g

The CM of the rod will be at:
x(CM) = (1/M)*Int[x, dm, 0 , M]

Now
M = (1/24)L^2 + 3L
so
dm = [(1/12)x + 3]dx

Thus:
x(CM) = (1/M)*Int[x, dm, 0 , M] = (1/M)*Int[x*[(1/12)x + 3], dx, 0, L]

= (1/42)*Int[(1/12)*x^2 + 3x, dx, 0, 12]

= (1/42)*[(1/12)*(1/3)*12^3 + 3*(1/2)*12^2]

= 44/7 cm

So your answers check out.

-Dan