A cylindrical rod is 50in in length and 4in in diameter and has a varying density. Its density function is given by f(x) = 40 - 0.01x^2, in which the density decreases as you move out along the rod. The end of the rod where the density if 40 is considered to be where x = 0. At what value along the rod will the center of mass of this rod be located?
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the center of mass will be given by the weighted average of the density by the position, so it will be given by the following integral:
int (0 to 50) [ x f(x) dx ] / int (0 to 50) [ f(x) dx ] =
int (x(40-0.01x^2)dx) = 20x^2 - 0.01/4x^4
int (f(x)dx) = 40x - 0.01/3x^3
- evaluating between 0 and 50 gives:
20*50^2 - 0.01/4*50^4 = 34,375
40*50 - 0.01/3*50^3 = 1,583
so the center of mass is at: 21.7in, and 2in above the ground (due to the diameter of the cylinder)
(note that the center of mass is a bit smaller than 25in which would be the expectation if the density was constant along the rod)
int (0 to 50) [ x f(x) dx ] / int (0 to 50) [ f(x) dx ] =
int (x(40-0.01x^2)dx) = 20x^2 - 0.01/4x^4
int (f(x)dx) = 40x - 0.01/3x^3
- evaluating between 0 and 50 gives:
20*50^2 - 0.01/4*50^4 = 34,375
40*50 - 0.01/3*50^3 = 1,583
so the center of mass is at: 21.7in, and 2in above the ground (due to the diameter of the cylinder)
(note that the center of mass is a bit smaller than 25in which would be the expectation if the density was constant along the rod)
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Physics??? What kind of incompetent would give a physics equation without units?
Density is given as units of length² !?!?
Total Crαp
And BTW thanks for telling me that f(0) = 40 - 0 = 40. The math there was sooo confusing!!!!
Finally WTF are the units for X ?!?!!? Sloppy sh!t. Epic fail. Tell your teacher s/he (or the book?) is an idiot.
We may assume that the units for x in the equation are inches, otherwise we can't solve the problem.
(Actually, that's not true. We just need to assume some unit of length for x to solve the problem. Centimeters or light-years would do as well, although the densities would not be realistic; feet would do fine.)
For an object whose mass is a function of length (1 dimensional), the center of mass will be the point, C, at which the mass between 0 and C is half of the total.
BTW "density if 40" ??? have you ever heard of checking your work?
A thin disk will have mass proportional to density
So we know at point x density = D = f(x) = 40-0.01x²
So we know ∫Ddx is proportional to the mass (from 0 to x) or mass = k ∫Ddx {o to x}
If you are unable to integrate (take the antiderivative) of this function, then I weep for you.
Hint ½ = (mass {0 to C}) ÷ (mass total)
Density is given as units of length² !?!?
Total Crαp
And BTW thanks for telling me that f(0) = 40 - 0 = 40. The math there was sooo confusing!!!!
Finally WTF are the units for X ?!?!!? Sloppy sh!t. Epic fail. Tell your teacher s/he (or the book?) is an idiot.
We may assume that the units for x in the equation are inches, otherwise we can't solve the problem.
(Actually, that's not true. We just need to assume some unit of length for x to solve the problem. Centimeters or light-years would do as well, although the densities would not be realistic; feet would do fine.)
For an object whose mass is a function of length (1 dimensional), the center of mass will be the point, C, at which the mass between 0 and C is half of the total.
BTW "density if 40" ??? have you ever heard of checking your work?
A thin disk will have mass proportional to density
So we know at point x density = D = f(x) = 40-0.01x²
So we know ∫Ddx is proportional to the mass (from 0 to x) or mass = k ∫Ddx {o to x}
If you are unable to integrate (take the antiderivative) of this function, then I weep for you.
Hint ½ = (mass {0 to C}) ÷ (mass total)
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Density=Mass/Volume. Density*Volume=Mass
Volume=cross sectional area*length=pi*2^2*50=200pi.
Density=40-0.01x^2.
Mass=200pi(40-0.01x^2)
Mass=8000pi-2x^2.
Volume=cross sectional area*length=pi*2^2*50=200pi.
Density=40-0.01x^2.
Mass=200pi(40-0.01x^2)
Mass=8000pi-2x^2.