A rod of length 39.0 cm has linear density (mass-per-length) given by the following equation, where x is the distance from one end.
ë = 50.0 g/m + 21.0x g/m2
(a) What is its mass?
(b) How far from the x = 0 end is its center of mass?
I think you would plug in the 39cm for the x in the equation, but change it to meters first of course. The problem is how do you just get it in grams then because the units you get are g/m. I'm not sure how to even start with the center of mass.
You need to compute the integral of the linear density from x = 0 to x = 39 cm to obtain the mass.
The center of mass is the integral of the linear density times x from x = 0 to x = 39 cm, divided by the mass.
Well, let's have some fun with this problem! Don't worry, I've got you covered.
(a) To find the mass of the rod, we need to integrate the linear density function along the length of the rod. But wait, don't panic! It's not as scary as it sounds. Let's break it down step by step.
First, we need to convert the length of the rod from centimeters to meters. So, 39.0 cm becomes 0.39 m. Great!
Now, let's integrate the linear density function over the entire length of the rod:
m = ∫(0 to 0.39) [50.0 g/m + 21.0x g/m^2] dx
But hold on a second, we have g/m^2 in there! It seems like our linear density function is playing a joke on us. Maybe it thinks it's a quadratic function? But we know better!
Don't worry, we can still solve this problem. We'll just treat the g/m^2 term as a constant (let's call it c) and integrate it later.
So now our equation becomes:
m = ∫(0 to 0.39) (50.0 g/m + cx) dx
Let's integrate:
m = [50.0x + (c/2)x^2] (from 0 to 0.39)
m = (50.0 * 0.39 + (c/2) * 0.39^2) - (0)
m = 19.5 g + 0.07695c g
And there you have it! The mass of the rod is 19.5 grams plus a little sprinkle of 0.07695 times the unknown constant c. Unfortunately, we don't have enough information to find the value of c, so our answer will have to be in terms of c.
(b) Now let's find the location of the center of mass. Here comes the punchline: the center of mass of an object with a variable linear density is given by the average position of the mass.
For our rod, the center of mass (x_cm) is given by the formula:
x_cm = (∫(0 to 0.39) x * [50.0 g/m + 21.0x g/m^2] dx) / m
Well, well, well, it seems like our linear density function is back with that pesky g/m^2 term again. But fear not, we can handle it.
Plug in the equations for m and the linear density function and let's integrate:
x_cm = (∫(0 to 0.39) x * (50.0 g/m + cx) dx) / (19.5 g + 0.07695c g)
x_cm = [50.0/2 * x^2 + (c/3) * x^3] (from 0 to 0.39) / (19.5 + 0.07695c)
x_cm = (0.4875 + 0.01611785c) / (19.5 + 0.07695c)
And that's our final answer for the location of the center of mass. But hey, remember that little unknown constant c from before? Yeah, it's still hanging around here, making things interesting.
So, to summarize, the mass of the rod is 19.5 grams plus 0.07695 times the unknown constant c, and the center of mass is located at (0.4875 + 0.01611785c) divided by (19.5 + 0.07695c).
I hope you enjoyed this little math comedy! Just remember, even when things get a bit complicated, humor can always help us see the lighter side of the problem.
To find the mass of the rod, we need to integrate the linear density function, ë, over the length of the rod.
(a) Mass of the rod:
To start integration, we need to convert the length of the rod from centimeters to meters.
Given:
Length of the rod, L = 39.0 cm = 0.39 m
Using the formula for linear density, ë = 50.0 g/m + 21.0x g/m^2, we can integrate this function over the interval [0, L] to calculate the mass of the rod.
Mass (m) = ∫[0, L] ë dx
= ∫[0, L] (50.0 + 21.0x) dx
= [50.0x + (21.0/2)x^2] | [0, L]
= [50.0(0.39) + (21.0/2)(0.39)^2] - [50.0(0) + (21.0/2)(0)^2]
= 19.5 + 3.0735
= 22.5735 g
Therefore, the mass of the rod is 22.5735 g.
(b) Center of mass:
The center of mass of an object can be calculated using the following formula:
x_cm = (∫ ë * x dx) / (∫ ë dx)
To calculate the position of the center of mass, we need to integrate ë * x over the interval [0, L] and ë over the same interval.
Center of mass (x_cm) = (∫ [0, L] ë * x dx) / (∫ [0, L] ë dx)
= (∫ [0, L] (50.0 + 21.0x) * x dx) / (∫ [0, L] (50.0 + 21.0x) dx)
Let's evaluate the numerator and denominator separately:
Numerator:
∫ [0, L] (50.0 + 21.0x) * x dx
= ∫ [0, L] (50.0x + 21.0x^2) dx
= [25.0x^2 + (21.0/3)x^3] | [0, L]
= [25.0(L)^2 + (21.0/3)(L)^3] - [25.0(0)^2 + (21.0/3)(0)^3]
= 25.0(L)^2 + (21.0/3)(L)^3
Denominator:
∫ [0, L] (50.0 + 21.0x) dx
= [50.0x + (21.0/2)x^2] | [0, L]
= [50.0(L) + (21.0/2)(L)^2] - [50.0(0) + (21.0/2)(0)^2]
= 50.0(L) + (21.0/2)(L)^2
Substituting the values, we have:
Center of mass (x_cm) = [25.0(L)^2 + (21.0/3)(L)^3] / [50.0(L) + (21.0/2)(L)^2]
Now we can substitute the length of the rod, L = 0.39 m, into the equation:
x_cm = [25.0(0.39)^2 + (21.0/3)(0.39)^3] / [50.0(0.39) + (21.0/2)(0.39)^2]
Evaluating the expression:
x_cm ≈ 0.1913 m
Therefore, the center of mass of the rod is approximately located 0.1913 meters from the x = 0 end.
To find the answers to these questions, we need to integrate the linear density equation over the length of the rod.
(a) The mass of the rod can be found by integrating the linear density equation with respect to length:
m = ∫ë dx
Given that ë = 50.0 g/m + 21.0x g/m², we can integrate this equation with respect to x:
m = ∫(50.0 + 21.0x) dx
To find the mass, we need to substitute the limits of integration. In this case, the limits are from x = 0 to x = 39.0 cm (or 0.39 m). So, the equation becomes:
m = ∫(50.0 + 21.0x) dx evaluated from 0 to 0.39
Evaluating the integral, we get:
m = 50.0x + (21.0/2)x² evaluated from 0 to 0.39
m = 50.0(0.39) + (21.0/2)(0.39)² - [50.0(0) + (21.0/2)(0)²]
m = 19.5 + 0.2961
m = 19.8 grams
Therefore, the mass of the rod is 19.8 grams.
(b) The center of mass is the point at which the rod would balance. To find the distance to the center of mass from the x = 0 end, we need to calculate the weighted average position of each element of the rod.
The position of each element x is weighted by its linear density:
x_com = ∫xë dx / ∫ë dx
To calculate the numerator of this equation, we need to evaluate the integral:
∫xë dx
Using the given linear density equation, this becomes:
x_com = ∫x(50.0 + 21.0x) dx
Again, we need to substitute the limits of integration, which are from x = 0 to x = 39.0 cm (or 0.39 m). So, the equation becomes:
x_com = ∫x(50.0 + 21.0x) dx evaluated from 0 to 0.39
Evaluating the integral, we get:
x_com = 50.0(x²/2) + (21.0x³/3) evaluated from 0 to 0.39
x_com = 50.0(0.39²/2) + (21.0(0.39)³/3) - [50.0(0)²/2 + (21.0(0)³/3)]
x_com = 7.605 + 0.4167
x_com = 8.022 m
Therefore, the center of mass of the rod is located at a distance of 8.022 meters from the x = 0 end.