Calculate Center of Mass for 1.05cm Baseball Bat with Linear Density Formula

[HandcuffedKitty]

A baseball bat of length 1.05 cetimeters has a lenear density given by l=.950 + 1.050x^2/l^2 find the x cooridinate of the center of mass in centimeters.

How am I supposed to work this problem?

 

[cookiemonster]

Center of mass is a weighted average. Each part's contribution, all other things being equal, increases proportional to distance from the origin as one moves away from the origin. Just combine "average" and "weighted" to get:

\overline{x} = \frac{\int^a_bx\lambda(x)dx}{\int^a_b\lambda(x)dx}

where [a,b] is the region describing the bat (0 to 1.05 in your case) and \lambda(x) is the linear density function in terms of x. You'll notice that the top half of the fraction is the "weighted" part, which you're dividing by the total mass to get the "average" part.

cookiemonster

 

[M98Ranger]

To calculate the center of mass for a 1.05cm baseball bat with a linear density formula, we can use the following steps:

Step 1: Determine the total mass of the baseball bat. We can do this by integrating the linear density function over the length of the bat. In this case, the length of the bat is 1.05cm, so the total mass would be:

m = ∫(0.950 + 1.050x^2/l^2) dx from x = 0 to x = 1.05

m = (0.950x + 0.350x^3/l^2) from x = 0 to x = 1.05

m = (0.950*1.05 + 0.350*1.05^3/l^2) - (0.950*0 + 0.350*0^3/l^2)

m = 0.9975 + 0.3864/l^2

Step 2: Determine the x coordinate of the center of mass. This can be done using the formula:

xcm = ∫x*(linear density function) dx / m from x = 0 to x = 1.05

xcm = (x*(0.950 + 1.050x^2/l^2) dx) / (0.9975 + 0.3864/l^2) from x = 0 to x = 1.05

xcm = (0.950x^2/2 + 1.050x^4/4l^2) / (0.9975 + 0.3864/l^2) from x = 0 to x = 1.05

xcm = (0.950*1.05^2/2 + 1.050*1.05^4/4l^2) / (0.9975 + 0.3864/l^2) - (0.950*0^2/2 + 1.050*0^4/4l^2) / (0.9975 + 0.3864/l^2)

xcm = (1.100625 + 1.264625/l^2) / (0.9975 + 0.3864/l^2)

xcm = (1.100625 +