Finding Equation for Linearly Changing Density

[rmfw]

Homework Statement

I need to find the Kinetic energy of a bar rotating about its center of mass.

I know the bar as length 3b and it's center of mass is located at 2b, the bar density changes linearly along it's length.

Homework Equations

T=1/2 W^2 I

The Attempt at a Solution

So I was trying to find I for this setup, which requires me to find the equation for density (λ).

I know the density changes linearly along it's length so it must be similar to an equation of the type: λ = k x , with k being a constant and x being position.

Now to find k I did the following equations:

∫λ dx = ∫ k x dx = m/2 (limits of integration are from 0 to 2b)

∫λ dx = ∫ k x dx = m/2 (limits of integration are from 2b to 3b)

The problem is that I get a false statement this way, making it impossible to find the equation for density.


This is very basic stuff but it's giving me a headache since I need to move forward on the problem but I can't due to this niche, help? thanks

 

[TSny]

In general, it is not true that half the mass will be on one side of the CM and the other half on the other side of the CM.

See here, near the bottom of the page, for finding the CM of a continuous distribution.

 

[TSny]

Note, "varying linearly" could be interpreted more generally as saying that $λ = a + kx$ where $a$ is some constant. But, you should be able to use the integral formula for $x_{cm}$ and the fact that $x_{cm} = 2b$ to show $a = 0$.

 

[Chestermiller]

What is your general equation for the center of mass if the linear density λ(x) varies with with x? First state your equation for the total mass in terms of λ(x).

Chet

 

[HalfLostAndSearching]

As a scientist, it is important to carefully analyze and understand the problem before attempting to solve it. In this case, the bar's density is changing linearly along its length, which means that the density is not a constant value. Therefore, we cannot use the equation λ = kx to represent the density. Instead, we need to consider the density as a function of position, denoted as ρ(x).

To find the equation for ρ(x), we can use the given information that the bar's length is 3b and its center of mass is located at 2b. This means that the bar's density must be highest at the center (2b) and lowest at the ends (0 and 3b). We can represent this as a linear function, such as ρ(x) = mx + b, where m is the slope and b is the y-intercept.

To find the values of m and b, we can use the given information that the bar's density is m/2 at 2b (center of mass) and 0 at 0 and 3b (ends). This gives us two equations: ρ(2b) = m(2b) + b = m/2 and ρ(0) = m(0) + b = 0. Solving these equations, we get m = -1/2b and b = 0.

Therefore, the equation for the bar's density is ρ(x) = -1/2bx. Now, we can proceed to find the moment of inertia, I, using the equation I = ∫ρ(x)x^2dx. With the given limits of integration (0 to 3b), we can solve this integral and find the value of I.

Finally, we can use the equation T = 1/2ω^2I to calculate the kinetic energy of the bar rotating about its center of mass.

In conclusion, it is important to carefully consider the given information and use appropriate equations to solve a problem. Linearly changing density requires us to use a function to represent the density, rather than a constant value. I hope this helps you to move forward with your problem. Good luck!