X and y coordinates, integration, semicircular plate (masteringphysics)

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 18K views
kottur
Messages
53
Reaction score
0

Homework Statement



Use equations x[itex]_{cm}[/itex]=[itex]\frac{1}{M}[/itex][itex]\int x dm[/itex] and y[itex]_{cm}[/itex]=[itex]\frac{1}{M}[/itex][itex]\int y dm[/itex] to calculate the x- and y-coordinates of the center of mass of a semicircular metal plate with uniform density [itex]\rho[/itex] and thickness t. Let the radius of the plate be R. The mass of the plate is thus M=[itex]\frac{1}{2}[/itex][itex]\rho\pi[/itex]a[itex]^{2}t[/itex].

Use the coordinate system indicated in the figure.

YF-08-51.jpg


1. Calculate the x-coordinate of the center of mass of a semicircular metal plate. Express your answer in terms of the variables a, ρ and t.

2. Calculate the y-coordinate of the center of mass of a semicircular metal plate. Express your answer in terms of the variables a, ρ and t.

Homework Equations



I think these:

[itex]\vec{r_{cm}}[/itex]=[itex]\frac{m_{1}\vec{r_{1}}+m_{2}\vec{r_{2}}+...}{m_{1}+m_{2}}[/itex]

But instead of the sum I need to integrate, right?
Does this equation work in 3D?

The Attempt at a Solution



I'm not sure how to use the equation and what information to use where.

To find x-coordinate:

x[itex]_{cm}[/itex]=[itex]\frac{Mx_{cm}}{M}[/itex]=x[itex]_{cm}[/itex] ??

y[itex]_{cm}[/itex]=[itex]\frac{My_{cm}}{M}[/itex]
 
Physics news on Phys.org
Well I can tell you a couple of things. Because of symmetry, you don't need to use the z coordinate, you already know the z coordinate of centre of mass. I would also say the same thing for the x coordinate. So the only coordinate that you need to iron out is the y coordinate.

EDIT: You will have to put dm in terms of something else I believe.
 
Last edited:
But how do I find the x coordinate in terms of rho, a and t?
 
kottur said:
Is x=0 by symmetry?

yes sir, because you go from -R to R.

The best day to find y is ysqrt(r^2-y^2) and do a substitution
 
I got the answer [itex]\frac{4a}{3\pi}[/itex] from a friend but I want to know how to get there!

How does that work with [itex]y=y\sqrt{r^{2}-y^{2}}[/itex]?

I haven't seen that in my textbook.
 
darn i forgot that textbook kind of sucks :(. maybe forget that method since you won't be able to reference it easily
 
Thanks anyway... :) :/
 
dm can be written in terms of rho dV. this rho will cancel out which gives you the clue that you're headed in the right direction. You just have to perform dV properly.