# Relativity/Magnitude Question

1. Nov 2, 2009

### raychelle93

Problem:

At the Large Hadron Collider, two beams of protons collide head-on. Each beam has a speed of .700c relative to the earth. What is the magnitude of the velocity of one beam of protons relative to the other one?

>>I have no idea what equations to use
>>I could really use a breakdown on the problem as well

>>Could someone at least start me off with the equations I need?

Thanks for any help xxxxx

2. Nov 2, 2009

### tiny-tim

Hi raychelle93!

You should have been given the velocity combination equation w = (u + v)/(1 + uv/c2)

3. Nov 14, 2009

### raychelle93

Thank you. I was able to get the problem done correctly

I have two other problems that I need help getting started on:

Moments of inertia
-A rigid body consists of 2 point masses m1= 1kg at a position vector r1= (1,2,3) m and m2 = 2kg at a position vector r2- (0,1,0) m. Calculate the moments of inertia of this body about the x, y and z axes.

-Four uniform solid spheres of equal mass M = 100 g and radius R = 3 cm are arranged in a square and rigidly connected by four rods of equal mass m = 30 g and length L = 10cm

a) calculate the moments of inertia of the system about the axis AB through the centers of the opposite sides of the square.
b) calculate the moments of inertia of the system about the axis A'B' through the vertices of the square.

(Figure is/looks like a square with A' at the top left, nothing marked at top right, M marked at bottom left and B' marked at bottom right. There is a dotted line in the center of the square marked A B and a diagonal dotted line across the square from points A' to B'. L is from M to B' and 2R is from the top of M to the bottom of M (imagine each vertices to be a sphere, therefore 2R is from the top of the sphere to the bottom).

Center of Mass
-Three point masses 30 g each are placed at the vertices of an equilateral triangle ABC and rigidly connected by three rods of length 10 cm. The Rods AB and AC have equal mass 50 g while the mass of the rod BC is 20 g.

Calculate the distance from the center of mass to the vertex A.

(Figure of Triangle has A vertex at top, C at the bottom left and B at the bottom right)

I have no idea how to get started on the Moments of Inertia problem, in fact if you could explain what exactly the problem is looking for that'd be great; but for the center of mass problem I know the equation is m1x1+m2x2+m3x3/m1+m2+m3 but I don't know what x is and how to use the mass of the rods in the equation...

Again any help would be extremely useful

4. Nov 15, 2009

### tiny-tim

Hi raychelle93!
Come on, you must know something about moment of inertia about an axis …

You can replace a system of masses {mi} at points {xi} by a single mass m at a single position x.

Then mx = ∑mixi, and of course m = ∑mi.

So x = (∑mixi)/(∑mi).

5. Nov 15, 2009

### raychelle93

Alright well I know that the moment of inertia depends on the axis in terms of location...and I think it has something to do with rotational motion like kinetic energy. In my textbook it says the definition of moment of inertia is ∑ mi ri ^2. I also know that there's different moments of inertia for various bodies so for a solid sphere I could use the equation Ir= 2/5MR^2 no idea what to do with all that information though or if that will even help me...

So do I have to find x first before I use the equation that I had posted?

6. Nov 15, 2009

### tiny-tim

Hi raychelle93!

(try using the X2 tag just above the Reply box )

For the moment of inertia of the two point masses, just use the definition (∑ mr2).

For the moment of inertia of the spheres, use the formula you mentioned (2/5 mr2) and the parallel axis theorem (look it up if you don't know it ).

And for the centre of mass, once you've found x, you can calculate its distance from A.

(there is a more "direct" way of finding that distance … can you see what it is? … but I don't think it's any quicker)

7. Nov 15, 2009

### raychelle93

Thanks; I was looking for that!

I've tried to work it out like this: mi12 + 02m2= x ; mi22 + 12m2=y ; mi32 + 02m2= z
I came up with the coordinates 1,6,9. Is this okay?

I'm going to work on that question last cuz I'm still somewhat confused...the parallel axis theorem is ( I think) the relationship of the moment of inertia of a body of mass about an axis through its center of mass--> I'=I+Md2 ?

I still don't understand how to find x using the equation you gave me. I know what mi is (the three point masses), but what is xi?

8. Nov 15, 2009

### tiny-tim

Hi raychelle93!
Sorry … completely wrong

The r in ∑ mr2 is the (perpendicular) distance from the axis.

Try again.
That's right … when you look up the moment of inertia of eg a sphere in tables, obviously they only give you the moment of inertia about an axis through the centre …

so you need the parallel axis theorem to give you the moment of inertia about a different axis.
Each xi is the vector position of mass mi (if you don't like vectors, you can use coordinates instead … then it's (∑mixi,∑miyi,∑mizi)/∑mi)

9. Nov 15, 2009

### raychelle93

Hello and thanks for being so patient

But I'd have to look at whether the points lie on the axes? But they do, don't they? In which case wouldn't r= 0?
How can I use these position vectors to help me? I thought the vector r was r and could just be substituted into the equation...

I hate that question lol...i'm working on it!

I tried to work something out and got 10 meters for both xcm and ycm: (30 grams x 10 cm) + (30 grams x 10 cm) + (30 grams x 10 cm)/30+30+30 [this is the three point masses]
and then I used the same equation for the rods: (50 grams x 10 cm) + (50 grams x 10 cm)+ (20 grams x 10 cm)/ 50+50+20
The coordinates would be (10,10) though...if this is correct so far, how would I use it to find the center of mass to vertex A?