Assignment 9

# Assignment 9 - PHYS 230 Assignment 9 Fall 2007 1. French...

This preview shows pages 1–3. Sign up to view the full content.

PHYS 230 Fall 2007 Assignment 9 1. French 1.24 A uniform rod of mass M and length 2 L spins with angular velocity ω ± c/L about its centre. In Special Relativity its angular momentum and kinetic energy are given by ± = ML 2 ω 3 ˆ 1 + A ω 2 L 2 c 2 + ··· ! K = ML 2 ω 2 6 ˆ 1 + B ω 2 L 2 c 2 + ··· ! What are the values of A and B ? You will need: (1 - α ) - 1 2 ² 1 + 1 2 α + 3 8 α 2 + ··· , where α ± 1. We need the SR deﬁnitions of momentum and energy, namely p = γm 0 v ; and E = γm 0 c 2 as well as the pre-Einstein deﬁnition of angular momentum, namely ± = mrv = rp (where we have assumed that r and v are mutually perpendicular). Since the answers are expressed as series expansions in powers of v 2 /c 2 , we will expand γ immediately to give γ ² 1 + 1 2 v 2 /c 2 + 3 8 v 4 /c 4 + ··· Then, the angular momentum of the stick is ± total ² Z L - L [1 + 1 2 v 2 /c 2 ] rvμdr where μ is the rest-mass per unit length, μ = M 0 / 2 L . The velocity v is simply v = ωr . Thus ± total ² [ ωM 0 / 2 L ] Z L - L r 2 dr + [ ω 3 M 0 / 4 Lc 2 ] Z L - L r 4 dr or ± total ² 1 3 M 0 L 2 ω + 1 10 M 0 L 4 ω 3 /c 2 so that A = 3 / 10 . Notice that the ﬁrst term is just the “pre-Einstein”, non-relativistic value. Next: the energy. Here the ﬁrst term will be the rest-mass energy M 0 c 2 . So, to get the two required terms, we will need to keep the next term in the expansion of γ . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Thus: E ± Z L - L [1 + 1 2 v 2 /c 2 + 3 8 v 4 /c 4 ] c 2 μdr E ± M 0 c 2 + 1 2 ( M 0 /L ) ω 2 Z L - L r 2 dr + 3 8 ( M 0 /L ) ω 4 Z L - L r 4 dr/c 2 E ± M 0 c 2 + 1 6 M 0 L 2 ω 2 + 3 40 M 0 L 4 ω 4 /c 2 so that B = 9 / 20 . This time, notice that the second term is the non-relativistic one. 2. Do all of these calculations using only the parallel axis theorem, and no other mathe- matics. Especially not any calculus. The parallel axis theorem applies to the moment of inertia of a two-dimensional object rotating about an axis perpendicular to its plane. It states that: the moment of inertia for rotation about a given axis is I = I C + MD 2 where M is the mass, I C is the moment of inertia for rotation about the centre of mass, and D is the distance which separates the given axis from the axis through the centre of mass. (By construction, these two axes are parallel to each other.) Figure out the moments of inertia, for axes perpendicular to their planes and through their respective centres of mass, of the following “simple” objects. (a) A uniform stick of mass M , length 2 L . Hint: a stick of length 2 L is two sticks, each of length L , placed end-to-end. What is the ratio of the moments of inertia for two sticks, one twice the length of the other? Think of this problem as follows: there are two unknowns – the moment of inertia of the stick itself, about its own CofM, and the moments of inertia – around their own CofM’s – of the two half-size sticks from which it is constructed. Call these I big and I small . What are the two relationships between these two unknowns?
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.

### Page1 / 11

Assignment 9 - PHYS 230 Assignment 9 Fall 2007 1. French...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online