Vectors, Matrices, and Quadratic Forms: (a) (3 points) The biggest mystery that

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Vectors, Matrices, and Quadratic Forms: (a) (3 points) The biggest mystery that we'll try to make sense of in Chapter 10 is why the angular momentum and angular velocity vectors aren't always in the same direction. It will turn out that the angular momentum vector L is equal to the moment of inertia matrix I multiplied by the angular velocity vector ü. In order to see why these two vectors aren't always perpendicular to each other, suppose that w = x+9+z (we're picking a really simple example where all of the components are equal to 1) and the moment of inertia matrix is I-0 2 0 0 0 1 What is Iü? Hint: Write ü as a column vector. (b) (2 points) Is Id parallel to ? Why or why not? (c) (5 points) Suppose that we write the kinetic energy of a rotating rigid body as where Iz, Iy, and I2 are the moments of inertia about the three axes. If we were to instead write KE as I, what would the elements of I be? Write it out as a 3x 3 matrix. NOTE THAT THIS IS NOT THE SAME I THAT WE HAD IN PART A! IF YOU WRITE OUT THE MATRIX FROM PART A YOU WILL GET THIS PART WRONG!!!

Explanation / Answer

given

I = (1 0 0)(0 2 0)(0 0 1)

now, w = i + j + k ( where i, j and k are unit vectors along x, y and z directions respectively)

hence

Iw = (1 0 0)(1)

(0 2 0)(1)

(0 0 1)(1)

Iw = (1 2 1)T

hence

Iw = i + 2j + k

b. Iw.w = (i + 2j + k).(i + j + k) = (1 + 2 + 1) = 4

Iw.w = |Iw||w|cos(theta) = 4

sqrt(1 + 4 + 1)*sqrt(1 + 1 + 1)*cos(theta) = 4

sqrt(2*3*3)*cos(theta) = 4

cos(theta) = 4/3*sqrt(2)

theta = 19.47122063449069 deg

hence Iw is not parallel to w

this is because I is not a scalar but a tensor, due to which Iw does no longer remain parallel to w, if I was a scalar then Iw would be parallel to it

c. KErot = 0.5(Ixx*wx^2 + Iyy*wy^2 + IzZ*wz^2)

Ixx = 1

Iyy = 2

Izz = 1

wx = 1

wy = 1

wz = 1

hence

KE rot = 0.5(1*1^2 + 2*1^2 + 1*1^2) = 0.5(1 + 2 + 1) = 2 J

if KE = 0.5 w I w

then

KE = 0.5 (1) (A B C) (1)

(1) (D E F) (1)

(1) (G H I) (1)

KE = 0.5 (1)(A + B + C)

(1)(D + E + F)

(1)(G + H + I)

KE = 0.5 *(A + B + C + D + E + F + G + H + I)

comparing we see

A = Ixx

B = C = D = 0

E = Iyy

F = G = H = 0

I = Izz

hence

KE = 0.5 (1) (Ixx 0 0 ) (1)

(1) (0 Iyy 0 ) (1)

(1) (0 0 Izz) (1)

KE = 0.5 (1)(Ixx Iyy Izz)

(1)

(1)

KE = 0.5*(Ixx + Iyy + Izz) = 0.5(1 + 2 + 1) = 2 J

hence elements of I are (Ixx 0 0 ), (0 Iyy 0), (0 0 Izz)

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