Question

a. The moment of inertia of the reaction wheels has a value RI and the moment of inertia of the telescope about its symmetry axis, and about an axis perpendicular to its symmetry axis is represented by Iz and Ix , respectively. Find an expression for the final speed of the appropriate reaction wheel wRf to cause the telescope to change its direction at a rate of wt for a nominal (i.e. idling) rotational speed of the flywheels of wRi . Assume the initial rotational speed of the telescope is zero.

b. The moment of inertia of a solid cylindrical body (which is a passable approximation of a space telescope) is given by Iz=(1/2)mR^2 and Ix=(m/12)(R^2+H^2). Consider a space telescope of mass 1000 kg, radius 1.20 m and height 4.50 m and a moment of inertia of the reaction wheels of 3x10^-3 kgm^2 and an idling speed of the flywheels of 200 rad/s. Calculate the changes in the rotational speed required to cause a rotation of 0.100 degree per second for rotation
i. about the symmetry axis and
ii. about the axis perpendicular to the symmetry axis.

Answer 1

(b) Using a conservation of angular momentum, we have

L₁ = L₂

I_T$\omega$_T = I_W$\omega$_W

(i) about the symmetry axis -

I_T = I_z = (1/2) m R²

[(0.5) (1000 kg) (1.2 m)²] $\omega$_T = (3 x 10^-3 kg.m²) (200 rad/s)

(720 kg.m²) $\omega$_T = 600 kg.m²/s

$\omega$_T = 0.833 rad/s

The changes in the rotational speed required to cause a rotation of 0.1 degree per second for rotation which will be given as :

$\Delta$$\omega$ = (0.8333 rad/s) - (0.0017 rad/s)

$\Delta$$\omega$ = 0.831 rad/s

(ii) about the axis perpendicular to the symmetry axis -

I_T = I_x = (1/12) m (R² + H²)

(1/12) (1000 kg) [(1.2 m)² + (4.5 m)²] $\omega$_T = (3 x 10^-3 kg.m²) (200 rad/s)

(1807.5 kg.m²) $\omega$_T = 600 kg.m²/s

$\omega$_T = 0.3319 rad/s

The changes in the rotational speed required to cause a rotation of 0.1 degree per second for rotation which will be given as :

$\Delta$$\omega$ = (0.3319 rad/s) - (0.0017 rad/s)

$\Delta$$\omega$ = 0.33 rad/s