Question

The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with the fixed spatial line a–a. Due to an injection error, the vehicle’s angular velocity vector ω is misaligned 5◦ (the wobble angle) from the symmetry axis. Calculate to three significant figures the maximum angle φ between fixed line a–a and the axis of the cylinder.

Answer 1

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

$A=\frac{1}{12}m(3r^2+l^2)$

Replacing the values we have,

$A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)$$A=197.9kg.m^2$

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

$c=\frac{1}{2}mr^2$

$c=\frac{1}{2}(500)(0.5)^2$

$c=62.5kg.m^2$

Finally we need to find the required angle between the fixed line a-a (I attached an image )

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}$

Replacing the values that we have,

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}$

$\Phi = 2tan^{-1}(\sqrt{0.076634})$

$\Phi = 2tan^{-1}(0.2768)$

$\Phi = 2(15.47)$

$\Phi = 30.95\°$