The motion of a particle of mass \(m\) in space is subject to the two constraint equations $$ x^{2}+y^{2}+z^{2}=1 \quad \text { and } \quad x \sin (\omega t)-y \cos (\omega t)=0 $$ where \(\omega\) is constant. Show that the particle is moving on a circle that is rotating with constant angular speed \(\omega\) about a vertical diameter. Show that its position can be specified parametrically by $$ x=\sin q \cos (\omega t), \quad y=\sin q \sin (\omega t), \quad z=\cos q $$ and express the kinetic energy as a function of \(q, \dot{q}\), and \(t\).

We have been given the following constraint equations: 1. \(x^2+y^2+z^2=1\) 2. \(x\sin(\omega t)-y\cos(\omega t)=0\) The first constraint equation represents a sphere with radius 1 and centered at the origin, while the second constraint equation involves \(\sin\) and \(\cos\) functions with respect to time. We need to combine these constraints to determine the particle's motion.

We propose the position of the particle can be specified parametrically by: \(x=\sin{q}\cos{(\omega t)}\) \(y=\sin{q}\sin{(\omega t)}\) \(z=\cos{q}\) where \(q\) represents the angle and \(\omega\) is the constant angular speed. Let's plug these parametric equations back into our constraint equations:

Substituting the parametric equations into the constraint equations: 1. \((\sin{q}\cos{(\omega t)})^2 + (\sin{q}\sin{(\omega t)})^2 + (\cos{q})^2 = 1\) 2. \((\sin{q}\cos{(\omega t)})\sin{(\omega t)} - (\sin{q}\sin{(\omega t)})\cos{(\omega t)} = 0\) Simplifying the first equation, we get: \(\sin^2{q}\cos^2{(\omega t)} + \sin^2{q}\sin^2{(\omega t)} + \cos^2{q} = 1\) Since \(\sin^2{\theta}+\cos^2{\theta}=1\), the equation reduces to: \(\sin^2{q} + \cos^2{q} = 1\), which is true. Now, let's simplify the second equation: \(\sin{q}\cancelto{\cos^2{(\omega t)}}{(\sin^2{(\omega t)} + \cos^2{(\omega t)})} - \sin{q}\sin{(\omega t)}\cos{(\omega t)} = 0\) The constraint equations hold true for the proposed parametric equations, confirming that the particle is moving on a rotating circle with constant angular speed \(\omega\).

To determine kinetic energy, we first differentiate the position parametric equations with respect to time: \(\dot{x}=\sin q(-\omega \sin(\omega t)) + \cos q\cos(\omega t) \cdot \dot{q} = \omega\sin{q}\cos{q}(\omega t) - \cos{q}\sin{(\omega t)}\dot{q}\) \(\dot{y}=\sin q (\omega\cos(\omega t)) + \cos q \sin(\omega t) \cdot \dot{q} = \omega\sin{q}\sin{q}(\omega t) + \cos{q}\cos{(\omega t)}\dot{q}\) \(\dot{z}=-\sin q \cdot \dot{q}\) Now, recalling that the kinetic energy is given by \(K=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)\), we can substitute the derivatives of position and simplify: \(K=\frac{1}{2}m[\omega^2\sin^2{q}\cos^2{(\omega t)} + 2\omega\sin{q}\cos{q}\cos{(\omega t)}\sin{(\omega t)}\dot{q} + \cos^2{q}\sin^2{(\omega t)}\dot{q}^2 \\ + \omega^2\sin^2{q}\sin^2{(\omega t)} - 2\omega\sin{q}\cos{q}\sin{(\omega t)}\cos{(\omega t)}\dot{q} + \cos^2{q}\cos^2{(\omega t)}\dot{q}^2 + \sin^2{q}\dot{q}^2]\) Simplifying further: \(K=\frac{1}{2}m[\omega^2\sin^2{q}(\cos^2{(\omega t)}+\sin^2{(\omega t)}) + \cos^2{q}(\sin^2{(\omega t)}+\cos^2{(\omega t)})\dot{q}^2 + \sin^2{q}\dot{q}^2]\) Since \(\sin^2{\theta}+\cos^2{\theta}=1\), the equation can be further simplified: \(K = \frac{1}{2}m[\omega^2\sin^2{q} + (\cos^2{q}+\sin^2{q})\dot{q}^2]\) And finally, we have the kinetic energy as a function of \(q\), \(\dot{q}\), and \(t\): \(K = \frac{1}{2}m[\omega^2\sin^2{q} + \dot{q}^2]\)