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In a simple pendulum angular velocity $ω$ is related to angle $θ$ as $$θ = wΔt$$ But what would be the relationship between the angular velocity and angles comprising a spherical pendulum described by angles $θ$ and $\phi$? enter image description here

Qmechanic
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Lenny White
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  • If the motion is constrained to a plane then we can still define an angle in the same way. But if the motion is haphazard, then you can only talk about angular displacement if you have displacement as a function of time. – Superfast Jellyfish Mar 07 '20 at 17:47

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enter image description here

So the $3\times 3$ Rotation Matrix R is a function of $\varphi$ and $\theta$

With

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Where equation (1) and (2) are the pendulum differential equation . With the solution of $\varphi(t)$ and $\theta(t)$ you can obtain the component of the angular velocity $\vec \omega$ equation (3).

Example

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Eli
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  • I'm sorry I don't quite get this. If $R′=R w$ then are we deriving $w=R^{−1} R'$? We can solve for $φ, φ', θ, θ'$. But how do we solve the equation $w=f(φ,φ′,θ,θ′)$? Do we create matrix $R′$ from $φ′,θ′$ and matrix $R$ from $φ,θ$? – Lenny White Mar 08 '20 at 01:38
  • You can use thus equation $\dfrac{dR}{dt}=R\tilde\omega $ to obtain the angular velocity, where tilde omega is a skew tensor, check this equation with one rotation $\varphi$ about the z axes you get $\omega_z=\dot\varphi$ – Eli Mar 08 '20 at 08:35
  • See also https://physics.stackexchange.com/questions/67053/velocity-in-a-turning-reference-frame – Eli Mar 08 '20 at 20:24