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I have a question on finding the horizontal velocity component of a rotating bar in a (periodic) circular motion. Consider a bar rotating with angular motion: $$ \theta(t) = A \sin (kt) $$

where $A$ is amplitude, $k$ is a constant, and $t$ is time.

rotating bar M

The bar rotates from initial position $M$ to $M'$ with angular velocity $\omega = \dot{\theta}$. Let's consider a horizontal axis $p$ intersecting the rotating bar at a fixed distance $d$. At $M'$, the transverse velocity is given as: $$ v_t = \omega d' $$

where $d' = d / \cos \theta$.

Here, to find the horizontal velocity component $v_x$ of the bar at M', I have expressed it as (including time derivative of $\theta$):

$$ v_x = v_t \cos \theta $$ $$ v_x = k A \cos (kt) (d/\cos\theta) \cos\theta $$ $$ v_x = k A d \cos (kt) $$

To verify this, I tried a second method by defining the displacement of $q$ along $p$ axis as following:

$$ s = d \tan \theta $$

and performs time derivative of s to get the horizontal velocity $v_x'$ as followings:

$$ v_x' = \frac{ds}{dt} = k A d \cos(kt) [\tan^2(\theta) + 1] $$

The results are different by a factor of $(1+\tan^2(\theta))$. Is there a missing component in the derivation of $v_x$ ?

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    $\begingroup$ In your second method remember that $d$ and $\theta$ are both functions of time, so $\frac{ds}{dt} = \frac{dd}{dt}\tan \theta + d(1+\tan^2 \theta)\frac{d\theta}{dt}$. I think you have omitted the first term. $\endgroup$
    – gandalf61
    Commented Apr 10, 2019 at 10:32
  • $\begingroup$ @gandalf61 I think I forgot to mention that $d$ is a fixed distance while $d'$ is not. $\endgroup$
    – Karsun
    Commented Apr 10, 2019 at 10:50
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    $\begingroup$ If $d$ is fixed then $q'$ has a radial veloctiy $v_r$ as well as a tangential velocity $v_t$. So it is no longer true that $v_x=v_t\cos\theta$. Instead $v_x=v_t\cos\theta + v_r\sin\theta$ and your second method gives the correct answer. $\endgroup$
    – gandalf61
    Commented Apr 10, 2019 at 11:11
  • $\begingroup$ @gandalf61 thanks, your intuition is right. $\endgroup$
    – Karsun
    Commented Apr 10, 2019 at 12:56

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In your first version you assume $d'$ is constant when writing $v_t=\omega d'$. The motion of $q'$ is a the compound motion of the point rotating with the rod, and also the point sliding further up on the rod. By taking the horizontal component of $v_t$ you ignore the contribution of sliding up on the rod. So your second method is correct.

Alternatively, you could aso say $$ v_x = v_{t,x} + v_{r,x} = kAd \cos kt + \frac{d (d')}{dt} \sin\theta = kAd \cos kt + \frac{\sin^2\theta}{\cos^2\theta} kAd \cos kt = kAd \cos(kt)(1 + \tan^2 \theta)$$ where $v_t$ and $v_r$ are the tangential and radial velocities, and the $x$ subscript means their $x$-component.

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