Horizontal velocity component on a periodic circular motion

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.

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 $$v_x$$ ?

• 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. Apr 10 '19 at 10:32
• @gandalf61 I think I forgot to mention that $d$ is a fixed distance while $d'$ is not. Apr 10 '19 at 10:50
• 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. Apr 10 '19 at 11:11
• @gandalf61 thanks, your intuition is right. Apr 10 '19 at 12:56

1 Answer

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.