It is asked to find the velocity and acceleration of the plataform:

Determine the velocity and acceleration of platform $P$ as a function of the angle of cam $C$ if the cam rotates with a constant angular velocity $\omega$. The pin connection does not cause interference with the motion of $P$ on $C$. The platform is constrained to move vertically by the smooth vertical guides.

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Solution is simple. I tried two approaches, each one giving a different answer.

Solution 1:

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By simple geometry, the $y$ variable relates to the $\theta$ variable as:

$$y = r+r\cos(\theta) = r(1+\sin \theta) $$

Differentiating twice (noticing $\dot{\theta}=\omega$ is constant, we get the velocity and acceleration asked:

$$ \dot{y} = r\dot{\theta}\cos\theta = r\omega \cos\theta $$ $$ \ddot{y} = -r\omega^2 \sin\theta $$

Solution 2:

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The cam rotates about the fixed point O (the pin connection). Since the cam rotates about a fixed axis, any point $P on the body travels along a circular path about the fixed axis. Its velocity and acceleration are:

$$ \vec{v}_P = \vec{\omega} × \vec{r}_P $$

$$ \vec{a}_P = \vec{\alpha} × \vec{r}_P + \vec{\omega} × (\vec{\omega} × \vec{r}_P) = -\omega^2 \vec{r}_P $$

Last inequality follows for planar motion and, since $\omega$ is constant, $\vec{\alpha}=\vec{0}$.

The idea for the second solution is to calculate the velocity and acceleration of the point of contact of the cam and the platform. Since the platform is constrained to move vertically, the answer for the question is only the vertical component of the velocity and acceleration of this point of contact.

Position vector is: $ \vec{r}_P = -r\cos\theta \hat{i} + r(1+\sin\theta) \hat{j} $ The angular velocity vector is: $ \vec{\omega} = -\omega \hat{k} $


$$ \vec{v}_P = \left( -\omega \hat{k} \right) × \left( -r\cos\theta \hat{i} + r(1+\sin\theta) \hat{j} \right) \\ \vec{v}_P = \omega r (1+\sin\theta) \hat{i} + \omega r \cos\theta \hat{j}$$

$$ \vec{a}_P = -\omega^2 \vec{r}_P = r\omega^2\cos\theta \hat{i} - r\omega^2(1+\sin\theta) \hat{j} $$

Vertical component of velocity is $v_y = r \omega \cos\theta $, which is exactly the same answer as before. For acceleration, however, vertical component is $a_y = - r\omega^2(1+\sin\theta) $, which is not the same answer as before.

What is wrong in the second solution? What am I missing here?

  • $\begingroup$ solution 1 in the very 1st equation: height is r sin theta. $\endgroup$
    – Tiger Guy
    Oct 5, 2022 at 21:30


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