Rotational Motion: Slider Acceleration Explained

This problem asks us to determine the component of acceleration of point P, located on a slider that moves along a rotating link AR. The link AR rotates about a fixed point A.

We are provided with the following variables:

• Angular velocity of the link: $\omega$
• Angular acceleration of the link: $\alpha$
• Radial distance of point P from A: $r$
• Linear velocity of the slider on the link: $v$
• Linear acceleration of the slider on the link: $f$

It's standard to interpret $v$ as the velocity of the slider along the link (radial velocity, $\frac{dr}{dt}$) and $f$ as the acceleration of the slider along the link (radial acceleration, $\frac{d^2r}{dt^2}$ or $\frac{dv}{dt}$).

Acceleration Components: Point P on Rotating Link

The absolute acceleration of point P ($\vec{a}_P$) can be found by considering its motion relative to the rotating link and the motion of the link itself. We can use a coordinate system fixed to the link AR, with unit vectors $\hat{u}_r$ along the link (from A to P) and $\hat{u}_\theta$ perpendicular to the link.

The acceleration $\vec{a}_P$ has multiple components:

• Centripetal acceleration: Due to the rotation of the link, directed towards the center of rotation A. Its magnitude is $r\omega^2$.
• Tangential acceleration (link): Due to the angular acceleration of the link, perpendicular to the link. Its magnitude is $r\alpha$.
• Acceleration relative to the link: This is the acceleration of the slider along the link, given as $f$.
• Coriolis acceleration: This arises because the slider is moving ($v$) along the rotating link. It's perpendicular to both the axis of rotation and the velocity $v$. Its magnitude is $2\omega v$.

Deriving Perpendicular Acceleration Component

We can derive the absolute acceleration $\vec{a}_P$ using the formula for acceleration in a rotating frame:

In this case:

• The origin A is fixed, so $\vec{a}_{origin} = 0$.
• The frame rotating with the link has angular velocity $\vec{\omega}_{frame} = \omega \hat{k}$ and angular acceleration $\vec{\alpha}_{frame} = \alpha \hat{k}$ (assuming rotation about the z-axis, $\hat{k}$).
• The position vector relative to the origin is $\vec{r}_{rel} = r \hat{u}_r$.
• The velocity relative to the frame is $\vec{v}_{rel} = v \hat{u}_r$ (velocity along the link).
• The acceleration relative to the frame is $\vec{a}_{rel} = f \hat{u}_r$ (acceleration along the link).

Let's calculate each term:

1. Centripetal acceleration: $\vec{\omega}_{frame} \times (\vec{\omega}_{frame} \times \vec{r}_{rel}) = \omega \hat{k} \times (\omega \hat{k} \times r \hat{u}_r) = \omega^2 r (\hat{k} \times \hat{u}_\theta) = -r\omega^2 \hat{u}_r$. This acts radially inward.
2. Tangential acceleration (link): $\vec{\alpha}_{frame} \times \vec{r}_{rel} = \alpha \hat{k} \times r \hat{u}_r = \alpha r (\hat{k} \times \hat{u}_r) = r\alpha \hat{u}_\theta$. This acts tangentially.
3. Coriolis acceleration: $2\vec{\omega}_{frame} \times \vec{v}_{rel} = 2 (\omega \hat{k}) \times (v \hat{u}_r) = 2\omega v (\hat{k} \times \hat{u}_r) = 2\omega v \hat{u}_\theta$. This also acts tangentially.
4. Acceleration relative to the frame: $\vec{a}_{rel} = f \hat{u}_r$. This acts radially along the link.

Summing the components:

The component of acceleration perpendicular to the link AR is the tangential component, along $\hat{u}_\theta$. Its magnitude is:

Confirming the Correct Answer

The derived magnitude of the acceleration component perpendicular to the link AR is $2\omega v + r\alpha$. This matches the expression provided in option 3.