What is the direction of P relative to A?

[EastWindBreaks]

Homework Statement

Homework Equations

The Attempt at a Solution

does Vpa has the same direction as Vp? i thought Vp is tangent to the circular path that point P creates( perpendicular to vector Rpa) but from the figure, Vp doesn't seem to be tangent to the path, but Vpa does...

 

[Orodruin]

I suggest that you write down expressions for $\vec R_A$ and $\vec R_{PA}$ and differentiate them with respect to time. This will give you the velocity $\vec v_P = \dot{\vec R_{P}}$.

Also, consider the case when $\theta$ is constant. What happens to $\vec v_{PA}$ then?

 

[EastWindBreaks]

I suggest that you write down expressions for $\vec R_A$ and $\vec R_{PA}$ and differentiate them with respect to time. This will give you the velocity $\vec v_P = \dot{\vec R_{P}}$.

Also, consider the case when $\theta$ is constant. What happens to $\vec v_{PA}$ then?

$\vec V_{PA}$ = $\vec ω_P$ × $\vec R_{PA}$ so,when $\theta$ is a constant, $\vec V_{PA}$ = 0?
$\vec V_A$ = $\vec V_P$ - $\vec V_{PA}$ = $\vec V_P$ - $\vec ω_P$ X $\vec R_{PA}$

 

[EastWindBreaks]

I suggest that you write down expressions for $\vec R_A$ and $\vec R_{PA}$ and differentiate them with respect to time. This will give you the velocity $\vec v_P = \dot{\vec R_{P}}$.

Also, consider the case when $\theta$ is constant. What happens to $\vec v_{PA}$ then?

ok, so basically, $\vec V_P$ = $\vec V_A$ + $\vec V_{PA}$， its a vector sum, so V_P is perpendicular to $\vec R_{PA}$, and $\vec V_{PA}$ is not, but why $\vec V_A$ is not 0 since point A is fixed?

 

[Orodruin]

The point A is not fixed. The link is moving, as per the title of the image you attached.

 

[EastWindBreaks]

The point A is not fixed. The link is moving, as per the title of the image you attached.

oops, my bad, missed " translating base"... so if point A is fixed, and link AP is still rotating, then $\vec V_P$ = $\vec V_{PA}$, correct?

 

[Orodruin]

Yes

 

[EastWindBreaks]

Yes

thank you