Derive the magnitude and direction of the linear acceleration

[saltine]

[Solved] Centripetal Acceleration

Homework Statement

Derive the magnitude and direction of the linear acceleration at the tip of the stick, ignore the effect of gravity.

Homework Equations

r = length of stick
a = dv/dt

The Attempt at a Solution

I first find the x and y components of v, then I took the derivative to find ax and ay:

vx = rω cos(θ)
vy = rω sin(θ)

ax = rω ( -sin(θ) dθ/dt ) = rω ( -sin(θ) (-ω) ) = rω² sin(θ)
ay = rω ( cos(θ) dθ/dt ) = rω ( cos(θ) (-ω) ) = -rω² cos(θ)

I recognized that ω is -dθ/dt since it is in the opposite direction. This shows that the magnitude of a is rω². If I were to draw the vector on the graph, it would also show that a is pointing toward the hinge.

But how would I show that a is always pointing toward the hinge?

Is this argument valid:

Let q be the a vector from the tip to the center, then

qx = sin(θ)
qy = -cos(θ)

Now, a • q = rω²cos(ф) = rω², where ф is the angle between the two vectors and ф=0 in this case. Therefore a and q are in the same direction.

Is there a simpler way to show this?


- Thanks

 

[naresh]

Your solution is perfect.

Since the components of the acceleration vector are proportional to sin θ and -cosθ, you can also draw a right triangle to show that the acceleration vector points towards the center.