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.

 

[thomate1]

for providing the equations and your attempt at a solution. Your approach is correct, and your argument for the direction of the linear acceleration is valid. Another way to show that the linear acceleration is always pointing towards the hinge is to use the equation for centripetal acceleration, which is given by a = ω2r, where ω is the angular velocity and r is the radius of the circular motion. Since the stick is rotating at a constant angular velocity, the magnitude of the linear acceleration will also be constant at rω2. This means that the direction of the linear acceleration will always be towards the center of the circular motion, which is the hinge in this case.