Resultant linear acceleration at the rim of a disk

[fuddyduddy]

Homework Statement

A uniform disk with radius 0.390m and mass 27.0kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to θ(t)=( 1.50rad/s)t+( 9.00rad/s²)t² .

What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100rev ?

Homework Equations

tangential acceleration = r * α

The Attempt at a Solution

The velocity equation should be the derivative of the position equation, so
θ(t) = 1.5t + 9t² → ω(t) = 1.5 + 18t
Derivative of velocity equation should be acceleration so
ω(t) = 1.5 + 18t → α(t) = 18 rad/s²

Now a_t = rα = (0.39m)(18 rad/s²) = 7.02 m/s²

Where did I go wrong?

 

[BvU]

What is the exercise asking for, precisely ?

 

[fuddyduddy]

Oh! My fault! It's asking:
What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100rev ?

 

[BvU]

Can't find a flaw in your work. Why do you think it's wrong ?

 

[fuddyduddy]

Because mastering physics says it is incorrect :/

 

[BvU]

Well, then it definitely must be wrong... So either we both make the same calculation error or we both have a wrong perception of what's going on.

Let's bet on the latter: we both thought the tangential acceleration would be the answer. In other words: we calculated the acceleration of a particle that thus far has been attached to the disk and isn't any more. That can well have been our misconception:
perhaps the writer of the exercise wants the linear acceleration of a particle that stays in the same place on the rim of the disk ...

I can't help but feel a litlle bit tricked :(

 

[fuddyduddy]

What would be the difference? Isn't the angular acceleration constant and the radius constant, thus making the linear acceleration constant for any point on the disk at that radius?

 

[BvU]

Tangential acceleration is constant indeed. But apparently, linear acceleration $\ne$ tangential acceleration.

The particle experiences a force to undergo this angular acceleration and also a force to stay in a circular trajectory on the rim of the disc.

What I suspect the exercise means is: the linear acceleration squared is radial squared plus tangential squared

And I must admit that linear acceleration is defined as $\ddot { \vec x }$

 

[fuddyduddy]

Ah I see what you mean now. We both misinterpreted the problem, thanks for working through it with me!