A block is rotated with $ω=a\cdot t$ [closed]

Question

A block of mass $m$ lies on a table. The coefficient of friction between the table and the mass is $μ$. The table is rotated around where the following relation holds: $$ω=a\cdot t$$ (where t is time). I.e the angular velocity grows linearly.

I need to find the time $t$ at which the block would slip away from the starting point.

It is obvius that the block "slides off" when $\omega >\sqrt{\dfrac{μg}{r}}$. Substituting $ω$ with $at$ gives $$t=\sqrt{\dfrac{μg}{a^2 r}}$$ And that is my answer. Is that correct? The answer in my book says $$t=\sqrt[4]{\dfrac{(μg)^2-(ar)^2}{(a^2r)^2}}$$ Why is that expression so complicated?

Answer 1

There are two components of acceleration. One tangential and one radial. The friction force must account for both of these components.

$$ a_R = \dot{v}= \alpha r $$ $$ a_T = \frac{v^2}{r} = \omega^2 r$$

Your friction condition is

$$ \sqrt{ a_R^2 + a_T^2 } \le \mu g $$