Man on Platform: Min Value of μs for Radial Motion to be Straight

[bznm]

Homework Statement

A platform rotates in counterclockwise with angular velocity w.
A man walks frm the center of the platform to the border with constant radial velocity v' wrt the platform.
$\mu_s$ is the static friction coefficient.

Calculate the minimum value for $\mu_s$ such that the radial motion is straight.
What about a', value of the man acceleration wrt the platform?

Homework Equations

$a_0=a'+a_{cc}+a_c$

where $a_0$ absolute acceleration, a'=man acceleration wrt the patform, $a_{cc}= 2 w x v'$, $a_c=-w^2 r u_r$

$u_r$= unit vector with radial direction
$u_t$= unit vector with tangent direction

The Attempt at a Solution

absolute velocity : $v_0=v' u_r+wr w_t$
absolute acceleration: $dv_0/dt=2v' w u_t-w^2r u_r$

I want that the man goes straight on respect with an observer on the paltform, so I "cancel" the Coriolis acceleration:
$\mu_s =2v'w/g$

$a'=a_0-a_{cc}-a_c+a_{friction}=2v'wu_t$

But I have obtained the Coriolis acceleration! ... Something went wrong. Please, help me!

 

[J Hann]

Doesn't the man experience a "fictitious" force which is the Coriolis force?
What happens if you look at the problem from the actual inertial frame?
Then the angular momentum is I w (where w is the angular velocity).
What is the torque required to produce the change in angular momentum due to
the man walking on the platform?

 

[haruspex]

$a_0=a'+a_{cc}+a_c$

Agreed, if you mean that vectorially. And you know a', yes? So what is the net acceleration?

$\mu_s =2v'w/g$

No, that assumes the only acceleration of the man is Coriolis. There is also centripetal/centrifugal.