A cylinder rests on a horizontal rotating disc, as shown in the figure. Find at what angular velocity, `omega`, the cylinder falls off the disc, if the distance between the axes of the disc and cylinder is `R`, and the coefficient of friction `mugtD//h` where `D` is the diameter of the cylinder and It is its height.

Answer

`sqrt(D)/(hR)`

Explanation

The centripetal force that keeps the cylinder at rest on the disc is the frictioal force `f`. According to a non inertial observer on disc a pseudo force the cylinder reacts with an equal and oppostite force, `F`, which sometimes is reffered to as the centrifugal force.

`F=momega^(2)R`
where `M` is the mass of the cylinder.
The cylinder can fall off either by slipping away or by tilting about point `P`, depending of whichever takes place first. the critical agular speed `w_(1)` for slipping occurs when `F` equals `f:F=f`
`Momega_(1)^(2)R+mugM`
where `g` is the gravitational acceleration. Hence `omega_(1)=sqrt((mug)/R)`
`F` ties to rotaste the cylinder about `P`, but the weight `W` opposes it. The rotatiion becomes pssible, when the torque caused by `W`.
`Fh/2=W D/2implies F=W D/h`
`Momega_(2)^(2)R=Mg D/h`
giving `omega_(2)=sqrt(D/(hR))`
Since we are given `mugtD/h`, we see that `omega_(1)gtomega_(2)` and the cylinder falls off by rolling over at `omega=omega_(2).`