Question

A cylinder rests on a horizontal rotating disc, as shown in the figure. If the distance between the axis of the disc and cylinder is \(R\), and the coefficient of friction \(\mu >\frac{D}{h}\), where \(D\) is the diameter of the cylinder and h is its height. The angular velocity, \(\omega\) at which the cylinder falls off the disc will be

Answer 1

FBD is draw from with respect to person standing on disc or we can say non-inertial frame of reference
The centripetal force that keeps the cylinder at rest on the disc is the frictional force \(f\)
According to a non-inertial observer on disc, a pseudo force of the cylinder reacts with an equal and opposite force, \(F\), which sometimes is referred to as the centrifugal force and
\(M\) is the mass of the cylinder.

Formula used: Centrifugal force \(F=M\omega^2 R\)
The cylinder can fall off either by slipping away or by tilting about point \(P\), depending on whichever takes place first.

First we will find minimum angular speed required for sliding of cylinder
The critical angular speed \(\omega_1\), for slipping occurs when \(F\) equals \(f\)
\(F=f\)
\(M\omega^2_1R=\mu gM\)
where g is the gravitational acceleration.
Hence,
\(\omega_1=\sqrt{\dfrac{\mu g}{R}}\)
Now we will find minimum angular speed required to topple the cylinder

\(F\) tries to rotate the cylinder about \(P\), but the weight \(W\) opposes it. The rotation becomes possible, when the torque created by \(F\) is large enough to take over the opposing torque caused by \(W\)
Taking torque about point of contact \(P\)
\(F \dfrac{h}{2} (clockwise)=W \dfrac{D}{2}(anti-clockwise)\)
\(F=W\frac{D}{h}\)
\(D\omega_2^2 R=Mg \dfrac{D}{h}\)
\(\therefore \omega_2=\sqrt{\dfrac{Dg}{hR}}\)
Since we are given that \(\mu>\dfrac{D}{h}\)
we see that
\(\sqrt{\dfrac{\mu g}{R}}>\sqrt{{Dg}{hR}}\)
\( \omega_1>\omega_2\)
and the cylinder falls off by rolling over at
\(\omega=\omega_2\)