Question

A uniform cylinder of mass M, radius R is rotating about its own axis with a speed of n r.p.s. It is gently placed against a corner as shown in Fig.. Coefficient of freection between walls and cylinder is μ. The number of revolutions completed by cylinder before coming to rest is

(1) \(\frac{n^2R(μ^2+1)}{16\pi^2μg(μ+1)}\)

(2) \(\frac{n^2R(μ^2+1)}{32\pi^2μg(μ+1)}\)

(3) \(\frac{n^2R(μ^2+1)}{48\pi^2μg(μ+1)}\)

(4) \(\frac{n^2R(μ^2+1)}{32\pi^2μg(μ-1)}\)

Answer 1

 (2) \(\frac{n^2R(μ^2+1)}{32\pi^2μg(μ+1)}\)

The forces acting are shown in Fig. There is no translational motion therefore.

N₁ + f₂ =  Mg  .....(i)

and N₂ = μN₁ .....(ii)

From Eqns (i) and (ii) we have

Taking torque of forces about O.

\(\tau\) is in clockwise direction.

Let α be the angular retardation. Then