Question

A uniform disc of radius r_(0) lies on a smooth horizontal plane. A similar disc spinning with the angular velocity omega_(0) is carefully lowered onto the first disc. How soon do both discs spin with the same angular-velocity if the friction coefficient between them is equal to mu?

Answer 1

Given:

A uniform disc of radius r0 = R lies on a smooth horizontal plane.

A similar disc spinning with the angular velocity W0 is carefully lowered onto the first disc.

friction coefficient between them is equal to U

To Find:

How soon do both discs spin with the same angular-velocity.

Solution:

By applying conservation of angular momentum,

• IW0 = Iw + Iw =  2Iw
• W = W0/2
• W is the angular velocity with which both the disc spin together.

Due to the frictional force, the torque acting in different parts of the disc will be different.

Hence we can consider the disc as a collection of a lot of rings.

Hence to find the torque , consider a elemental ring of radius r and width dr .

Now ,  mass of this ring element = $\frac{m}{\pi R^{2} }$ x 2πrdr

Therefore,

Torque due to the frictional force on this element is ,

• dT = u x dm x g x r
• dT = 2umgr²dr / R²
• Total torque = T =2umg/R² x  $\int\limits^R_0 {r^{2} } \, dr$  = 2umgR/3 , a constant

Angular acceleration,

• α = Torque/moment of inertia =
• α = (2umgR/3) / (mR²/2) = 4ug/3R , a constant.

We know ,

Final angular velocity - initial angular velocity =  angular acceleration x t

• Wo/2 = αt
• t = Wo/2α = 3RWo/8ug

The time taken for both the discs to spin with the same angular velocity is t = $\frac{3r_0\omega_0}{8\mu g}$