Question

Hello good morning everyone

Can someone please also help me in this question?(Question 249)

Thank you so much !

Answer is : 3wR/4kg


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Answer 1

Answer:

It is the moment of friction force which brings the disc to rest The force of friction is applied to each section of the disc, and since these sections lie at different distances from the axis, the moments of the forces of friction differ from section to section. To find Nz, where z is the axis of rotation of the disc let us partition the disc into thin rings (Fig.). The force of friction acting on the considered element dfr=k(2πrdrσ)g, (where σ is the density of the disc)

Explanation:

HOPE IT WILL HELP YOU BRO

Answer 2

Here's your answer!

• Let's first assume the elemental area, dr.

• Let the area of that elemental area be dA.

Now, dA = 2πr.dr

And, elemental mass per unit area, dm = 2πr.dr × M/πr².

» dm = 2Mr.dr / πr²

And, dmg or dN = mg / πR² × 2πr.dr

» dmg or dN = 2Mgr.dr / R²

Now, as coefficient of friction is given as k.

Let's, consider it to be elemental.

» df = (dN) k

» df = 2MgK.dr / R²

[ Rest in the attachments ]

Hope it helps!