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A small coin of mass 40g is placed on th...

A small coin of mass `40g` is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration `alpha = 2 rad//s^(2)` . The coefficient of static friction between the coin and the disc is `mu_(s) = 3//4` and coefficient of kinetic friction is `mu_(k) = 0.5`. The coin is placed at a distance `r =1m` from the centre of the disc. The magnitude of the resultant force on the coin exterted by the disc just before it starts slipping on the disc is : `(Take g = 10 m//s^(2))`

Text Solution

Verified by Experts

we draw the `FBD` of the coin
`f_(T)=` tangential component of frictional force on coin
`f_(R )=` radial component of frictional force on coin `s`
`f_(T)=malphar`
`f_(r )=momega_(f)^(2)r`
`sqrt(f_(R )^(2)+f_(T)^(2)) le mu_(g)mg`
`2piN=(omega_(f)^(2))/(2alpha)`
Solving the equations, we get
`N=(((mu_(g)g)/(r )-alpha^(2))^(1//2))/(4pialpha)`
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