Question

An uniform circular disc rolls down without slipping on an inclined plane of inclination 45º. The minimum co-efficient of friction to support pure rolling is
1
3
WIN
c
VIN
CIN
2.
31​

Answer 1

Given info : An uniform circular disc rolls down without slipping on an inclined plane of inclination 45º.

To find : The minimum coefficient of friction to support pure rolling is ...

solution : first find general formula so that we can find it easily.

let angle of inclination is θ , mass of disc is m, radius of disc is r.

for rotational motion,

torque = I × α

⇒r × F = I × α

here F = frictional force = μmgcosθ

⇒μmgcosθr/I = α .......(1)

for linear motion,

mgsinθ - μmgcosθ = ma

⇒gsinθ - μgcosθ = a .......(2)

for pure rolling, a = αr ......(3)

from equations (1) , (2) and (3),

μmgcosθr²/I = gsinθ - μgcosθ

⇒μcosθ[mr²/I + 1 ] = sinθ

⇒μ = tanθ/[mr²/I + 1] , this is required formula.

now moment of inertia of disc, I = 1/2 mr²

θ = 45°

so, μ = tan45°/[mr²/(1/2mr²) + 1] = 1/3

Therefore the minimum coefficient of friction to support pure rolling is 1/3.