Question

A sphere of mall 5 kg rolls without slipping on an inclined plane of inclination 30∘. Choose the correct statement(s).

- Force of friction acting on sphere is 5g7.
- Linear acceleration of sphere down the plane is 5g14.
- For pure rolling, μ>0.165.
- None of these

Solution

The correct options are

**A** Force of friction acting on sphere is 5g7.

**B** Linear acceleration of sphere down the plane is 5g14.

**C** For pure rolling, μ>0.165.

N= Normal force

f= friction force

mg= weight of sphere

Let the linear acceleration of the sphere down the plane be a.

So, equation of linear motion for the centre of mass

mgsinθ−f=ma.....(1)

∵ Sphere rolls without slipping

So α=ar

where α=Angular acceleration,

a=Linear acceleration

For Rotational motion,

τ0=Iα

where τ0= Torque about centre of mass

⇒fr=25mr2(ar)

⇒f=25ma......(2)

From eq (1) & (2)

mgsinθ=75ma

⇒a=57gsinθ=57×g×sin30∘=514g

⇒a=5g14

and f=27mgsinθ=27×5×g×12=5g7

f=5g7

For pure rolling,

fmax≥fkinetic

⇒μmgcosθ≥27mgsinθ

⇒μ≥27tan θ=0.165

i.e μ>0.165

Options A, B & C are correct.

N= Normal force

f= friction force

mg= weight of sphere

Let the linear acceleration of the sphere down the plane be a.

So, equation of linear motion for the centre of mass

mgsinθ−f=ma.....(1)

∵ Sphere rolls without slipping

So α=ar

where α=Angular acceleration,

a=Linear acceleration

For Rotational motion,

τ0=Iα

where τ0= Torque about centre of mass

⇒fr=25mr2(ar)

⇒f=25ma......(2)

From eq (1) & (2)

mgsinθ=75ma

⇒a=57gsinθ=57×g×sin30∘=514g

⇒a=5g14

and f=27mgsinθ=27×5×g×12=5g7

f=5g7

For pure rolling,

fmax≥fkinetic

⇒μmgcosθ≥27mgsinθ

⇒μ≥27tan θ=0.165

i.e μ>0.165

Options A, B & C are correct.

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