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Question

A spherical cavity of radius R2 is removed from a solid sphere of radius R and the sphere is placed on a rough horizontal surface as shown in the figure. The sphere is given a gentle push and the friction is large enough to prevent slipping. Find the time period of oscillation of the sphere.

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Solution

The correct option is **D** 2π√177R10g

From the figure, we can deduce that position of centre of mass of solid sphere (x1,y1)=(0,0) units

Position of centre of mass of cavity (x2,y2)=(0,R2) units

Then, position of centre of mass of the sphere with cavity is given by

ycm=ρV1y1−ρV2y2ρ(V1−V2)

⇒43π(R3−R38)ρycm=43πR38×ρ×R2

⇒78ycm=−R16

⇒ycm=x(say)=−R14 units

Moment of inertia of the cavitied sphere about an axis ⊥ to plane of the figure, through point of contact (P) is calculated as follows-

Let M= mass of sphere with cavity

From definition ,

M=43πR3(1−18)ρ⇒ρ=3M4π R3×87 units

Mass of the removed sphere of radius R2 is m

⇒m=ρ43π×R38=M7 units

Mass of sphere without cavity M0=m+M=8M7 units

∴ Required moment of inertia

I=(M.I of complete sphere without cavity about an axis through P) - (M.I of the cavity about an axis through P)

I=75M0R2−[25m(R2)2+m(3R2)2]

=75×8M7×R2−[4720M7R2]=177140MR2

Now, the sphere is displaced gently by θ.

Restoring torque about point P is given by

τ=Mgxsinθ

When θ is very small sinθ≈θ

⇒τ=−R14Mgθ

But we know that, τ=Iα

∴Iα=−R14Mgθ

⇒177140MR2 α=−R14Mgθ

⇒α=−140177×14gRθ

Comparing this with α=−ω2θ, we get

ω=√10177gR

⇒ Time period of oscillation

T=2πω=2π√177R10g

Thus, option (d) is the correct answer.

From the figure, we can deduce that position of centre of mass of solid sphere (x1,y1)=(0,0) units

Position of centre of mass of cavity (x2,y2)=(0,R2) units

Then, position of centre of mass of the sphere with cavity is given by

ycm=ρV1y1−ρV2y2ρ(V1−V2)

⇒43π(R3−R38)ρycm=43πR38×ρ×R2

⇒78ycm=−R16

⇒ycm=x(say)=−R14 units

Moment of inertia of the cavitied sphere about an axis ⊥ to plane of the figure, through point of contact (P) is calculated as follows-

Let M= mass of sphere with cavity

From definition ,

M=43πR3(1−18)ρ⇒ρ=3M4π R3×87 units

Mass of the removed sphere of radius R2 is m

⇒m=ρ43π×R38=M7 units

Mass of sphere without cavity M0=m+M=8M7 units

∴ Required moment of inertia

I=(M.I of complete sphere without cavity about an axis through P) - (M.I of the cavity about an axis through P)

I=75M0R2−[25m(R2)2+m(3R2)2]

=75×8M7×R2−[4720M7R2]=177140MR2

Now, the sphere is displaced gently by θ.

Restoring torque about point P is given by

τ=Mgxsinθ

When θ is very small sinθ≈θ

⇒τ=−R14Mgθ

But we know that, τ=Iα

∴Iα=−R14Mgθ

⇒177140MR2 α=−R14Mgθ

⇒α=−140177×14gRθ

Comparing this with α=−ω2θ, we get

ω=√10177gR

⇒ Time period of oscillation

T=2πω=2π√177R10g

Thus, option (d) is the correct answer.

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