Question

A uniform disc of radius r is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillation. What should be the distance of the hole from the center for it to have the minimum time period?

Answer 1

Answer ⇒ T = 2π√(r√2/g)

Explanation ⇒ Let the distance between the centre of mass to the point of suspension be x.  

Moment of Inertia of the disc about the point of suspension = 1/2 × mr² + mx²    [This result came from Parallel axis theorem.]

= m(r²/2+x²)

Now, We know when disc will be suspended motion cannot be linear. Thus, the motion will be angular and that will be angular S.H.M.

∴  Using the formula,

$T = 2\pi \sqrt{\frac{i}{mgl'} }$, where i is the moment of inertia of the disc about its end.

T = 2π√{m(r²/2+x²)/mgx}

T = 2π√{(r²/2+x²)/gx}

On Squaring both sides of the equation,

T² = 4π²r²/2gx + 4π²x/g  

Always remember this condition in physics, For x to be minimum dT/dx = 0

Thus, Differentiating both sides of the equations w.r.t. x.  w.r.t. x

d(T²)/dx = d(4π²r²/2gx)/dx + d(4π²x/g)/dx

2T × dT/dx = 4π²r²/2g d(1/x)/dx + 4π²/g dx/dx

∴ dT/dx =  [4π²r²/2g d(1/x)/dx + 4π²/g dx/dx]/2T

Now, this is equal to zero.

∴ 4π²r²/2g × d(1/x)/dx + 4π²/g dx/dx = 0

∴ r²/2 × -1/x² + 1 = 0

r²/2x² = 1

∴ x² = r²/2

∴ x = r/√2

Now, substitute this value of x in Equation of T².

∴ T² = 4π²r²/(2gr/√2) + 4π²r/√2g

∴ T² = 4π²r/√2g + 4π²r/√2g

∴ T² = 8π²r/√2g =4√2π²r/g

∴ T = 2π√(r√2/g)

Hope it helps.