Rotational Inertia Question?

Your answers (A,B, ... E) in your question don't match the ones in your link. You've swapped the letter around!

If T is the tension in the string attached to mass m, then by considering the mass alone we get:

T=ma.

T produces a torque (moment) on the disk = TR₂ clockwise = maR₂ clockwise.

F produces a torque = FR₁ anticlockwise

Total torque (anticlockwise) = FR₁ - maR₂

Consider the disc alone and work out its angular acceleration (α):

torque = moment of inertia x angular acceleration (τ=Iα)

FR₁ - maR₂ = Iα

α = (FR₁ - maR₂)/I (equation 1)

Linear acceleration and angular acceleration are related by a=αr, so:

a = αR₂

α = a/R₂ (equation 2)

From equations 1 and 2:

a/R₂ = (FR₁ - maR₂)/I

Then do some messy algebra.

a = (FR₁R₂ - maR₂²)/I

a = FR₁R₂/I - maR₂²/I

a + maR₂²/I = FR₁R₂/I

a(1 + mR₂²/I) = FR₁R₂/I

a = (FR₁R₂/I)/(1 + mR₂²/I)

Multiply numerator and denominator of right hand side by I:

a = (IFR₁R₂/I)/(I + ImR₂²/I)

a = FR₁R₂/(I + mR₂²)

This is answer B in your typed list and answer C in your link.