Let the masses of blocks A and B be 4.50 kg and 2.00 kg, respectively, the moment of inertia of the wheel about its axis be 0.400 kg⋅m2, and the radius of the wheel be 0.100 m. Find the magnitude of linear acceleration of block A if there is no slipping between the cord and the surface of the wheel.

Due to friction force that rotates the wheel, tension T₁ acting on 4.5 kg will be different from tension T₂ acting on 2 kg

For motion of 4.5 kg

4.5 g - T₁ = 4.5 a

For motion of 2.00 kg

T₂ - 2g = 2 a

4.5 g - T₁ + T₂ - 2g = 6.5 a

2.5 g + (T₂ - T₁) = 6.5 a

For rotational motion of wheel

I α = (T₁ - T₂) R, I is moment of inertia of wheel, α is angular acceleration

(T₁ - T₂) = I α / R

2.5 g - I α / R = 6.5 a

2.5 g = I α / R + 6.5 a

2.5 g = I a / R² + 6.5 a

2.5 g = (I / R² + 6.5) a

a = 2.5 g / (I / R² + 6.5)

= 2.5 x 9.8 /.4 /.1² + 6.5)

= 24.5 / 46.5

=.5269 m/s²

52.7 m / s²