Question

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.

Answer 1

Step-by-step explanation:

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