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Question

The moment of inertia of the pulley system as shown in the figure is 3 kgm2.The radii of bigger and smaller pulleys are 2m and 1 m respectively. As the system is released from rest, find the angular acceleration of the pulley system. (Assume that there is no slipping between string and pulley and string is light)
[Take g=10 m/s2]
742331_b2c053407fe5485494f04f6bcb2071b5.png

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Solution

Given:
Mass M1=6Kg, M2=3Kg
I=3Kgm2
Let α is the angular Acceleration of the Pulley

From FBD of 6kg block
6gT1=6a
60T1=6(α×2)
60T1=12α .....(1)

FBD of 3Kg block
T230=3α ......(2)

Resultant torque: τ=Iα
T1×2T2×1=I×α
2T1T2=3α .........(3)

On solving above 3 equation
α=30rad/s2

1530054_742331_ans_77995c561c0846f5982829e3e7a513ef.png

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