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Byju's Answer

Standard XII

Physics

Moment of Inertia of a Rod

The moment of...

Question

The moment of inertia of the pulley system as shown in the figure is $3 k g - m^{2}$ .The radii of bigger and smaller pulleys are $2 m$ 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 $g = 10 m / s^{2}$ ]

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Solution

Given:
Mass $M_{1} = 6 K g, M_{2} = 3 K g$
$I = 3 K g - m^{2}$
Let $α$ is the angular Acceleration of the Pulley

From FBD of 6kg block
$6 g - T_{1} = 6 a$
$60 - T_{1} = 6 (α \times 2)$
$60 - T_{1} = 12 α$ .....(1)

FBD of 3Kg block
$T_{2} - 30 = 3 α$ ......(2)

Resultant torque: $τ = I α$
$T_{1} \times 2 - T_{2} \times 1 = I \times α$
$2 T_{1} - T_{2} = 3 α$ .........(3)

On solving above 3 equation
$α = 30 r a d / s^{2}$

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Given:Mass M1=6Kg, M2=3Kg I=3Kg−m2Let α is the angular Acceleration of the PulleyFrom FBD of 6kg block6g−T1=6a60−T1=6(α×2)60−T1=12α .....(1)FBD of 3Kg blockT2−30=3α ......(2)Resultant torque: τ=IαT1×2−T2×1=I×α2T1−T2=3α .........(3)On solving above 3 equation α=30rad/s2