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A block of mass m rests on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose radius is R, to a hanging bucket with mass m . The system is released from rest, and the block is observed to move a distance d over a time interval of T. What is the moment of inertia of the pulley about its axis? Make sure to give your answer in terms of the given quantities.

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Answer:

Explanation:

Let the tension in the cord be Tā‚ and Tā‚‚ Ā .

for motion of block placed on horizontal table

Tā‚ = m a Ā , a is acceleration of the whole system .

for motion of hanging bucket of mass m

mg - Tā‚‚ = ma

adding the two equation

mg + Tā‚- Tā‚‚ = 2ma

for rotational motion of the pulley

torque = moment of inertia x angular acceleration

(Tā‚‚ - Tā‚) R = I x Ī± , I is moment of inertia of pulley , Ī± is angular acceleration .

(mg - 2ma ) R = I x Ī±

(mg - 2ma ) R = I x a / R

(mg - 2ma ) RĀ² = I x a

mgRĀ² = Ā 2ma RĀ² + I x a

a = mgRĀ² / (2m RĀ² + I )

Since body moves by distance d in time T

d = 1/2 a TĀ²

a = 2d / TĀ²

mgRĀ² / (2m RĀ² + I ) = 2d / TĀ²

mgRĀ²TĀ² = 2d x (2m RĀ² + I )

mgRĀ²TĀ² - Ā 4dm RĀ² = Ā 2dI

m RĀ² ( gTĀ² - 4d ) = 2dI

I = Ā m RĀ² ( gTĀ² - 4d ) ] / 2d .

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