Question

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.

Answer 1

Answer:

Explanation:

Let the tension in the cord be T₁ and T₂  .

for motion of block placed on horizontal table

T₁ = m a  , a is acceleration of the whole system .

for motion of hanging bucket of mass m

mg - T₂ = ma

adding the two equation

mg + T₁- T₂ = 2ma

for rotational motion of the pulley

torque = moment of inertia x angular acceleration

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

(mg - 2ma ) R = I x α

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

(mg - 2ma ) R² = I x a

mgR² =  2ma R² + I x a

a = mgR² / (2m R² + I )

Since body moves by distance d in time T

d = 1/2 a T²

a = 2d / T²

mgR² / (2m R² + I ) = 2d / T²

mgR²T² = 2d x (2m R² + I )

mgR²T² -  4dm R² =  2dI

m R² ( gT² - 4d ) = 2dI

I =  m R² ( gT² - 4d ) ] / 2d .