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Moment of Inertia

A block of mass m = 2.1 kg is attached to a string that is wrapped around the circumference of a wheel of radius R = 8.2 cm. The wheel rotates freely about its axisand the string wraps around its circumference without slipping. Initially the wheel rotates with an angular speed (omega) causing the block to rise with a linear speedv = 0.36 m/s .Find the moment of inertia of the wheel if the block rises to a height of h= 7.5 cm before momentarily coming to rest.
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Answer #1
Mass of the block, m = 2.1 kg

Initial speed,v = 0.36 m/s

Radius of wheel is r = 0.082 m

Height rised by block, h = 0.075 m


From conservation of energy


Total initial energy = Total final energy


(1/2)mv2 + (1/2) I ω2 = mgh


(1/2)mv2 + (1/2) I ( v/r )2 =mgh


mv2 + I(v/r )2 = 2mgh


I= [ 2mgh - mv2 ] / ( v/r)2

= [ 2 * 2.1 * 9.8 * 0.075 - 2.1 * 0.36^2] / ( 0.36 / 0.082 )^2


Moment of inertia,I = 0.146kg.m2

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