Question

A chord is wound around a circumference of a wheel of diameter 0.3m ,the axis of the wheel is horizontal .A mass 0.5Kg is attached at the end of the chord and it is allowed to fall from rest .If the weight falls 1.5m in 4sec then what is the angular accelaration of the wheel and what is the moment of inertia of the wheel?

Answer 1

now as given

the distance moved by the wheel = 1.5 m

in time = 4 s

so, the linear velocity will be

v = 1.5m /4s

or

v = 0.375 m/s

now,

the angular velocity will be

w = v/r

here r = d/2 = 0.3/2 = 0.15 m

or

w = 0.375 / 0.15

thus,

w = 2.5 rad/s

so,

the angular acceleration of the wheel will be

a = w/t = 2.5 / 4

or

a = 0.625 rad/s²

and

the moment of inertia will be

I = T/a

here

torque T = r x F = r x mg

or

T = 0.15 x (0.5x9.8)

so,

T = 0.735 Nm

and

so,

I = 0.735/0.625

thus, moment of inertia

I = 1.176 kg.m²