Question

A string is wrapped on a wheel of moment of inertia 0.20 kg.m² and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in figure, Find the acceleration of the block ?

Answer 1

Given in the question :-

Assume , Acceleration = a

Tension = T

Now for the block , we know that

$2g-T = 2a$

or $T = 2g-2a$

Now put the value in the formula

Torque on the pulley = F × r

T × 0.10

=0.10 T N-m

Now,

= 0.10(2 g -2 a)

= 0.20 ( g-a)

On the pulley, MI is

I = 0.20 kg-m²

Now for finding the angular acceleration

α = Torque/M.I.

α = T/I

or,

$a/r = 0.20(g-a)/0.20 =g-a$

$a = (gr-ar)$

$a = gr/(1+r)$

$a=9.8*0.10/(1+0.1)$

a = 98/110

a = 0.89 m/s²

Hope it Helps :-)

Answer 2

This is the answer!