Question

The pulley shown in figure has a radius 10 cm and moment of inertia 0.50 kg-m² about its axis. Assuming the inclined planes to be friction less, calculate the acceleration of the 4.0 kg block?

Answer 1

Given in the question :-

Assume acceleration of 4kg block = a

Tension = T

Now if we can take tension in 2 kg = T'

So,

$4g.sin45^0 - T = 4a$

$T = 4g/\sqrt{2} -4a$

Also, $T'-2g.sin45^0= 2a$

as the acceleration is same

Now,$T' = 2a + 2g/\sqrt{2}$

Therefore the Net torque

= ( T- T' ) × 0.10

=0.10 × {( 4g/√2-4a)-(2a + 2g/√2)}

$=0.10 * (2g /\sqrt{2}- 6a)$

Now we have , Angular acceleration

α =a/r

α = a/0.10

Hence, from Torque = Iα

0.10(2g/√2-6a) = Iα

0.10(2g/√2-6a) = 0.50 × a/0.10

= 0.50 × a/0.10 = 5a

50a= 2g/√2-6a

56a = 2g/√2

= 2× 9.8×√2/2

=9.8√2

Hence a = 9.8 × 1.41/56

= 0.2474 m/s²

Hope it Helps :-)