Question

Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m², Find the tension in the part of the string joining the pulleys.?

Answer 1

Given in the question :-

Assume that Tension = T'.

Therefore, Torque at the big pulleys = T' × 0.10 N-m

Now for Angular acceleration,

$\alpha = a/r$

α=0.10 × T'/0.20

α=T'/2

or

$a = rT'/2$

a =0.10 × T'/2

a = 0.05 T'

Now Net torque at small side of pulley

$= (T-T') r'$

=(T-T') × 0.05 N-m

∴Angular acceleration

$\alpha = a/r'$

= 0.05 (T-T')/I'

0.05T'/0.05

=0.05(T-T')/0.10

T' =0.5T - 0.5T'

$1.5T' =0.5 T$ -----(i)

Now for the block

$2g -T = 2a$

T = 2g-2a

= 2g-2 × 0.05 T'

= 2g-0.10 T'

Putting this value in eq. (i), we get

$1.5T' = 0.5 (2g-0.10T')$

1.5T' = 9.8-0.05T'

1.55T' =9.8

T' = 9.8/1.55

T' = 6.32 N

Hope it Helps :-)