Question

The descending pulley shown in figure, has a radius 20 cm and moment of inertia 0.20 kg-m², The fixed pulley is light and the horizontal plane friction less. Find the acceleration of the block if its mass is 1.0 kg.?

Answer 1

Given in the question :-

Assume that , Tension = T (Plane) and T'(Side)

Now we know the formula,

T =ma

T = 1.0 × a

T= a

Now acceleration ( pulley) = a/2

and$\alpha = a/2r$

α = a/0.4 m /s²

Therefore , Torque (pulley)

$I= (T'-T) * 0.20$

I = 0.20 kg-m²

as we know,

α = T/I then

a/0.4 = (T'-T)*0.20/0.20

T-T' = -a/0.4

T-T' = -2.5 a ----(a)

Let the mass of the pulley be M .

$Mr^2/2 = I$

M = 2 × 0.20/0.2× 0.2

M =10 kg

Now.

$Mg -T-T' = Ma/2$

$T+T' =Mg-\frac{Ma}{2}$

T+T' =10*9.8 - 10a/2

T+T' =98 - 5a ----(b)

adding these a and b eq. value , we get

$2T = 98 - 5a - 2.5a$

2a = 98 -7.5a

9.5a = 98

a = 98/9.5

a = 10.31 m/s²

Hope it Helps :-)