Question

The descending pulley shown in figure has a radius $20cm$ and moment of inertia $0.20kg-m^2.$ The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is $1.0kg.$

Answer 1

Tension = T (Plane) and T'(Side)

Now we know the formula,

$T=ma$

$T=1.0\times a=a$

Now acceleration ( pulley)$=\frac{a}{2}$

$\alpha =\frac{a}{0.4}m/s^2$

Therefore , Torque (pulley)

$I=0.20kg-m^2$

$\alpha =\frac{T}{I}$ then

$\frac{a}{0.4}=(T^{\prime }-T)\times \frac{0.20}{0.20}$

$T-T^{\prime }=-\frac{a}{0.4}$

$T-T^{\prime }=-2.5a$----(a)

Let the mass of the pulley be M

$M=2\times \frac{0.20}{0.2\times 0.2}=10kg$

$T+T^{\prime }=10\times 9.8-10\frac{a}{2}$

$T+T^{\prime }=98-5a$ ----(b)

adding these a and b eq. value , we get

$2a=98-7.5a$

$9.5a=98$

$a=\frac{98}{9.5}$

$a=10.31m/s²$