Question

The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0⋅20 kg-m2. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1⋅0 kg.
Figure

Answer 1

The acceleration of the block if its mass is 1.0 Kg is 10.3 ms².

Explanation:

Step 1:

A is light pulley and B is the pulley that descends I = 0.20 kg-m²

Radius (r) = 20 cm                              

Radius converting centimeter to meter  :

          $r=\frac{20}{100} m=0.20 \mathrm{m}$

Step 2:

Block’s mass = 1 kg.

The equation follows ,

       $\mathrm{T}_{1}=\mathrm{m}_{1} \mathrm{a}$  -----> eqn ( 1 )

       $\left(T_{2}-T_{1}\right) r=l \alpha$  -----> eqn ( 2 )

       $\mathrm{M}_{2} \mathrm{g}-\mathrm{m}_{2} \frac{a}{2}=\mathrm{T}_{1}+\mathrm{T}_{2}$  -----> eqn ( 3 )

       $\left(T_{2}-T_{1}\right)=\frac{t a}{2 R^{2}}=\frac{5 a}{2}$

And,        

          $T_{1}=a$  ( ∵ $=\frac{a}{2 R}$ )

          $T_{2}=\frac{7}{2 a}$

Step 3:

On substituting the values ,

                          $\mathrm{M}_{2 \mathrm{B}}=\mathrm{m}_{2} \frac{\mathrm{a}}{2}+\frac{7 \mathrm{a}}{2}+a$

                       $\frac{2 t}{r^{2} g}=\frac{2 t}{r^{2}} \frac{a}{2}+\frac{9 a}{2} \quad\left(\frac{1}{2}=m r^{2}\right)$

                          98 = 5a + 4.5a

                          9.5a = 98

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

                                $a=10.3 \mathrm{ms}^{2}$          

Thus, the acceleration is 10.3ms².                              

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

10.3 m per second square