Question

The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0⋅5 kg-m2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4⋅0 kg block.
Figure

Answer 1

The acceleration of the 4.0 Kg block is  0.247 m/s².

Explanation:

Step 1:

Given values in the question,  

Assume, 4 kg acceleration block = a

                                        Tension = T

Step 2:

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

So,  

        $4 g \cdot \sin 45^{\circ}-T=4 a$

        $4 g \cdot \frac{1}{\sqrt{2}}-T=4 a$

        $4 g \cdot \frac{1}{\sqrt{2}}-4 a=T$

Also,

        $T^{\prime}-2 g \cdot \sin 45^{\circ}=2 a$

As the acceleration is same.

        $T^{\prime}=2 a+2 g \cdot \sin 45^{\circ}$

        $T^{\prime}=2 a+2 g \cdot \frac{1}{\sqrt{2}}$

Step 3:

Now,

So the Overall Torque  $=\left(T-T^{\prime}\right) \times 0.10$

                                     $=0.10 \times\left\{\left(4 g \cdot \frac{1}{\sqrt{2}}-4 a\right)-\left(2 a+2 g \cdot \frac{1}{\sqrt{2}}\right)\right\}$

                                     $=0.10\left(2 g \frac{1}{\sqrt{2}}-6 a\right)$

Now we have to accelerate angular,

                                          $\alpha=\frac{a}{r}$

                                          $\alpha=\frac{a}{0.10}$

Step 4:

Hence, from Torque = Iα

     $0.10\left(2 g \frac{1}{\sqrt{2}}-6 a\right)=1 \alpha$  

     $0.10\left(2 g \frac{1}{\sqrt{2}}-6 a\right)=0.50 \times \frac{a}{0.10}$

            $=0.50 \times \frac{a}{0.10}=5 a$        

                           $50 a=2 g \frac{1}{\sqrt{2}}-6 a$

                   $50 a+6 a=2 g \frac{1}{\sqrt{2}}$

                    $56 a=2 g \frac{1}{\sqrt{2}}$

                          $=\frac{2 \times 9.8 \times \sqrt{2}}{2}$

                          $= 9.8\sqrt{2}$  

Hence,            $a=9.8 \times \frac{1.41}{56}$

                       $a$ $=0.2474 \mathrm{m} / \mathrm{s}^{2}$

Therefore, the acceleration is 0.247 m/s².      

Answer 2

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

0.247 m/s will be the answer