Answer:
Accelerations of both the sides is 0.6125 [tex]m/s^{2}[/tex], A moves downwards whereas B moves upwards.
[tex]\alpha[/tex]=6.125 rad/[tex]s^{2}[/tex]
Tension on side A = 4.5 × g= 44.1 m/[tex]s^{2}[/tex]
Tension on side B= 2.0 × g= 19.6 m/[tex]s^{2}[/tex]
Explanation:
As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.
Observe the figure attached, let the tension with Block A be [tex]T_{2}[/tex] and the tension attached with Block B be [tex]T_{1}[/tex] .
Tensions will be only be due to the weight of the blocks as no other force is present.
[tex]T_{2}[/tex] = 4.5 × g= 44.1 m/[tex]s^{2}[/tex]
[tex]T_{1}[/tex] = 2.0 × g= 19.6 m/[tex]s^{2}[/tex]
Now, lets make a torque equation about the center of the wheel and find the alpha
[tex]T_{2}[/tex]×R- [tex]T_{1}[/tex]×R= MI( Moment of Inertia of Wheel)× Alpha
where, R= Radius of the wheel=0.100m and
Alpha([tex]\alpha[/tex])= Angular acceleration of the wheel
MI of the wheel= 0.400 kg/[tex]m^{2}[/tex]
[tex](44.1-19.6)R=0.400\alpha[/tex]
[tex]\alpha = \frac{24.5 * 0.100}{0.400}[/tex]
[tex]\alpha[/tex]=6.125 rad/[tex]s^{2}[/tex]
Acceleration = R ×[tex]\alpha[/tex]
= 0.1 * 6.125
=0.6125 [tex]m/s^{2}[/tex]
Accelerations of both the sides is 0.6125 [tex]m/s^{2}[/tex], A moves downwards whereas B moves upwards.
