Let the masses of blocks a and b be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.400 kg⋅m2, and the radius of the wheel be 0.100 m . find the magnitude of linear acceleration of block a if there is no slipping between the cord and the surface of the wheel. a. a of block a
b. a of block b
c. alpha of pulley
d. tension side a
e. tension side b

Accelerations of both the sides is 0.6125 , A moves downwards whereas B moves upwards.

=6.125 rad/

Tension on side A = 4.5 × g= 44.1 m/

Tension on side B= 2.0 × g=  19.6 m/

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 and the tension attached with Block B be .

Tensions will be only be due to the weight of the blocks as no other force is present.

= 4.5 × g= 44.1 m/

= 2.0 × g=  19.6 m/

Now, lets make a torque equation about the center of the wheel and find the alpha

×R- ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m  and

           Alpha()= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/

=6.125 rad/

Acceleration = R ×

                    = 0.1 * 6.125

                    =0.6125

Accelerations of both the sides is 0.6125 , A moves downwards whereas B moves upwards.