Question

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

Answer 1

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$, A moves downwards whereas B moves upwards.

$\alpha$=6.125 rad/$s^{2}$

Tension on side A = 4.5 × g= 44.1 m/$s^{2}$

Tension on side B= 2.0 × g=  19.6 m/$s^{2}$

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 $T_{2}$ and the tension attached with Block B be $T_{1}$ .

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

$T_{2}$ = 4.5 × g= 44.1 m/$s^{2}$

$T_{1}$ = 2.0 × g=  19.6 m/$s^{2}$

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

$T_{2}$×R- $T_{1}$×R= MI( Moment of Inertia of Wheel)× Alpha

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

           Alpha($\alpha$)= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/$m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$=6.125 rad/$s^{2}$

Acceleration = R ×$\alpha$

                    = 0.1 * 6.125

                    =0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$, A moves downwards whereas B moves upwards.