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**1. Homework Statement**

The system shown in the diagram contains two blocks, of masses 1.9 kg and 5.7 kg, connected by a light string over a pulley of radius 0.15 m and rotational inertia 2.8 kg m 2 . The block of mass 5.7 kg is free to slide on a horizontal frictionless surface and the pulley is free to rotate on a frictionless fixed horizontal axle passing through its center. Find the magnitude of the angular acceleration α of the pulley once the system is released from rest. The string connecting the blocks does not slip on the pulley. Enter your answer in rad/s 2 and use g = 9.8 m/s 2 .

The system shown in the diagram contains two blocks, of masses 1.9 kg and 5.7 kg, connected by a light string over a pulley of radius 0.15 m and rotational inertia 2.8 kg m 2 . The block of mass 5.7 kg is free to slide on a horizontal frictionless surface and the pulley is free to rotate on a frictionless fixed horizontal axle passing through its center. Find the magnitude of the angular acceleration α of the pulley once the system is released from rest. The string connecting the blocks does not slip on the pulley. Enter your answer in rad/s 2 and use g = 9.8 m/s 2 .

**2. Homework Equations**

τ=Torque

t=tension

A=linear acceleration

a

r=radius

τ=I*a

A=r*a

F=A*m

τ=Torque

t=tension

A=linear acceleration

a

_{Θ}=Angular accelerationr=radius

τ=I*a

_{Θ}A=r*a

_{Θ}F=A*m

**3. The Attempt at a Solution**

box with mass m is box 1

box with mass 3m is box 2

ok so, drawing a FBD of both boxes we get. F

and F

m*g-m*a=3*m*a. So we get a=g/4.

Now since we know acceleration we can use it in a equation.

τ=r*t.where t= (m*g-m*a) or (3*m*a).

rearranging the formula τ=I*a

we substitute the variables in. we get τ/I=a

and we know all the variables there substituting them we get. 0.15*(3*1.9*2.45)/2.8 = a

and the final answer is a

but when I put it down as the answer it marks me incorrect. Could someone help me out to see where I went wrong ? thank you :)

box with mass m is box 1

box with mass 3m is box 2

ok so, drawing a FBD of both boxes we get. F

_{1}=m*g-t=m*a. t=m*g-m*a.and F

_{2}=3*m*a=t. Since tension and acceleration is the same throughout.m*g-m*a=3*m*a. So we get a=g/4.

Now since we know acceleration we can use it in a equation.

τ=r*t.where t= (m*g-m*a) or (3*m*a).

rearranging the formula τ=I*a

_{Θ}.we substitute the variables in. we get τ/I=a

_{Θ}. = r*(3*m*a)/I = a_{Θ}.and we know all the variables there substituting them we get. 0.15*(3*1.9*2.45)/2.8 = a

_{Θ}.and the final answer is a

_{Θ}= 0.748125.but when I put it down as the answer it marks me incorrect. Could someone help me out to see where I went wrong ? thank you :)