Rotational Motion of blocks and string

[tdusffx]

Two blocks, m1= 1kg and m2 = 2kg, are connected by a light string as shown in the figure (the figure is just shows a pulley with 2 blocks on each sides). If the radius of the pulley is 1m and its moment of inertia is 5kg*m^2, the acceleration of the system in g is:

I have no idea how to do this problem...I tried U1 + K1 = Uf + Kf but it doesn't work because I have very few given information...

also, I couldn't aplly the (t1-t2)R = I(alpha) because its frictionless..so i don't know where to go from here.

 

[Doc Al]

Newton's 2nd law

also, I couldn't aplly the (t1-t2)R = I(alpha) because its frictionless..

Sure you can. The pulley is frictionless, not massless.

You need to analyze the forces on each mass and the pulley. Apply Newton's 2nd law to each to get three (connected) equations. Solve them together and you can figure out the acceleration (and the two tensions).

 

[tdusffx]

so it doesn't matter if I make T1 positive or T2...the problem did not state whether its counter clockwise nor clockwise

and so the equation for the pulley is (t1-t2)R = I(alpha)

T1 = M1g
T2 = M2g

then I just plug and the numbers?

((19.6-9.8)(1))/5 = alpha

 

[Doc Al]

so it doesn't matter if I make T1 positive or T2...the problem did not state whether its counter clockwise nor clockwise

Doesn't matter.

and so the equation for the pulley is (t1-t2)R = I(alpha)

This is OK.

T1 = M1g
T2 = M2g

then I just plug and the numbers?

No. While the forces on M1, for example, are T1 and M1g, these forces are not equal. (If they were equal, the mass would not accelerate.)

Instead you have to carefully apply Newton's 2nd law to each mass (and the pulley) and then combine the equations.