block 3M slides downward without friction. The string connecting blocks 3M and M is ideal (that is, its mass can be neglected and it is not stretchable). The pulley, a uniform solid disc of mass M and radius r , rotates without slipping. Find the acceleration of block 3M.

(2*g/9)(3sin(theta)-1)

Hi enjoy,I got (2*g/5)(3*sin(theta-1) which is not in options.. Can you please show me how to get 2*g/9? I appreciate your help.

In question no.2 a) torque would be 2*F*r/2? I am confused that from top view, which force we have to use?

a) alpha= 2F/mr ccw

b) alpha= 2F/mr cw
c) alpha= 8F/3mr ccw

To find the acceleration of block 3M, we can start by analyzing the forces acting on the system.

Let's consider the forces acting on each individual block:
- For block M, there are two forces: the tension force T in the string and the gravitational force Mg.
- For block 3M, there are three forces: the tension force T in the string, the gravitational force 3Mg, and the normal force N exerted by the pulley.

Now, let's use Newton's second law (F = ma) to find the equations of motion for each block:

For block M:
1. In the vertical direction:
- T - Mg = Ma, where A is the acceleration of block M.
Since T = Mg - Ma, we can substitute this into the equation above:
- (Mg - Ma) - Mg = Ma
- -Ma - Mg = Ma
- -2Ma = Mg
- 2A = g
- A = g/2

For block 3M:
2. In the vertical direction:
- T - 3Mg = 3Ma, where a is the acceleration of block 3M.
Since T = 3Mg - 3Ma, we can substitute this into the equation above:
- (3Mg - 3Ma) - 3Mg = 3Ma
- -3Ma - 3Mg = 3Ma
- -6Ma = 3Mg
- 6a = 3g
- a = g/2

Thus, the acceleration of block 3M is g/2, the same as the acceleration of block M.