A light string that is attached to a large block of mass $4m$ passes over a pulley with negligible rotational inertia and is wrapped around a vertical pole of radius $r$, as shown above. The system is released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each end. The

rotational inertia of the pole and the rod are negligible.

(a) Determine the rotational inertia of the rod-and-block apparatus attached to the top of the pole.

(b) Determine the downward acceleration of the large block.

(c) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare with the value 4mgD ? Check the appropriate space below.

** Greater than 4mgD** Equal to 4mgD

____ Less than 4mgD

Justify your answer.

a) The answer is simply $I=2mL^{2}$.

b) Using $4mg-T=4ma$ and $I\alpha = I\alpha/r =Tr$ we can find $a=2gr^{2}/(2r^{2}+L^{2})$.

c) One way is to use conservation of energy to say that the total kinetic energy of the three blocks should equal the change in potential energy $4mgD$. I also tried to use energy conservation and kinematics. The total kinetic energy of the three blocks seems to be $1/2 mv^{2}+1/2mv^{2}+4mv^{2}= 5mv^{2}$ (assuming all masses have the same linear velocity).

Then since I know the acceleration from b) I can find that $v=2gr^{2}/(2r^{2}+L^{2})t$ and so $D=gr^{2}/(2r^{2}+L^{2})t^{2}$. Solving for $t$ and replacing into $v$ I can find the velocity at distance $D$ and then plug into the kinetic energy $5mv^{2}$. Doing this however I dont get $4gmD$. Is my assumption that the linear velocities of all blocks is the same wrong? Or what is wrong with the reasoning?