Pulleys and rotational motion problem

Two blocks are joined by a light string that passes over the pulley shown above, which has radius $R$ and moment of inertia $I$ about its center. $T_1$ and $T_2$ are the tensions in the string on either side of the pulley and $α$ is the angular acceleration of the pulley. Which of the following equations best describes the pulley's rotational motion during the time the blocks accelerate?
(A) $m_2$$g$$R$ = $I$$α (B) T_2$$R$ = $I$$α (C) (T_2 - T_1)R = I$$α$
(D) ($m_2$ – $m_1$)$g$$R = I$$α$

The correct answer is C. The reasoning is that the net torque is equal to $T$$_2R - T_1R. What I'm having trouble understanding is why the torque produced by m_2 is simply T$$_2$R. $m_2$ is accelerating downwards. Wouldn't the free body diagram of $m_2$ be $mg$ pointing downwards and $T_2$ pointing upwards? Thus, shouldn't the net force acting on the block be equal to $mg - T_2$, and consequently, the torque caused by $m_2$ be equal to $R(m_2g - T_2)$? Why are we allowed to ignore the weight of the block?

$T_2$ has all the relevant information in it. It and $T_1$ are the only forces actually acting on the Pulley, $T_2$ will have some dependence on $m_2$, but since the question defines a tension force acting from the weight to the pulley for you, there's no reason to not just use that.

So, with the knowledge that all we need are $T_1$ and $T_2$, and the radius to calculate the torque, $$\tau_1 = T_1 R$$ $$\tau_2 = T_2 R$$ And so the net torque is: $\tau_{net} = (T_2 - T_1)R = I\alpha$

Separate free body diagrams are required for the block $m_2$ and the pulley. That is why they are called free body diagrams : we isolate them from each other and consider only the forces acting directly on each.

The forces which act directly on the pulley are the tensions $T_1, T_2$ in the string. The string transmits these forces between the blocks and the pulley. The weight of the block does not itself act on the pulley, so we ignore it when considering the forces on the pulley.

The weight of the block is not being ignored altogether. It affects the value of $T_2$. We take account of it when we look at the block as a free body and write an equation of motion for it : $m_2g-T_2=m_2a$.

The fact that the string is inextensible and does not slip against the pulley also relates the accelerations $a$ of $m_2$ and $m_1$ to the angular acceleration $\alpha$ of the pulley.

• "Tension" is a force within string molecules. It doesn't affect the pulley. The pulley exerts a reaction force on the string: and thus an equal force is exerted on the pulley. this is the force exerted on the pulley. – satan 29 Jun 11 at 19:08