There is a block of mass 1 on a frictionless surface attached to a string. The string goes over a frictionless pulley. The pulley is a hoop with a rotational inertia of I=MR^2. The other end of the string is attached to a hanging block of mass 2. The hanging block is then released. To find the acceleration of the system, why could one not treat Mass 1 as a point particle on the hoop, and create a point particle moment of inertia for the Mass 1 block, and add then that "I" to that of the hoop? And then one could have the Fg of mass 2 as the only force causing a torque on the pulley.
To find the acceleration of the system in this scenario, you cannot simply treat Mass 1 as a point particle on the hoop and add its moment of inertia to that of the hoop. This is because the block of Mass 1 has a physical size and shape that cannot be ignored.
In order to analyze the system correctly, you need to consider the rotational dynamics of the pulley and the linear dynamics of both the hanging block and the block on the surface.
To derive the expression for the acceleration of the system, you can start by considering the forces acting on the hanging block. The only force acting on it is the force of gravity (Fg = m2 * g), where m2 is the mass of the hanging block and g is the acceleration due to gravity.
Next, consider the forces acting on the block on the surface. The tension in the string provides an upward force (T) and the gravitational force provides a downward force (Fg = m1 * g), where m1 is the mass of the block on the surface.
Since the string is assumed to be ideal (massless and frictionless), the tension in the string will be the same throughout its length. Therefore, the tension in the string at the hanging block will also be T.
Now, let's consider the rotational dynamics of the pulley. The torque on the pulley is given by the tension in the string multiplied by the radius of the pulley (T * R). This torque is responsible for the angular acceleration of the pulley.
The moment of inertia of the pulley is given by I = MR^2, where M is the mass of the pulley and R is its radius.
The angular acceleration of the pulley (α) can be related to the linear acceleration of the blocks by the equation α = a/R, where a is the linear acceleration of the hanging block and the block on the surface. Since the string length does not change, the linear acceleration of both blocks will be the same and can be referred to as a.
Now, we can equate the torque on the pulley to the moment of inertia of the pulley times its angular acceleration: T * R = I * α.
Substituting the value of the moment of inertia of the pulley and rearranging the equation, we get: T = (M + m1) * a / R.
Also, the acceleration of the hanging block (a) is equal to the acceleration of the block on the surface due to the string being ideal.
Next, we can set up the equations for the linear dynamics of the blocks. Using Newton's second law, we have:
For the hanging block: Fg = m2 * g = m2 * a.
For the block on the surface: T - Fg = m1 * a.
Now, substitute the value of T from the pulley equation into the equation for the block on the surface. Simplifying the equation, we get:
(M + m1) * a / R - m1 * g = m1 * a.
Simplifying further, we find:
A = g * (m2 - m1) / (m1 + M).
Hence, the acceleration of the system in this setup is given by the expression above.
By analyzing the rotational dynamics of the pulley and the linear dynamics of the blocks, we arrive at the correct expression for the system's acceleration. Simply treating Mass 1 as a point particle on the hoop would ignore the physical size and shape of the block, leading to an incorrect conclusion.