# Forces (moment of inertia) Question

• Lord Dark
In summary: I get it :) ,, actually I didn't focus on what he wrote ,, thanks again ,, am bothering you this day (my last day :D)In summary, the pulley has a moment of inertia of 0.5*M*R2^2. The angular acceleration of the pulley is 0.5*M*R^2. The block A and block B each have an acceleration of -0.5*M*R^2. The tensions in the strings are 0.5*M*R. The block B hits the ground at a speed of Vb = 4.0*M*R*R^2.

## Homework Statement

hi everyone,, I need an explanation for this question:
A 1.0 kg block A is hanged by a light string from a pulley of mass
M = 1.0 kg. Another block B of mass 4.0 kg is attached to a mass-less rope wound around a mass-less shaft. The pulley has a radius 260.0 Rcm= and the shaft has a radius. The block B is released from rest when it is at height h = 4.0 m above the ground. 130.0 cm R=
The rotational inertia (moment of inertia) of the pulley with respect to its center of mass is.
I(com)=0.5*M*R2^2

(a) Find the angular acceleration of the pulley.
(b) Find the accelerations of the blocks.
(c) Determine the tensions in the strings.
(d) Find the speeds of the block A and block B just before the block B hits the ground.

## The Attempt at a Solution

Actually ,, I have the answer ,, but the problem I don't know how did he get the moment of inertia like that (in the attachment)

#### Attachments

Lord Dark said:
Actually ,, I have the answer ,, but the problem I don't know how did he get the moment of inertia like that (in the attachment)
You have the mass of the pulley and are told the equation to use to find its moment of inertia. Maybe you can restate your question.

I mean shouldn't I=I(com)+M1R1^2+M2R2^2 ??

Lord Dark said:
I mean shouldn't I=I(com)+M1R1^2+M2R2^2 ??
No, those masses are not rigidly attached to the pulley. The only thing rotating here is the pulley. Treat this as three separate objects: m1, m2, and m3.

then the teacher's answer is wrong ? and TbR1-TaR2=I(com)*alfa only ?? not like his answer ? (in the attachment)

Lord Dark said:
then the teacher's answer is wrong ? and TbR1-TaR2=I(com)*alfa only ?? not like his answer ? (in the attachment)
What makes you think that the teacher is doing anything other than TbR1-TaR2=I(com)*alfa? (Where I is just the rotational inertia of the pulley.) That's his equation #3.

(You could lump them all together, but I would not advise it. Do it like your teacher did. Write three separate equations, then combine them.)

aha ,, now I get it :) ,, actually I didn't focus on what he wrote ,, thanks again ,, am bothering you this day (my last day :D)