Rotation - Angular acceleration

[cupid.callin]

Homework Statement

The Attempt at a Solution

I need to know what will be angular acceleration of the pulley

i guessed that point A will have same acceleration as of the string and therefore of the block i.e. 2a

so angular acc., α = 2a/r

but it comes out that α = a/r

WHY???

 

[I like Serena]

i guessed that point A will have same acceleration as of the string and therefore of the block i.e. 2a

so angular acc., α = 2a/r

but it comes out that α = a/r

WHY???

It's true that the string at point A has the same acceleration as the block (2a).
However, it will still be that α = a/r.

To understand this, let's say that the block moves to the right by 2x.
Then the string at A will go downward by 2x.
However, the pulley itself will go down by x, and not by 2x.

Since the string goes down by 2x and the pulley goes down by x, the the pulley will rotate by an amount phi = x/r.
Note that the missing string part is found on the right of the pulley where the string length increases by x.

 

[cupid.callin]

To understand this, let's say that the block moves to the right by 2x.
Then the string at A will go downward by 2x.

wont the String at A will go down by x
and also by x on point diametrically opposite to A ???

But i still cant understand why α = a/r

 

[I like Serena]

wont the String at A will go down by x
and also by x on point diametrically opposite to A ???

But i still cant understand why α = a/r

I've drawn the following picture:

Here you have the situation before and after the block moves 2x to the right.
As you can see, point A lowers to A' which is 2x lower.
The pulley moves down by x, and rotates on its circumference by x.
This makes the pulley rotate by an amount φ = x/r (and not by 2x/r!).
With 'α' being the second derivative of φ, and with 'a' being the second derivative of 'x', this yields α = a/r (and not 2a/r!).

 

[cupid.callin]

I've drawn the following picture:
View attachment 33301

Here you have the situation before and after the block moves 2x to the right.
As you can see, point A lowers to A' which is 2x lower.
The pulley moves down by x, and rotates on its circumference by x.
This makes the pulley rotate by an amount φ = x/r (and not by 2x/r!).
With 'α' being the second derivative of φ, and with 'a' being the second derivative of 'x', this yields α = a/r (and not 2a/r!).

Hey thanks a lot "I like Serena"

You really helped me a lot

I had 4 questions stuck due to this problem !!!!

Thanks a Lot !!!!!!!!!!!!!!!!!!