0
$\begingroup$

Suppose an ideal string is passed from a pulley of any mass, (placed on a smooth horizontal surface), with sufficient friction so that the rope doesn't slide on it! Now if the pulley is attached to a block over an ideal pulley to another block hanging! Now the block hanging can translate downwards with any acceleration! So my question is why does the pulley has less acceleration in comparison to the hanging block?

the situation is shown in the image! Why is "a>á"?

$\endgroup$
5
  • $\begingroup$ Your question is not entirely clear. From the looks of your diagram, the pulley has an angular acceleration that is directly related to the acceleration of the block. $\endgroup$ Feb 16, 2020 at 13:48
  • $\begingroup$ I know that! I was just asking why the linear acceleration of the pulley on the table is less than the linear acceleration of the block hanging... $\endgroup$ Feb 16, 2020 at 14:05
  • $\begingroup$ Isn’t the pulley on the table fixed to an axel that is attached to the ground? The pulley does not move because the axel is placing a force on the center of the pulley. $\endgroup$ Feb 16, 2020 at 14:07
  • $\begingroup$ Think like it is on a massless block $\endgroup$ Feb 16, 2020 at 17:13
  • $\begingroup$ What is the triangle in your drawing? $\endgroup$ Feb 16, 2020 at 17:53

1 Answer 1

1
$\begingroup$

Because the acceleration of the mass is the same as the acceleration of a dot on the string. As the pulley moves, the string unwinds due to the rotation. You have $dx=dx'+Rd\theta$, so $a=a'+R\alpha$

enter image description here

The black square is a mark on the string just to show how much it moved: dx, which is the total displacement of the string, which is the same as that of the hanging block. dx' is how much the center of the pulley moved, and $d\theta$ how much it rotated as he moved.

$\endgroup$
3
  • $\begingroup$ Sir I didn't get that! Could you elaborate on the same! Which mass are referring to? Plz elaborate $\endgroup$ Feb 16, 2020 at 17:14
  • $\begingroup$ I meant the hanging block. I will post a drawing clarifying it soon $\endgroup$
    – user65081
    Feb 16, 2020 at 21:16
  • $\begingroup$ Thank you so very much sir! It's clear now! Thanks a lot again! $\endgroup$ Feb 17, 2020 at 3:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.