-3
$\begingroup$

So the problem Im dealing with has a corresponding diagram below where a mass is being spun in a circular path (at an increasing angular speed) causing the hanging weight to be lifted through the rope. There is a demonstration here by Prof Julius Miller at the 9 minutes and 6 second mark- https://www.youtube.com/watch?v=z3BSkMj1wLc. I wanted someone to explain why this phenmena occurred. Someone did reply and said that T (horizontal)=Mg where M is the hanging weight. But how can T possibly equal Mg? A horizontal force is not related to a vertical force.

expriemntla setup

$\endgroup$
3
  • $\begingroup$ A string can only provide force in its direction. The vertical down force is Mg which gives the tension in the string. So that string pulls with that much.. even after the bend. $\endgroup$
    – Al Brown
    Aug 14, 2021 at 0:48
  • $\begingroup$ Also. People have stopped helping you I think because, I noticed you never upvote anything they write or pick their answers. $\endgroup$
    – Al Brown
    Aug 14, 2021 at 0:49
  • 4
    $\begingroup$ You asked the same question before "Centrifugal" weight spinning causing vertical weight to lift $\endgroup$
    – ACB
    Aug 14, 2021 at 4:25

1 Answer 1

0
$\begingroup$

For $m$ to undergo circular motion, a centripetal force is required, which is provided by the tension. However, resulting from Newton's Third Law pairs, this tension will apply a force equal in magnitude on mass $M$.

If you spin $m$ at the right frequency, then the magnitude of the centripetal force provided by the tension force will be equal in magnitude to $Mg$.

Although tension is acting (almost*) horizontally on $m$, the (assumed massless) string is vertically connected to $M$, and tension will thus act along that vertical axis.

*I say almost, because provided that the point of rotation remains fixed, $m$ can never spin perfectly horizontally since a component of the tension force is needed to counteract $mg$.

Tension will always act in the direction the string is pointing:

enter image description here

$\endgroup$

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