rotating ball

Suppose I pull out a rope and rotate a small mass as in the diagram with its speed constantly increasing


$Q_1$ is the path of the rotating mass $m$
$Q_0$ is the center of the rotating mass
$R$ is the rope holding the rotating mass
$P$ is the path of inertia
$a_{cf}$ is the centrifugal acceleration
$a_{cp}$ is the centripetal acceleration
$F_{cf}$ is the centrifugal force
$F_{cp}$ is the centripetal force
$v$ is the velocity of the object along $Q_1$
$a$ is the acceleration of the object along $Q_1$

How do I get the acceleration $a$ of the mass $m$ along it's path or rather How do you get the acceleration of a rotating mass from its centripetal and centrifugal accelerations?

Edit: My initial calculation $$ a_{cp} = \omega ^2r \\ \text{ where } \omega \text{ is the angular speed } \\ a_{cp} = \omega (\omega r) \\ \text{ since } v = \omega r \\ a_{cp} = \omega v \\ \text{ also } v = at \\ a_{cp} = \omega (at) \\ \text{ from } \omega = \theta/t \\ a_{cp} = \omega t(a) \\ a_{cp} = \theta (a) \\ a = a_{cp}/\theta $$

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    $\begingroup$ You should show your own efforts too while asking a question. $\endgroup$ – Harshit Joshi Jan 5 '19 at 13:58
  • $\begingroup$ @harshit54 note taken. I made a comment on @CyborgOctopus answer showing my initial calculations. I'ld edit the question $\endgroup$ – LiNKeR Jan 5 '19 at 14:50

You have to be careful about how to interpret this diagram. If the centrifugal force is implied to be acting on the rope, that's fine. However, if you think of both forces as acting on the ball, it would not rotate at all because the centrifugal and centripetal forces balance each other out, causing its acceleration along the path to be zero. For the ball, centrifugal force only comes into play if you want to regard it as not accelerating. If you are in a frame of reference from which it is rotating in uniform circular motion, there is no centrifugal force on the ball, this force is only exerted by it on the rope. In this case, the net force on the ball is centripetal force ($\mathbf{F_{cp}}$) and the acceleration is $\mathbf{a} = \mathbf{a_{cp}}$. The acceleration of any object in uniform circular motion is always the inward-directed centripetal acceleration.

  • $\begingroup$ "The acceleration of any object in..." by uniform circular motion you mean a constant circular path ? $\endgroup$ – LiNKeR Jan 5 '19 at 8:29
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    $\begingroup$ Yeah, a circular path with a fixed size at constant speed. $\endgroup$ – CyborgOctopus Jan 5 '19 at 8:36
  • $\begingroup$ Initially I was thinking since $$ a_{cp} = \omega ^2r \\ \text{ where } \omega \text{ is the angular speed } \\ a_{cp} = \omega (\omega r) \\ \text{ since } v = \omega r \\ a_{cp} = \omega v \\ \text{ also } v = at \\ a_{cp} = \omega (at) \\ \text{ from } \omega = \theta/t \\ a_{cp} = \omega t(a) \\ a_{cp} = \theta (a) \\ a = a_{cp}/\theta $$ is this correct too or is it always $$ a = a_{cp} $$ $\endgroup$ – LiNKeR Jan 5 '19 at 10:13
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    $\begingroup$ @LiNKeR That's incorrect because you use the formula $v = at$, which assumes that the initial velocity is zero and the acceleration is constant. For something moving in a circle, the acceleration is always changing direction and it has a nonzero initial velocity. $a = a_{cp}$ is always right. $\endgroup$ – CyborgOctopus Jan 5 '19 at 18:39
  • $\begingroup$ Oh! that's true from $$v = u + at \\ ( u = 0 ) $$ big thanks @CyborgOctopus for clearing up things! $\endgroup$ – LiNKeR Jan 5 '19 at 23:46

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