I am wondering, when solving rigid body exercises, how can I express the relationship between linear and angular acceleration for a general case? E.g. what would be the linear acceleration in function of the angular one of a $1\, \mathrm m$ rod that is rotating through a fixed point $0.6\, \mathrm m$ away from its mass center? And what about the case of a yoyo?

Edit: I know the basic relation $a= \alpha R$, but I am confused as to how to choose $R$ and my textbook is not helping.


As you stated, the angular acceleration, tangential linear acceleration and distance between the reference point and the object are related using the following formula:

$$\vec{a} = \vec{\alpha} \times \vec{r}$$

$\vec{r}$ is merely the displacement vector between your choice of reference point and the object. The object needn't necessarily move in a circle for the formula to work.

The choice of reference point is arbitrary; you can choose any point. We often use the centre of mass or the centre of rotation as it simplifies the math but there is no rule which states that you should do your calculations about that point only.

  • $\begingroup$ Thank you, your explanation is really clear. I am still wondering, then, in a yoyo, how come we choose r to be the axle radius and not the actual yoyo radius then? $\endgroup$ – Bee Apr 23 '17 at 14:26
  • $\begingroup$ It depends on what motion you are talking about. The motion of the center of mass of the yoyo is not the same thing rotation about the center of mass. For example, the earth rotates around its own axis and it also rotates around the sun. These two rotations are different. It depends on what rotation you are talking about. If you are rotating your yoyo in a circular direction, then you have two rotational motions: one of the yoy around your hand and the other one being the rotation of yoyo along its axis. $\endgroup$ – Yashas Apr 23 '17 at 14:29
  • $\begingroup$ If you can provide a better description of how the yoyo is moving, then I can give you a better explanation. I am making assumptions based on the question you are asking (your comment to this answer) so it would be better if you could give a better description of the problem so that I can get rid of those assumptions (in my previous comment). $\endgroup$ – Yashas Apr 23 '17 at 14:30
  • $\begingroup$ I am basing myself on the exercise proposed and thoroughly explained here: wtfprofessor.com/rotational-motion It turns out I understand how to solve everything, but I am not exactly sure why a=α∗b and not a=α∗R $\endgroup$ – Bee Apr 23 '17 at 14:32
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    $\begingroup$ The string is providing the torque and the rate at which the string is unwrapping is given by $\omega$. The string wraps around the axle which has a radius of $b$. $\endgroup$ – Yashas Apr 23 '17 at 14:47

A point, whose position vector is $\vec r$, of a rigid body with angular velocity $\vec \omega$ has velocity $\vec v=\vec\omega\times\vec r$. By differentiating $\vec v$ with respect to time we obtain the acceleration $$\vec a=\vec\alpha\times\vec r+\vec\omega\times\vec v,$$ where $\vec\alpha=d\vec\omega/dt$ is the angular acceleration.

The first term, $\vec\alpha\times\vec r$, is parallel to the velocity vector and is normally called tangential acceleration. The second term, $\vec\omega\times\vec v$ is radially inwards and is called centripetal acceleration.

  • $\begingroup$ Lol, I totally forgot to mention about the centripetal acceleration in my answer. But I guess the OP was actually looking for the tangential acceleration. $\endgroup$ – Yashas Apr 23 '17 at 14:45
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    $\begingroup$ @Yashas Indeed I believe he is looking for the tangential acceleration but it is worth to mention the general case =) $\endgroup$ – Diracology Apr 23 '17 at 14:47

There are two fundamental equation for linear acceleration and angular acceleration; these are:

$m \ddot{x} = F, \theta \ddot{\alpha} = M$.

it holds further $M = (x-x_0) \times F$ for some center of mass coordinate $x_0$. If you combine these equations you will get a relation between linear and angular acceleration.

In the case of a yoyo you must add some kinematic conditions like

$z = R \phi$

with the yoyo radius $R$, the height coordinate $z$ and the rotation angle $\phi$.


Your question in incomplete. If the rod is rigid, every point in the rod is undergoing equal angular acceleration. But because every point in the rod is situated at a unique radius from the axis of rotation, each point will experience a unique value of linear acceleration given by the equation $$a = \alpha \times (radius)$$


protected by Qmechanic Apr 23 '17 at 14:05

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