**Problem Statement:**

A merry-go-round of radius R rotates with constant angular velocity **ω**. A boy throws a ball from its center directly to an observer *O’* sitting on the edge of the merry-go-round. The initial velocity of the ball with respect to *O’* is **v’**. Another observer *O* is at rest on the ground (see figure).

Calculate as a function of the givens:

- The Coriolis acceleration of the ball at the initial time.
- The centrifugal acceleration of the ball at the same time.
- What is the trajectory of the ball with respect to
*O*? And with respect to*O’*?

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**Solution:**

In this page you can find an explanation of both the Coriolis and the centrifugal acceleration.

The Coriolis acceleration is given by:

Where **ω** is the angular velocity of the merry-go-round and **v’** is the velocity of the ball with respect to *O’*, the rotating frame of reference (which is a non-inertial frame of reference since it rotates with the merry-go-round).

First we are going to determine the direction of the cross product **ω** ⨉ **v’**. The Coriolis acceleration will have opposite direction due to the minus sign included in its definition.

In order to make it easier to apply the right hand rule we have drawn a side view of the merry-go-round.

As you can see in the figure above, **ω** ⨉ **v’** is perpendicular to the screen and points inwards (direction –**k**).

The magnitude of the Coriolis acceleration is:

because **ω** and **v’** are perpendicular.

Finally, the Coriolis acceleration is given by:

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The centrifugal acceleration is:

Where **ω** is the angular velocity of the merry-go-round and **r** the position vector of the ball with respect to *O’*.

We determine the direction of **ω</strong > ⨉ r using the right hand rule:**

and then we multiply **ω** and **ω** ⨉ **r** using again the right hand rule:

The centrifugal acceleration has opposite direction because of the minus sign included in its definition.

Then we calculate the magnitude of the centrifugal acceleration in two steps. First we calculate the magnitude of the first cross product:

And then the second one:

Finally, the centrifugal acceleration is given by:

With respect to *O* the ball will follow a straight-line path and with respect to *O’* a spiral.