In a swing ride in an amusement park, the angle and speed of a seat in circular motion can be modelled by the banking angle equation:

$$\tan \theta=\frac{r\omega^2}{g}$$

Since tan of $90^{\circ}$ or $\pi/2$ rad is undefined, does this mean the seat can't be perpendicular to the vertical (the pole)? enter image description here

I would also appreciate it if someone could provide an explanation as to why the banking angle increases with increasing angular velocity.


2 Answers 2


In order to understand from where the equation for acceleration comes from, first draw the free-body diagram of a seat. In the diagram below, $\theta$ is the displacement angle, $G = mg$ is the body weight, and $T$ is the tension (pull) excerted by the rope.

In equilibrium, the seat does not move in vertical direction. First Newton's law for vertical component is:

$$G - T \cos \theta = 0$$

which equals:

$$T = \frac{mg}{\cos \theta} \tag 1$$

However, there is movement in horizontal direction (circular motion). Second Newton's law for horizontal component is:

$$T \sin \theta = m a \tag 2$$

where $a$ is acceleration towards the center of rotation (centripetal acceleration).

By combining Eqs. (1) and (2) we get

$$\tan \theta = \frac{a}{g} \tag 3$$

For circular motion, the centripetal acceleration is defined as:

$$a = \frac{v^2}{R}$$

where $v$ is speed which is constant, and $R$ is radius. The speed could also be expressed as:

$$v = \frac{2 \pi R}{T} = \omega R$$

where $T$ is time for one full revolution, and $\omega$ is circular frequency. The centripetal acceleration can also be defined as:

$$a = \omega^2 R \tag 4$$

By combining Eqs. (3) and (4) we obtain the final form

$$\boxed{\tan \theta = \frac{\omega^2 R}{g}}$$

You can never reach an angle of $\theta = 90^\circ$ as in that case there is no vertical component of the tension ($T \cos \theta$) which must cancel the weight ($G = mg$). Because of this the body would also move (fall) in the vertical direction.

Here I derive the expression for centripetal (radial) acceleration. Before we start, let's define position, velocity and acceleration vectors in two dimensions:

$$\vec{r} = x \hat{\imath} + y \hat{\jmath}, \qquad \vec{v} = v_x \hat{\imath} + v_y \hat{\jmath}, \qquad \vec{a} = a_x \hat{\imath} + a_y \hat{\jmath}$$

where $\hat{\imath}$ and $\hat{\jmath}$ are unit vectors that span horizontal plane.

Time derivative of displacement is velocity, while derivative of velocity is acceleration:

$$\frac{d}{dt}\vec{r} = \vec{v} = \dot{x} \hat{\imath} + \dot{y} \hat{\jmath}, \qquad \frac{d}{dt}\vec{v} = \vec{a} = \dot{v}_x \hat{\imath} + \dot{v}_y \hat{\jmath}$$

where $v_x = \dot{x}$, $v_y = \dot{y}$, $a_x = \dot{v}_x$, and $a_y = \dot{v}_y$.

We will also need equation for scalar product of two vectors $\vec{c} = (c_\imath, c_\jmath)$ and $\vec{e} = (e_\imath, e_\jmath)$:

$$\vec{c} \cdot \vec{e} = |\vec{c}| |\vec{e}| \cos \alpha = c_\imath e_\imath + c_\jmath e_\jmath$$

where $\alpha$ is angle between the two vectors, while $|\vec{c}|$ and $|\vec{e}|$ are vector length:

$$|\vec{c}|^2 = c_\imath^2 + c_\jmath^2$$

Finally, we define $|\vec{r}| = R$, $|\vec{v}| = v_0$, and $|\vec{a}| = a_0$.

Let's now start from the equation for position and velocity in circular motion:

$$x^2 + y^2 = R^2 \tag 5$$ $$v_x^2 + v_y^2 = v_0^2 \tag 6$$

where $R$ is radius of the motion which is assumed to be constant, and $v_0$ is speed of the motion which is assumed to be time-varying.

Take the time derivative of the Eq. (5):

$$x v_x + y v_y = 0 \quad \rightarrow \quad v_0 R \cos \phi = 0$$

where $\phi$ is angle between position and velocity vectors. We conclude that the angle is $\phi = 90^\circ$, i.e. position and velocity vectors are perpendicular!

Take the time derivative of the above equation:

$$x a_x + y a_y = -v_0^2 \quad \rightarrow \quad a_0 R \cos \rho = -v_0^2$$

where $\rho$ is angle between position and acceleration vectors. We conclude that the angle is in 2nd or 3rd quadrant, i.e. position and acceleration vectors point in opposite directions!

Take the time derivative of the Eq. (6):

$$v_x a_x + v_y a_y = v_0 \dot{v}_0 \quad \rightarrow \quad a_0 \cos \varepsilon = \dot{v}_0$$

where $\varepsilon$ is angle between velocity and acceleration vectors. Since $a_0 \cos \varepsilon$ is projection of vector $\vec{a}$ on vector $\vec{v}$, the tangential acceleration is defined as:

$$\boxed{a_{||} = a_0 \cos \varepsilon = \dot{v}_0}$$

In other words, the tangential acceleration changes the velocity magnitude. When the circular motion is uniform (constant speed), the tangential acceleration is zero.

Now we find relation between angles $\phi$, $\rho$ and $\varepsilon$:

$$\rho = \phi + \varepsilon = 90^\circ + \varepsilon \quad \rightarrow \quad \cos \rho = - \sin \varepsilon$$

The equation for radial acceleration becomes:

$$a_0 R \cos \rho = -v_0^2 \quad \rightarrow \quad a_0 \sin \varepsilon = \frac{v_0^2}{R}$$

Since $a_0 \sin \varepsilon$ is projection of vector $\vec{a}$ on axis perpendicular to vector $\vec{v}$ (which is in the opposite direction to position vector $\vec{r}$), this is also called radial acceleration:

$$\boxed{a_{\perp} = a_0 \sin \varepsilon = \frac{v_0^2}{R}}$$

In other words, the radial acceleration only changes direction of the velocity vector, but does not affect its magnitude!

This concludes derivation of expressions for tangential and radial accelerations.


enter image description here

Look at this free body diagram, from here you obtain

$$\tan \theta=\frac{m\omega^2r}{mg}$$

Thus if the angular velocity $\omega$ increases the angele $\theta$ increases


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