How does the direction of the acceleration change in non-uniform circular motion?

Question

If I consider an object moving with a non-uniform circular motion with constant angular acceleration, the tangential acceleration modulus remains constant, while the centripetal acceleration modulus changes ($a_c=Rω^2$). Does the angle that the total acceleration forms with the centripetal change? If so, does it mean that, when I make an object tied to a rope rotate and I increase the speed, the tension (and thus the rope) constantly changes its direction?

Answer 1

The acceleration written in polar coordinates is:

$$ \vec{a}= \left(\ddot{r} - r \dot{\theta}^2\right)\hat{r} + \left(r \ddot{\theta}-2\dot{r}\dot{\theta}\right)\hat{\theta} $$

In your case $r$ is constant so $\dot{r}$ and $\ddot{r}$ are both zero and the equation for the acceleration simplifies to:

$$ \vec{a}= - r \dot{\theta}^2\hat{r} + r \ddot{\theta}\,\hat{\theta} $$

If the object starts from rest at time zero the angular velocity is $\dot{\theta} = \ddot{\theta}t$ so we get:

$$ \vec{a}= - r \ddot{\theta}^2t^2\hat{r} + r \ddot{\theta}\,\hat{\theta} $$

So you can immediately see that the radial component increases as $t^2$ while the tangential component is constant. We can find the angle the acceleration makes with the tangent vector because its tangent is the ratio of the centripetal to the tangential component and this gives a nice simple result:

$$\tan\phi = \ddot{\theta}t^2 $$

So at time zero $\phi=90°$ i.e. the acceleration is purely tangential, at time $t = \sqrt{1/\ddot{\theta}}$ the angle is $\phi=45°$, and as time goes on the angle tends to zero i.e. the acceleration tends to being purely centripetal.

Answer 2

Consider the ordinary differential equation (ODE) for the acceleration in non-uniform circular motion with constant angular acceleration. The radial or centripetal acceleration ($a_c$) is given by $a_c = R\omega^2$, where $\omega$ is the angular velocity. The angular velocity $\omega$ is related to the angular displacement $\theta$ and time (t) by the ODE:

$$\frac{d\omega}{dt} = \alpha,$$

where $\alpha$ is the constant angular acceleration.

Now, let's express the radial acceleration $a_c$ in terms of $\theta$ and $t$. The relation between angular displacement, angular velocity, and time is given by:

$$\omega = \frac{d\theta}{dt}.$$

So,

$$a_c = R\left(\frac{d\omega}{dt}\right) = R\alpha.$$

Now, let's consider the total acceleration $a$:

$$a = \sqrt{a_t^2 + a_c^2}.$$

Substitute the expressions for $a_t$ and $a_c$:

$$a = \sqrt{(R\alpha)^2 + (R\omega)^2}.$$

Now, differentiate this expression with respect to time $t$ to get the ODE for the acceleration:

$$\frac{da}{dt} = \frac{R\alpha}{\sqrt{(R\alpha)^2 + (R\omega)^2}}\frac{d\alpha}{dt} + \frac{R\omega}{\sqrt{(R\alpha)^2 + (R\omega)^2}}\frac{d\omega}{dt}.$$

This ODE describes how the total acceleration changes with respect to time in non-uniform circular motion with constant angular acceleration.