# Keeping the change in angle constant

Say I have a ball under a plane that has been secured to the ground on the left, as shown in the diagram (the red part). When I push the ball to the left, the angle increases. If I wanted the change in angle to be constant, what velocity function would the ball have to follow?

I know the change in angle change starts off as being large and then continues to diminish because of the sin curve. So, I'm thinking the velocity might just be the cos function. Is this right?

• Such an "intuitive" answer has little chance to be right. – Yves Daoust Jun 7 '17 at 10:16
• Think about what must happen backwards in time as the angle gets small. At some finite time ago the angle was $0$. What does that mean for the position (and therefore the velocity) of the ball? Does the cosine fulfill this intuition? – Arthur Jun 7 '17 at 10:22
• You'll need to start by writing down some formulae to describe how the angle changes with the change in distance of the ball from the fixed point. Then you may be able to start working out at what rate the ball will have to accelerate/decelerate at to keep the angle change constant – lioness99a Jun 7 '17 at 10:23
• May the angle of the plane be $\theta$ and the location of the centre of the ball $s(t)$. The radius of the ball is $1$. Now it's easy to see that $$\tan{\frac{\theta}{2}} = \frac{1}{s(t)} \Rightarrow s(t) = \frac{1}{\tan{\frac{\theta}{2}}} = \frac{\sin{\theta}}{1-\cos{\theta}}$$ The time derivative is $$\frac{d s(t)}{dt} = \frac{d s(t)}{d \theta} \frac{d \theta}{dt} = \frac{1}{\cos{\theta}-1} \frac{d \theta}{dt}$$ This is the formula that you'd use to for the speed of the ball. We don't know $\theta$, so you have to solve the first equation for $\cos{\theta}$ ... – Matti P. Jun 7 '17 at 10:27

Establish the relation between the plane and the position of the ball, measured as the horizontal distance between the contact points of the plane and the ball to the ground.

By trigonometry,

$$\tan\frac\theta2=\frac rd$$ or $$\theta=2\arctan\frac rd.$$

If you want a constant angular speed $\dot\theta=\omega$,

$$\omega=2\frac{-\dfrac{r\dot d}{d^2}}{1+\dfrac{r^2}{d^2}},$$

i.e.

$$v=-\omega\frac{d^2+r^2}{2r}.$$

In particular, when $d=r$ the plane is vertical and $v=-\omega r$, which is the expected value.

Then as $\theta=\omega t$, with a suitable time origin,

$$d=r\cot\frac{\omega t}2,$$ and $$v=-\frac{\omega r}2\left(\cot^2\frac{\omega t}2+1\right)=-\frac{\omega r}2\sec^2\frac{\omega t}2.$$