Minimum speed at the top of a loop

Problem

A cart travels around a vertical loop of radius $r = 8\ \mathrm{m}$. Calculate the minimum speed at the highest point so that it doesn’t leave the track.

At the highest point the weight $m\vec g$ and the reaction $\vec N$ both point downward, i.e. towards the centre.

Solution

Forces at the highest point. At the top of the loop the centre of the circle is below the cart. Both the weight $m\vec g$ and the normal reaction $\vec N$ from the track point downward, i.e. both towards the centre. Their sum provides the centripetal force: $mg + N = \frac{mv^2}{r}$

Condition for leaving the track. Contact with the track requires $N \ge 0$: the track can only push (not pull) the cart. The minimum speed corresponds to the limiting case $N = 0$, in which the weight alone acts as the centripetal force: $mg = \frac{mv_\text{min}^2}{r} \;\Longrightarrow\; v_\text{min} = \sqrt{gr}$

The mass cancels: the minimum speed does not depend on how heavy the cart is.

Numerical substitution. $v_\text{min} = \sqrt{9{,}81 \cdot 8} = \sqrt{78{,}5} \approx 8{,}86\ \mathrm{m/s}$

$\ev{v_\text{min} = \sqrt{gr} \approx 8{,}86\ \mathrm{m/s} \approx 32\ \mathrm{km/h}}$

Below this speed $N$ would have to become negative (impossible): the cart leaves the track and falls. Above it, the track adds the push $N > 0$ needed to complete the loop.

Linked items

Topics: Dinamica Concepts: Forza centripeta · Accelerazione centripeta · Forza normale · Forza peso Objects: Giro della morte