Worked exercise — minimum height for the loop

Problem

An object starts from rest at a height $h$ on an inclined plane and must complete a vertical loop of radius $r$. Find the minimum $h$ (a) without friction and (b) with friction.

Solution

Condition at the top of the loop. At the highest point of the loop (height $2r$) the minimum speed for the body to stay in contact with the track is $v_{\min} = \sqrt{gr}$ (the weight supplies the entire centripetal force: $mg = mv^2/r$). This is the critical condition that fixes the minimum starting height.

(a) Without friction. Define:

• State A (start): $v = 0$, height $h$.
• State B (top of the loop): $v = \sqrt{gr}$, height $2r$.

Conservation of mechanical energy $E_{\text{tot}}^A = E_{\text{tot}}^B$:

$mgh = \tfrac{1}{2}m\,v_{\min}^2 + mg(2r) = \tfrac{1}{2}m(gr) + mg(2r)$

Dividing by $mg$:

$h = \tfrac{1}{2}r + 2r \quad\Rightarrow\quad \ev{h_{\min} = \tfrac{5}{2}\,r}$

(b) With friction. Friction $F_a$ acts along the descending ramp, of length $\ell = h/\sin\theta$ (with $\theta$ the ramp’s inclination). The dissipated energy is $E_{\text{term}} = F_a\,\ell$. The energy balance becomes:

$mgh = \tfrac{1}{2}m(gr) + mg(2r) + F_a\cdot\frac{h}{\sin\theta}$

Collecting the terms in $h$:

$mgh - F_a\,\frac{h}{\sin\theta} = \tfrac{5}{2}mgr \quad\Rightarrow\quad h\left(mg - \frac{F_a}{\sin\theta}\right) = \tfrac{5}{2}mgr$

$\ev{h_{\min} = \frac{\tfrac{5}{2}\,mgr}{mg - \dfrac{F_a}{\sin\theta}}}$

Friction increases the required height: more initial potential energy is needed to compensate for that dissipated as heat along the descent. In the limit $F_a \to 0$ we recover $h_{\min} = \tfrac{5}{2}r$.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Conservazione dell’energia meccanica · Energia cinetica · Energia potenziale gravitazionale · Energia termica da attrito · Attrito dinamico Competenze: Conservazione dell’energia · Impostazione simbolica Oggetti: Giro della morte · Piano inclinato